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Wentworth solutions chapter 2.pdf
2- !1
Solutions for Chapter 2 Problems
1. Vectors in the Cartesian Coordinate System
P2.1: Given P(4,2,1) and APQ=2ax +4ay +6az, find the point Q.
APQ = 2 ax + 4 ay + 6 az = (Qx-Px)ax + (Qy-Py)ay+(Qz-Pz)az
Qx-Px=Qx-4=2; Qx=6
Qy-Py=Qy-2=4; Qy=6
Qz-Pz=Qz-1=6; Qz=7
Ans: Q(6,6,7)
P2.2: Given the points P(4,1,0)m and Q(1,3,0)m, fill in the table and make a sketch of
the vectors found in (a) through (f).
a. AOP = (4-0)ax + (1-0)ay + (0-0)az = 4 ax + 1 ay.
!
!
(see Figure P2.2ab)
b. BOQ =(1-0)ax + (3-0)ay + (0-0)az = 1 ax + 3 ay.
!
!
Vector Mag Unit Vector
a. Find the vector A
from the origin to P
AOP = 4 ax + 1 ay 4.12 AOP = 0.97 ax + 0.24 ay
b. Find the vector B
from the origin to Q
BOQ = 1 ax + 3 ay 3.16 aOQ = 0.32 ax + 0.95 ay
c. Find the vector C
from P to Q
CPQ = -3 ax + 2 ay 3.61 aPQ = -0.83 ax + 0.55 ay
d. Find A + B A + B = 5 ax + 4 ay 6.4 a = 0.78 ax + 0.62 ay
e. Find C – A C - A = -7 ax + 1 ay 7.07 a = -0.99 ax + 0.14 ay
f. Find B - A B - A = -3 ax + 2 ay 3.6 a = -0.83 ax + 0.55 ay
Fig. P2.2ab
2- !2
(see Figure P2.2ab)
c. CPQ = (1-4)ax + (3-1)ay + (0-0)az = -3 ax + 2 ay.
!
!
(see Figure P2.2cd)
d. A + B = (4+1)ax + (1+3)ay + (0-0)az = 5 ax + 4 ay.
!
!
(see Figure P2.2cd)
e. C - A = (-3-4)ax + (2-1)ay + (0-0)az = -7 ax + 1 ay.
!
!
(see Figure P2.2ef)
f. B - A = (1-4)ax + (3-1)ay + (0-0)az = -3 ax + 2 ay.
!
!
(see Figure P2.2ef)
P2.3: MATLAB: Write a program that will find the vector between a pair of arbitrary
points in the Cartesian Coordinate System.
A program or function for this task is really overkill, as it is so easy to perform the task.
Enter points P and Q (for example, P=[1 2 3]; Q=[6 5 4]). Then, the vector from P toQ is
simply given by Q-P.
As a function we could have:
function PQ=vector(P,Q)
% Given a pair of Cartesian points
FigP2.2ef
Fig. P2.2cd
2- !3
% P and Q, the program determines the
% vector from P to Q.
PQ=Q-P;
Running this function we have:
>> P=[1 2 3];
>> Q=[6 5 4];
>> PQ=vector(P,Q)
PQ =
5 3 1
Alternatively, we could simply perform the math in the command line window:
>> PQ=Q-P
PQ =
5 3 1
>>
2. Coulomb’s Law, Electric Field Intensity, and Field Lines
P2.4: Suppose Q1(0.0, -3.0m, 0.0) = 4.0nC, Q2(0.0, 3.0m, 0.0) = 4.0nC, and Q3(4.0m,
0.0, 0.0) = 1.0nC. (a) Find the total force acting on the charge Q3. (b) Repeat the problem
after changing the charge of Q2 to –4.0nC. (c) Find the electric field intensity for parts (a)
and (b).
(a) where R13 = 4 ax + 3 ay =, R13 = 5m, a13 = 0.8 ax + 0.6 ay.
so
!
Similarly, ! so !
(b) with Q2 = -4 nC, F13 is unchanged but so
!
(c) !
Fig. P2.4
2- !4
Likewise, !
P2.5: Find the force exerted by Q1(3.0m, 3.0m, 3.0m) = 1.0 µC on Q2(6.0m, 9.0m, 3.0m)
= 10. nC.
!
R12 = (6-3)ax + (9-3)ay + (3-3)az = 3 ax + 6 ay m
!
! , so !
P2.6: Suppose 10.0 nC point charges are located on the corners of a square of side 10.0
cm. Locating the square in the x-y plane (at z = 0.00) with one corner at the origin and
one corner at P(10.0, 10.0, 0.00) cm, find the total force acting at point P.
We arbitrarily label the charges as shown in Figure P2.6. Then
ROP = 0.1 ax + 0.1 ay
ROP = 0.141 m
Fig. P2.5
2- !5
aOP = 0.707 ax + 0.707 ay.
!
!
!
and then the total (adjusting to 2 significant digits) is:
!
P2.7: 1.00 nC point charges are located at (0.00, -2.00, 0.00)m, (0.00, 2.00, 0.00)m, (0.00,
0.00, -2.00)m and (0.00, 0.00, +2.00)m. Find the total force acting on a 1.00 nC charge
located at (2.00, 0.00, 0.00)m.
Figure P2.7a shows the situation, but we need only find the x-directed force from one of
the charges on Qt (Figure P2.7b) and multiply this result by 4. Because of the problem’s
symmetry, the rest of the components cancel.
Fig. P2.6
Fig. P2.7b
Fig. P2.7a
2- !6
!
so!
The force from all charges is then
!
P2.8: A 20.0 nC point charge exists at P(0.00,0.00,-3.00m). Where must a 10.0 nC
charge be located such that the total field is zero at the origin?
For zero field at the origin, we must cancel the +az directed field from QP by placing Q at
the point Q(0,0,z) (see Figure P2.8). Then we have Etot = EP + EQ = 0.
So, !
and
!
So then
!
Thus, Q(0,0,2.12m).
3. The Spherical Coordinate System
P2.9: Convert the following points from Cartesian to Spherical coordinates:
a. P(6.0, 2.0, 6.0)
b. P(0.0, -4.0, 3.0)
c. P(-5.0,-1.0, -4.0)
Fig. P2.8
2- !7
(a) !
(b) ! !
(c) !
P2.10: Convert the following points from Spherical to Cartesian coordinates:
a. P(3.0, 30.°, 45.°)
b. P(5.0, π/4, 3π/2)
c. P(10., 135°, 180°)
(a)
!
(b)
!
(c)
!
P2.11: Given a volume defined by 1.0m ≤ r ≤ 3.0m, 0 ≤ θ ≤ 0°, 90° ≤ φ ≤ 90°, (a)
sketch the volume, (b) perform the integration to find the volume, and (c) perform the
necessary integrations to find the total surface area.
(a)
(b)
!
2- !8
So volume V = 14 m3.
(c) There are 5 surfaces: an inner, an outer, and 3 identical sides.
!
So Stotal = 35 m2.
4. Line Charges and the Cylindrical Coordinate System
P2.12: Convert the following points from Cartesian to cylindrical coordinates:
a. P(0.0, 4.0, 3.0)
b. P(-2.0, 3.0, 2.0)
c. P(4.0, -3.0, -4.0)
(a) !
(b) !
(c) !
P2.13: Convert the following points from cylindrical to Cartesian coordinates:
a. P(2.83, 45.0°, 2.00)
Fig. P2.11
2- !9
b. P(6.00, 120.°, -3.00)
c. P(10.0, -90.0°, 6.00)
(a)
!
(b)
!
(c)
!
P2.14: A 20.0 cm long section of copper pipe has a 1.00 cm thick wall and outer diame-
ter of 6.00 cm.
a. Sketch the pipe conveniently overlaying the cylindrical coordinate system, lining
up the length direction with the z-axis
b. Determine the total surface area (this could actually be useful if, say, you needed
to do an electroplating step on this piece of pipe)
c. Determine the weight of the pipe given the density of copper is 8.96 g/cm3
(a) See Figure P2.14
(b) The top area, Stop, is equal to the bottom area. We must also find the inner area, Sinner,
and the outer area, Souter.
!
!
2- !10
The total area, then, is 210π cm2, or Stot = 660 cm2.
(c) Determining the weight of the pipe requires the volume:
!
So Mpipe = 2820g.
P2.15: A line charge with charge density 2.00 nC/m exists at y = -2.00 m, x = 0.00. (a) A
charge Q = 8.00 nC exists somewhere along the y-axis. Where must you locate Q so that
the total electric field is zero at the origin? (b) Suppose instead of the 8.00 nC charge of
part (a) that you locate a charge Q at (0.00, 6.00m, 0.00). What value of Q will result in a
total electric field intensity of zero at the origin?
(a) The contributions to E from the line and point charge must cancel, or !
For the line: !
and for the point charge, where the point is located a distance y along the y-axis, we have:
!
Therefore:
!
(b)
Fig. P2.14
2- !11
!
P2.16: You are given two z-directed line charges of charge density +1 nC/m at x = 0, y =
-1.0 m, and charge density –1.0 nC/m at x = 0, y = 1.0 m. Find E at P(1.0m,0,0).
The situation is represented by Figure P2.16a. A better 2-dimensional view in Figure
P2.16b
is useful for solving the problem.
!
! , and !
So ETOT = 18 ay V/m.
P2.17: MATLAB: Suppose you have a segment of line charge of length 2L centered on
the z-axis and having a charge distribution ρL. Compare the electric field intensity at a
point on the y-axis a distance d from the origin with the electric field at that point assum-
ing the line charge is of infinite length. The ratio of E for the segment to E for the infinite
line is to be plotted versus the ratio L/d using MATLAB.
This is similar to MATLAB 2.3. We have for the ideal case
!
Fig. P2.16a
Fig. P2.16b
Fig. P2.15
2- !12
For the actual 2L case, we have an integration to perform (Equation (2.35) with different
limits):
!
Now we manipulate these expressions to get the following ratio:
!
In the program, the actual to ideal field ratio is termed “Eratio” and the charged line half-
length L ratioed to the distance d is termed “Lod”.
% M-File: MLP0217
%
% This program is similar to ML0203.
% It compares the E-field from a finite length
% segment of charge (from -L to +L on the z-axis)
% to the E-field from an infinite length line
% of charge. The ratio (E from segment to E from
% infinite length line) is plotted versus the ratio
% Lod=L/d, where d is the distance along the y axis.
%
% Wentworth, 12/19/02
%
% Variables:
% Lod the ratio L/d
% Eratio ratio of E from segment to E from line
clc %clears the command window
clear %clears variables
% Initialize Lod array and calculate Eratio
Lod=0.1:0.01:100;
Eratio=Lod./(sqrt(1+Lod.^2));
% Plot Eratio versus Lod
semilogx(Lod,Eratio)
grid on
xlabel('Lod=L/d')
ylabel('E ratio: segment to line')
2- !13
Executing the program gives Figure P2.17.
So we see that the field from a line segment of charge appears equivalent to the field from
an infinite length line if the test point is close to the line.
P2.18: A segment of line charge ρL =10 nC/m exists on the y-axis from the origin to y =
+3.0 m. Determine E at the point (3.0, 0, 0)m.
It is clear from a sketch of the problem in Figure P2.18a that the resultant field will be
directed in the x-y plane. The situation is redrawn in a temporary coordinate system in
Figure P2.18b.
We have from Eqn (2.34) !
For Eρ we have:
!
With ρ = 3, we then have Eρ = 21.2 V/m.
For Ez:
!
Fig. P2.17
2- !14
Thus we have ETOT = 21 aρ – 8.8 az V/m.
Converting back to the original coordinates, we have ETOT = 21 ax – 8.8 ay V/m.
5. Surface and Volume Charge
P2.19: In free space, there is a point charge Q = 8.0 nC at (-2.0,0,0)m, a line charge ρL =
10 nC/m at y = -9.0m, x = 0m, and a sheet charge ρs = 12. nC/m2 at z = -2.0m. Deter-
mine E at the origin.
The situation is represented by Figure P2.19, and the total field is ETOT = EQ + EL + ES.
!
!
!
So: Etot = 18 ax + 20 ay + 680 az V/m.
P2.20: An infinitely long line charge (ρL = 21π nC/m) lies along the z-axis. An infinite
area sheet charge (ρs = 3 nC/m2) lies in the x-z plane at y = 10 m. Find a point on the y-
axis where the electric field intensity is zero.
We have ETOT = EL + ES.
!
Fig. P2.19
Fig. P2.20
2- !15
!
so
!
Therefore, P(0, 7m, 0).
P2.21: Sketch the following surfaces and find the total charge on each surface given a
surface charge density of ρs = 1nC/m2. Units (other than degrees) are meters.
(a) –3 ≤ x ≤ 3, 0 ≤ y ≤ 4, z = 0
(b) 1 ≤ r ≤ 4, 180° ≤ φ ≤ 360°, θ = π/2
(c) 1 ≤ ρ ≤ 4, 180° ≤ φ ≤ 360°, z = 0
(a) !
(b) !
(c) !
P2.22: Consider a circular disk in the x-y plane of radius 5.0 cm. Suppose the charge
density is a function of radius such that ρs = 12ρ nC/cm2 (when ρ is in cm). Find the
electric field intensity a point 20.0 cm above the origin on the z-axis.
Fig. P2.21b&cFig. P2.21a
2- !16
From section 4 for a ring of charge of radius a,! Now we have
ρL=ρsdρ and where ρs = Aρ nC/cm2. Now the total field is given
by the integral:
!
This can be solved using integration by parts, where u = ρ, du = dρ,
! This leads to
!
Plugging in the appropriate values we arrive at E = 6.7 kV/cm az.
P2.23: Suppose a ribbon of charge with density ρs exists in the y-z plane of infinite
length in the z direction and extending from –a to +a in the y direction. Find a general
expression for the electric field intensity at a point d along the x-axis.
The problem is represented by Figure P2.23a. A better representation for solving the
problem is shown in Figure P2.23b.
We have where ρL = ρsdy. Then, since
FigP2.23a FigP2.23b
2- !17
!
the integral becomes
!
It may be noted that the ay component will cancel by symmetry. The ax integral is found
from the appendix and we have
!
P2.24: Sketch the following volumes and find the total charge for each given a volume
charge density of ρv = 1nC/m3. Units (other than degrees) are meters.
(a) 0 ≤ x ≤ 4, 0 ≤ y ≤ 5, 0 ≤ z ≤ 6
(b) 1 ≤ r ≤ 5, 0 ≤ θ ≤ 60°
(c) 1 ≤ ρ ≤ 5, 0° ≤ φ ≤ 90°, 0 ≤ z ≤ 5
(a) !
(b)
!
(c)
!
P2.25: You have a cylinder of 4.00 inch diam-
eter and 5.00 inch length (imagine a can of
tomatoes) that has a charge distribution that
varies with radius as ρv = (6 ρ) nC/in3 where ρ is in inches. (It may help you with the
Fig. P2.24b
Fig. P2.24a
2- !18
units to think of this as ρv (nC/in3)= 6 (nC/in4) ρ(in)). Find the total charge contained in
this cylinder.
!
P2.26: MATLAB: Consider a rectangular volume with 0.00 ≤ x ≤ 4.00 m, 0.00 ≤ y ≤
5.00 m and –6.00 m ≤ z ≤ 0.00 with charge density ρv = 40.0 nC/m3. Find the electric
field intensity at the point P(0.00,0.00,20.0m).
% MLP0226
% calculate E from a rectangular volume of charge
% variables
% xstart,xstop limits on x for vol charge (m)
% ystart,ystop
% zstart,zstop
% xt,yt,zt test point (m)
% rhov vol charge density, nC/m^3
% Nx,Ny,Nz discretization points
% dx,dy,dz differential lengths
% dQ differential charge, nC
% eo free space permittivity (F/m)
% dEi differential field vector
% dEix,dEiy,dEiz x,y and z components of dEi
% dEjx,dEjy,dEjz of dEj
% dEkx,dEky,dEkz of dEk
% Etot total field vector, V/m
clc
Fig. P2.24c
2- !19
clear
% initialize variables
xstart=0;xstop=4;
ystart=0;ystop=5;
zstart=-6;zstop=0;
xt=0;yt=0;zt=20;
rhov=40e-9;
Nx=10;Ny=10;Nz=10;
eo=8.854e-12;
dx=(xstop-xstart)/Nx;
dy=(ystop-ystart)/Ny;
dz=(zstop-zstart)/Nz;
dQ=rhov*dx*dy*dz;
for k=1:Nz
for j=1:Ny
for i=1:Nx
xv=xstart+(i-0.5)*dx;
yv=ystart+(j-0.5)*dy;
zv=zstart+(k-0.5)*dz;
R=[xt-xv yt-yv zt-zv];
magR=magvector(R);
uvR=unitvector(R);
dEi=(dQ/(4*pi*eo*magR^2))*uvR;
dEix(i)=dEi(1);
dEiy(i)=dEi(2);
dEiz(i)=dEi(3);
end
dEjx(j)=sum(dEix);
dEjy(j)=sum(dEiy);
dEjz(j)=sum(dEiz);
end
dEkx(k)=sum(dEjx);
dEky(k)=sum(dEjy);
dEkz(k)=sum(dEjz);
end
Etotx=sum(dEkx);
Etoty=sum(dEky);
Etotz=sum(dEkz);
Etot=[Etotx Etoty Etotz]
Now to run the program:
Etot =
2- !20
-6.9983 -8.7104 79.7668
>>
So E = -7.0 ax -8.7 ay + 80. az V/m
P2.27: MATLAB: Consider a sphere with charge
density ρv = 120 nC/m3 centered at the
origin with a radius of 2.00 m. Now, remove the top half of the sphere, leaving a hemi-
sphere below the x-y plane. Find the electric field intensity at the point P(8.00m,
0.00,0.00). (Hint: see MATLAB 2.4, and consider that your answer will now have two
field components.)
% M-File: MLP0227
%
% This program modifies ML0204 to find the field
% at point P(8m,0,0) from a hemispherical
% distribution of charge given by
% rhov=120 nC/m^3 from 0 < r < 2m and
% pi/2 < theta < pi.
%
% Wentworth, 12/23/02
%
% Variables:
% d y axis distance to test point (m)
% a sphere radius (m)
% dV differential charge volume where
% dV=delta_r*delta_theta*delta_phi
% eo free space permittivity (F/m)
% r,theta,phi spherical coordinate location of
% center of a differential charge element
% x,y,z cartesian coord location of charge %
element
% R vector from charge element to P
% Rmag magnitude of R
% aR unit vector of R
% dr,dtheta,dphi differential spherical elements
% dEi,dEj,dEk partial field values
% Etot total field at P resulting from charge
clc %clears the command window
clear %clears variables
% Initialize variables
2- !21
eo=8.854e-12;
d=8;a=2;
delta_r=40;delta_theta=72;delta_phi=144;
% Perform calculation
for k=(1:delta_phi)
for j=(1:delta_theta)
for i=(1:delta_r)
r=i*a/delta_r;
theta=(pi/2)+j*pi/(2*delta_theta);
phi=k*2*pi/delta_phi;
x=r*sin(theta)*cos(phi);
y=r*sin(theta)*sin(phi);
z=r*cos(theta);
R=[d-x,-y,-z];
Rmag=magvector(R);
aR=R/Rmag;
dr=a/delta_r;
dtheta=pi/delta_theta;
dphi=2*pi/delta_phi;
dV=r^2*sin(theta)*dr*dtheta*dphi;
dQ=120e-9*dV;
dEi=dQ*aR/(4*pi*eo*Rmag^2);
dEix(i)=dEi(1);
dEiy(i)=dEi(2);
dEiz(i)=dEi(3);
end
dEjx(j)=sum(dEix);
dEjy(j)=sum(dEiy);
dEjz(j)=sum(dEiz);
end
dEkx(k)=sum(dEjx);
dEky(k)=sum(dEjy);
dEkz(k)=sum(dEjz);
end
Etotx=sum(dEkx);
Etoty=sum(dEky);
Etotz=sum(dEkz);
Etot=[Etotx Etoty Etotz]
Now to run the program:
Etot =
579.4623 0.0000 56.5317
So E = 580 ax + 57 az V/m.
2- !22
6. Electric Flux Density
P2.28: Use the definition of dot product to find the three interior angles for the triangle
bounded by the points P(-3.00, -4.00, 5.00), Q(2.00, 0.00, -4.00), and R(5.00, -1.00,
0.00).
Here we use !
!
!
!
!
P2.29: Given D = 2ρ aρ + sin φ az C/m2, find the electric flux passing through the surface
defined by 2.0 ≤ ρ ≤ 4.0 m, 90.° ≤ φ ≤ 180°, and z = 4.0 m.
!
!
P2.30: Suppose the electric flux density is given by D = 3r ar –cos φ aθ + sin2θ aφ C/m2.
Find the electric flux through both surfaces of a hemisphere of radius 2.00 m and 0.00° ≤
θ ≤ 90.0˚.
!
Fig. P2.28
2- !23
!
!
!
!
7. Gauss’s Law and Applications
P2.31: Given a 3.00 mm radius solid wire centered on the z-axis with an evenly dis-
tributed 2.00 coulombs of charge per meter length of wire, plot the electric flux density
Dρ versus radial distance from the z-axis over the range 0 ≤ ρ ≤ 9 mm.
For a 1 m length, !
! , where L is the length of the Gaussian sur-
face. Note that this expression for Qenc is valid for both Gaussian surfaces.
GS1 (ρ < a):
!
so !
GS2 (ρ > a):
!
This is plotted with the following Matlab routine:
% M-File: MLP0231
%
Fig. P2.30
2- !24
% Gauss's Law Problem
% solid cylinder with even charge
%
% Variables
% rhov charge density (C/m^3)
% a radius of cylinder (m)
% rho radial distance from z-axis
% rhomm rho in mm
% D electric flux density (C/m^3)
% N number of data points
% maxrad max radius for plot (m)
clc;clear;
% initialize variables
rhov=70.7e3;
a=0.003;
maxrad=.009;
N=100;
bndy=round(N*a/maxrad);
for i=1:bndy
rho(i)=i*maxrad/N;
rhomm(i)=rho(i)*1000;
D(i)=rhov*rho(i)/2;
end
for i=bndy+1:N
rho(i)=i*maxrad/N;
Fig. P2.31
2- !25
rhomm(i)=rho(i)*1000;
D(i)=(rhov*a^2)/(2*rho(i));
end
plot(rhomm,D)
xlabel('radial distance (mm)')
ylabel('elect. flux density (C/m^2)')
grid on
P2.32: Given a 2.00 cm radius solid wire centered on the z-axis with a charge density ρv
= 6ρ C/cm3 (when ρ is in cm), plot the electric flux density Dρ versus radial distance
from the z-axis over the range 0 ≤ ρ ≤ 8 cm.
Choose Gaussian surface length L, and as usual we have
! valid for both Gaussian surfaces.
In GS1 (ρ < a): !
so !
For GS2 (ρ > a): !
This is plotted for the problem values in the following Matlab routine.
% M-File: MLP0232
%
% Gauss's Law Problem
% solid cylinder with radially-dependent charge
%
% Variables
% a radius of cylinder (cm)
% rho radial distance from z-axis
% D electric flux density (C/cm^3)
% N number of data points
% maxrad max radius for plot (cm)
clc;clear;
% initialize variables
a=2;
maxrad=8;
N=100;
bndy=round(N*a/maxrad);
for i=1:bndy
rho(i)=i*maxrad/N;
2- !26
D(i)=2*rho(i)^2;
end
for i=bndy+1:N
rho(i)=i*maxrad/N;
D(i)=(2*a^3)/rho(i);
end
plot(rho,D)
xlabel('radial distance (cm)')
ylabel('elect. flux density (C/cm^2)')
grid on
P2.33: A cylindrical pipe with a 1.00 cm wall thickness and an inner radius of 4.00 cm is
centered on the z-axis and has an evenly distributed 3.00 C of charge per meter length of
pipe. Plot Dρ as a function of radial distance from the z-axis over the range 0 ≤ ρ ≤ 10
cm.
! this is true for all the Gaussian surfaces.
GS1 (ρ < a): since Qenc = 0, Dρ = 0.
GS2(a < ρ < b):
!
Fig. P2.32
2- !27
!
So,
!
GS3(ρ > b):
Qenc = 3h,
A plot with the appropriate values is generated by the following Matlab routine:
% M-File: MLP0233
% Gauss's Law Problem
% cylindrical pipe with even charge distribution
%
% Variables
% a inner radius of pipe (m)
% b outer radius of pipe (m)
% rho radial distance from z-axis (m)
% rhocm radial distance in cm
% D electric flux density (C/cm^3)
% N number of data points
% maxrad max radius for plot (m)
clc;clear;
% initialize variables
a=.04;b=.05;maxrad=0.10;N=100;
bndya=round(N*a/maxrad);
bndyb=round(N*b/maxrad);
for i=1:bndya
rho(i)=i*maxrad/N;
rhocm(i)=rho(i)*100;
D(i)=0;
end
for i=bndya+1:bndyb
Fig. P2.33a
2- !28
rho(i)=i*maxrad/N;
rhocm(i)=rho(i)*100;
D(i)=(3/(2*pi*rho(i)))*((rho(i)^2-a^2)/(b^2-a^2));
end
for i=bndyb+1:N
rho(i)=i*maxrad/N;
rhocm(i)=rho(i)*100;
D(i)=3/(2*pi*rho(i));
end
plot(rhocm,D)
xlabel('radial distance (cm)')
ylabel('elect. flux density (C/m^2)')
grid on
P2.34: An infinitesimally thin metallic cylindrical shell of radius 4.00 cm is centered on
the z-axis and has an evenly distributed charge of 100. nC per meter length of shell. (a)
Determine the value of the surface charge density on the conductive shell and (b) plot Dρ
as a function of radial distance from the z-axis over the range
0 ≤ ρ ≤ 12 cm.
!
For all Gaussian surfaces,
of height h and radius ρ, we have:
Fig. P2.33b
2- !29
!
GS1 (ρ < a): Qenc = 0 so Dρ = 0
GS2 (ρ > a):
!
% M-File: MLP0234
%
% Gauss's Law Problem
% cylindrical shell of charge
%
% Variables
% a radius of cylinder (m)
% Qs surface charge density (nC/m^2)
% rho radial distance from z-axis (m)
% rhocm radial distance in cm
% D electric flux density (nC/cm^3)
% N number of data points
% maxrad max radius for plot (cm)
clc;clear;
% initialize variables
a=.04;Qs=398;maxrad=0.12;N=100;
bndy=round(N*a/maxrad);
for i=1:bndy
rho(i)=i*maxrad/N;
rhocm(i)=rho(i)*100;
D(i)=0;
end
for i=bndy+1:N
rho(i)=i*maxrad/N;
rhocm(i)=rho(i)*100;
D(i)=Qs*a/rho(i);
end
plot(rhocm,D)
xlabel('radial distance (cm)')
ylabel('elect. flux density (nC/m^2)')
Fig. P2.34a
2- !30
grid on
P2.35: A spherical charge density is given by ρv = ρo r/a for 0 ≤ r ≤ a, and ρv = 0 for r >
a. Derive equations for the electric flux density for all r.
! This is valid for each Gaussian sur-
face.
GS1 (r < a): !
So !
GS2 (r > a): !
P2.36: A thick-walled spherical shell, with inner radius 2.00 cm and outer radius 4.00
cm, has an evenly distributed 12.0 nC charge. Plot Dr as a function of radial distance
from the origin over the range 0 ≤ r ≤ 10 cm.
Here we’ll let a = inner radius and b = outer radius. Then
! This is true for each Gaussian surface.
The volume containing charge is
Fig. P2.34b
2- !31
!
So !
Now we can evaluate Qenc for each Gaussian surface.
GS1 (r < a): Qenc = 0 so Dr = 0.
GS2 (a < r < b): !
Inserting our value for ρv, we find
!
GS3 (r >b): Qenc = Q,
This is plotted for appropriate values using the following Matlab routine:
% M-File: MLP0236
% Gauss's Law Problem
% thick spherical shell with even charge
%
% Variables
% a inner radius of sphere (m)
% b outer radius of sphere (m)
% r radial distance from origin (m)
% rcm radial distance in cm
% D electric flux density (nC/cm^3)
% N number of data points
% maxr max radius for plot (m)
% Q charge (nC)
clc;clear;
% initialize variables
a=.02;b=.04;
Q=12;
maxrad=0.10;
N=100;
bndya=round(N*a/maxrad);
bndyb=round(N*b/maxrad);
for i=1:bndya
2- !32
r(i)=i*maxrad/N;
rcm(i)=r(i)*100;
D(i)=0;
end
for i=bndya+1:bndyb
r(i)=i*maxrad/N;
rcm(i)=r(i)*100;
D(i)=(Q/
(4*pi*r(i)^2))*(r(i)^3-a^3)/
(b^3-a^3);
end
for i=bndyb+1:N
r(i)=i*maxrad/N;
rcm(i)=r(i)*100;
D(i)=Q/(4*pi*r(i)^2);
end
plot(rcm,D)
xlabel('radial distance (cm)')
ylabel('elect. flux density (nC/m^2)')
grid on
P2.37: Given a coaxial cable with solid inner conductor of radius a, an outer conductor
that goes from radius b to c, (so c > b > a), a charge +Q that is evenly distributed
throughout a meter length of the inner conductor and a charge –Q that is evenly dis-
tributed throughout a meter length of the outer conductor, derive equations for the electric
flux density for all ρ. You may orient the cable in any way you wish.
We conveniently center the cable on the z-axis. Then, for a Gaussian surface of length L,
! valid for all Gaussian surfaces.
GS1: (ρ < a): !
!
Fig. P2.36
2- !33
GS2 (a < ρ < b): !
GS3 (b < ρ < c): !
!
so !
GS4 (ρ > c): Qenc = 0, Dρ = 0.
8. Divergence and the Point Form of Gauss’s Law
P2.38: Determine the charge density at the point P(3.0m,4.0m,0.0) if the electric flux
density is given as D = xyz az C/m2.
!
ρv(3,4,0)=(3)(4)=12 C/m3.
P2.39: Given D = 3ax +2xyay +8x2y3az C/m2, (a) determine the charge density at the
point P(1,1,1). Find the total flux through the surface of a cube with 0.0 ≤ x ≤ 2.0m, 0.0
≤ y ≤ 2.0m and 0.0 ≤ z ≤ 2.0m by evaluating (b) the left side of the divergence theorem
and (c) the right side of the divergence theorem.
(a) !
(b) !
!
!
2- !34
!
!
!
!
!
(c) !
P2.40: Suppose D = 6ρcosφ aφ C/m2. (a) Determine the charge density at the point (3m,
90°, -2m). Find the total flux through the surface of a quartered-cylinder defined by 0 ≤
ρ ≤ 4m, 0 ≤ φ ≤ 90°, and -4m ≤ z ≤ 0 by evaluating (b) the left side of the divergence
theorem and (c) the right side of the divergence theorem.
(a) !
!
(b) !
note that the top, bottom and outside integrals yield zero since there is no component of
D in the these dS directions.
!
!
So, !
2- !35
(c)
!
P2.41: Suppose D = r2sinθ ar + sinθcosφ aφ C/m2. (a) Determine the charge density at
the point (1.0m, 45°, 90°). Find the total flux through the surface of a volume defined by
0.0 ≤ r ≤ 2.0 m, 0.0° ≤ θ ≤ 90.°, and 0.0 ≤ φ ≤ 180° by evaluating (b) the left side of
the divergence theorem and (c) the right side of the divergence theorem.
The volume is that of a quartered-sphere, as indicated in Figure P2.41.
(a)
!
(b) !
!
!
!
Summing these terms we have Q = 4(π2 – 1)C = 35.5C.
(c)
!
2- !36
9. Electric Potential
P2.42: A sheet of charge density ρs = 100 nC/m2 occupies the x-z plane at y = 0. (a) Find
the work required to move a 2.0 nC charge from P(-5.0m, 10.m, 2.0m) to M(2.0m, 3.0m,
0.0). (b)Find VMP.
(a) !
!
Notice that we are only concerned with movement in the y-direction. We then have:
!
(b) !
P2.43: A surface is defined by the function 2x + 4y2 –ln z = 12. Use the gradient equa-
tion to find a unit vector normal to the plane at the point (3.00m,2.00m,1.00m).
Let ! then
!
Fig. P2.41
2- !37
At (3,2,1),
!
P2.44: For the following potential distributions, use the gradient equation to find E.
(a) V = x+y2z (V)
(b) V = ρ2sinφ(V)
(c) V = r sinθ cosφ (V).
(a) !
(b) !
(c)
!
P2.45: A 100 nC point charge is located at the origin. (a) Determine the potential differ-
ence VBA between the point A(0.0,0.0,-6.0)m and point B(0.0,2.0,0.0)m. (b) How much
work would be done to move a 1.0 nC charge from point A to point B against the electric
field generated by the 100 nC point charge?
(a) !
The potential difference is only a function of radial distance from the origin. Letting ra =
6m and rb = 2m, we then have
!
(b) !
P2.46: MATLAB: Suppose you have a pair of charges Q1(0.0, -5.0m, 0.0) = 1.0 nC and
Q2(0.0, 5.0m, 0.0) = 2.0 nC. Write a MATLAB routine to calculate the potential VRO
moving from the origin to the point R(5.0m, 0.0, 0.0). Your numerical integration will
involve choosing a step size ΔL and finding the field at the center of the step. You should
try several different step sizes to see how much this affects the solution.
2- !38
% M-File: MLP0246
%
% Modify ML0207 to calculate the potential
% difference going from the origin (O) to the point
% R(5,0,0) given a pair of point charges
% Q1(0,-5,0)=1nC and Q2(0,5,0)=2nC.
%
% The approach will be to break up the distance
% from O to R into k sections. The total field E will
% be found at the center of each section (located
% at point P) and then dot(Ep,dLv) will give the
% potential drop across the kth section. Total
% potential is found by summing the potential drops.
%
% Wentworth, 1/7/03
%
% Variables:
% Q1,Q2
the point charges, in nC
% k number of numerical integration steps
% dL magnitude of one step
% dLv vector for a step
% x(n) x location at center of section at P
% R1,R2 vector from Q1,Q2 to P
% E1,E2 electric fields from Q1 & Q2 at P
% Etot total electric field at P
% V(n) portion of dot(Etot,dL) at P
clc %clears the command window
clear %clears variables
% Initialize variables
k=64;
Q1=1;
Q2=2;
dL=5/k;
dLv=dL*[1 0 0];
% Perform calculation
for n=1:k
x(n)=(n-1)*dL+dL/2;
R1=[x(n) 5 0];
R2=[x(n) -5 0];
Rmag1=magvector(R1);
Rmag2=magvector(R2);
2- !39
E1=9*Q1*R1/Rmag1^3;
E2=9*Q2*R2/Rmag2^3;
Etot=E1+E2;
V(n)=dot(Etot,dLv);
end
Vtot=sum(-V)
Now running the program:
Vtot =
-1.5817
So VRO = -1.6 V.
P2.47: For an infinite length line of charge density ρL = 20 nC/m on the z-axis, find the
potential difference VBA between point B(0, 2m, 0) and point A(0, 1m, 0).
!
P2.48: Find the electric field at point P(0.0,0.0,8.0m) resulting from a surface charge
density ρs = 5.0 nC/m2 existing on the z = 0 plane from ρ = 2.0 m to ρ = 6.0 m. Assume
V = 0 at a point an infinite distance from the origin.
(Method 1)
For a ring of charge it was previously found that
!
We can then break up our disk into differential rings (see Figure P2.48), each contributing
dE as:
!
So we then have
!
2- !40
This is easy to integrate if we let u = ρ2 + h2, then du = 2 ρ dρ, and we have
!
Solving, we arrive at
!
Upon inserting the appropriate values we find E = 48 V/m az.
(method 2)
Find an expression for potential and then evaluate the gradient at the point.
!
!
Now we let h = z and !
!
Plugging in the values we find E = 48 V/m az.
Fig. P2.48
2- !41
P2.49: Suppose a 6.0 m diameter ring with charge density 5.0 nC/m lies in the x-y plane
with the origin at its center. Determine the potential difference Vho between the point
h(0.0,0.0,4.0)m and the origin. (Hint: first find an expression for E on the z-axis as a
general function of z.)
For the ring of charge, replacing h with z, we have
!
!
Letting u = a2 + z2, du = 2z dz, we have
!
Replacing u and evaluating from 0 to h,
!
10. Conductivity and Current
P2.50: A columnular beam of electrons from 0 ≤ ρ ≤ 1 mm has a charge density ρv =-0.1
cos(πρ/2) nC/mm3 (where ρ is in mm) and a velocity of 6 x 106 m/sec in the +az direc-
tion. Find the current.
Let’s let where ρo = -0.1 nC/mm3. Then we’ll let u = uoaz, where uo =
6x109 mm/s. Notice we convert the units to mm. Now,
!
and with dS = ρ dρ dφ az we then have
!
This becomes
where A = 2πρouo.
Fig. P2.49
2- !42
Now we can integrate by parts, or ! where u = Aρ, du = Adρ,
! and !
We then have
!
To evaluate, we first find A = 2π(-0.1x10-9)(6x109)=3.77, and then
I = 2.40-1.53=0.87A.
I = 0.87A.
P2.51: Two spherical conductive shells of radius a and b (b > a) are separated by a mate-
rial with conductivity σ. Find an expression for the resistance between the two spheres.
First find E for a < r < b, assuming +Q at r = a and –Q at r = b. From Gauss’s law:
!
Now find Vab:
!
Now can find I:
!
Finally, !
P2.52: The typical length of each piece of jumper wire on a student’s protoboard is 5.0
cm. Assuming AWG-20 (wire diameter 0.812 mm) copper wire, (a) determine the resis-
tance for this length of wire. (b) Determine the power dissipated in the wire for 10. mA
of current.
2- !43
(a) !
so R = 1.7 mΩ
(b) !
P2.53: A densely wrapped coil of AWG-22 (0.644 mm diameter) copper magnet wire is
150 m long. The wire has a very thin insulative sheath. Determine the resistance for this
length of wire.
!
so R = 7.9Ω
P2.54: Determine an expression for the power dissipated per unit length in coaxial cable
of inner radius a, outer radius b, and conductivity between the conductors σ if a potential
difference Vab is applied.
From Eqn(2.84) we have !
Now for a given potential difference Vab we have
!
P2.55: Find the resistance per unit length of a stainless steel pipe of inner radius 2.5 cm
and outer radius 3.0 cm.
!
so we have !
so R/L = 1.0 mΩ/m
2- !44
P2.56: A nickel wire of diameter 5.0 mm is surrounded by a 0.50 mm thick layer of sil-
ver. What is the resistance per unit length for this wire? Assuming 1.0 m of this wire
carries 1.0 A of current, determine the power dissipated in the nickel portion and in the
silver portion of the wire.
We can treat this wire as two resistors in parallel. We have
!
!
!
To find the power dissipated, we first find the potential difference:
!
then
!
11. Dielectrics
P2.57: A material has 12.0 V/m ax field intensity with permittivity 194.5 pF/m. Deter-
mine the electric flux density.
!
P2.58: MATLAB: A 20 nC point charge at the origin is embedded in Teflon (εr = 2.1).
Find and plot the magnitudes of the polarization vector, the electric field intensity and the
electric flux density at a radial distance from 0.1 cm out to 10 cm.
We use the following equations:
!
% M-File: MLP0258
%
% Plot E, P and D vs distance r from a point
% charge Q at the origin with a dielectric.
2- !45
%
% Variables
% Q charge (C)
% eo free space permittivity (F/m)
% r radial distance (m)
% Chi electric susceptibility
% E electric field intensity(V/m)
% D electric flux density (C/m^2)
% P polarization vector (C/m^2)
% initialize variables
Q=20e-9;
er=2.1;
eo=8.854e-12;
Chi=er-1;
% perform calculations
r=0.001:.001:0.100;
rcm=r.*100;
E=Q./(4*pi*r.^2);
P=Chi*eo*E;
D=er*eo*E;
% plot data
subplot(2,1,1)
loglog(rcm,P,'--k',rcm,D,'-k')
legend('P','D')
ylabel('C/m^2')
grid on
subplot(2,1,2)
loglog(rcm,E)
ylabel('V/m')
xlabel('radial dis-
tance (cm)')
grid on
Fig. P2.58
2- !46
P2.59: Suppose the force is very carefully measured between a pair of point charges sep-
arated by a dielectric material and is found to be 20 nN. The dielectric material is re-
moved without changing the position of the point charges, and the force has increased to
100 nN. What is the relative permittivity of the dielectric?
!
P2.60: The potential field in a material with εr = 10.2 is V = 12 xy2 (V). Find E, P and D.
!
!
!
!
P2.61: In a mineral oil dielectric, with breakdown voltage of 15 MV/m, the potential
function is V = x3 – 6x2 –3.1x (MV). Is the dielectric likely to breakdown, and if so,
where?
!
! so from 6x – 12 = 0 we find the maximum electric field
occurs at x = 2m.
At x = 2m, we have E = -12+24+3.1 = 15.1 MV/m, exceeding the breakdown voltage.
12. Boundary Conditions
P2.62: For y < 0, εr1 = 4.0 and E1 = 3ax + 6πay + 4az V/m. At y = 0, ρs = 0.25 nC/m2. If
εr2 = 5.0 for y > 0, find E2.
E1 = 3ax + 6πay + 4az V/m (g) E2 = 3ax + 20.7ay + 4az V/m
(a) EN1 = 6πay (f) EN2 = DN2/5εo = 20.7ay
2- !47
(e) !
!
P2.63: For z ≤ 0, εr1 = 9.0 and for z > 0, εr2 = 4.0. If E1 makes a 30° angle with a normal
to the surface, what angle does E2 make with a normal to the surface?
Refer to Figure P2.63.
!
also
!
Therefore
! and after routine math we find !
Using this formula we obtain for this problem θ2 = 14°.
!
P2.64:
A plane defined by 3x + 2y + z = 6 separates two dielectrics. The first dielectric,
on the side of the plane containing the origin, has εr1 = 3.0 and E1 = 4.0az V/m. The other
dielectric has εr2 = 6.0. Find E2.
We first use gradient to find a normal to the planar surface.
Let F = 3x + 2y + z – 6 = 0.
(b) ET1 = 3ax + 4az (c) ET2 = ET1 = 3ax + 4az
(d) DN1 = εr1εoEN1 = 24πεo ay (e) DN2 = 0.92 ay
Fig. P2.63
2- !48
!
Now we can work the boundary condition problem.
!
!
!
!
Finally we have ! .
P2.65: MATLAB: Consider a dielectric-dielectric charge free boundary at the plane z =
0. Construct a program that will allow the user to enter εr1 (for z < 0), εr2, and E1, and
will then calculate E2. (Just for fun, you may want to have the program calculate the an-
gles that E1 and E2 make with a normal to the surface).
% M-File: MLP0265
%
% Given E1 at boundary between a pair of
% dielectrics with no charge at boundary,
% calculate E2. Also calculates angles.
%
clc
clear
% enter variables
disp('enter vector quantities in brackets,')
disp('for example: [1 2 3]')
er1=input('relative permittivity in material 1: ');
er2=input('relative permittivity in material 2: ');
a12=input('unit vector from mtrl 1 to mtrl 2: ');
E1=input('electric field intensity vector in mtrl 1: ');
% perform calculations
En1=dot(E1,a12)*a12;
Et1=E1-En1;
Et2=Et1;
Dn1=er1*En1; %ignores eo since it will factor out
Dn2=Dn1;
En2=Dn2/er2;
2- !49
E2=Et2+En2
% calculate the angles
th1=atan(magvector(Et1)/magvector(En1));
th2=atan(magvector(Et2)/magvector(En2));
th1r=th1*180/pi
th2r=th2*180/pi
Now run the program:
enter vector quantities in brackets,
for example: [1 2 3]
relative permittivity in material 1: 2
relative permittivity in material 2: 5
unit vector from mtrl 1 to mtrl 2: [0 0 1]
electric field intensity vector in mtrl 1: [3 4 5]
E2 =
3 4 2
th1r =
45
th2r =
68.1986
P2.66: A 1.0 cm diameter conductor is sheathed with a 0.50 cm thickness of Teflon and
then a 2.0 cm (inner) diameter outer conductor. (a) Use Laplace’s equations to find an
expression for the potential as a function of ρ in the dielectric. (b) Find E as a function of
ρ. (c) What is the maximum potential difference that can be applied across this coaxial
cable without breaking down the dielectric?
(a) Since V is only a function of ρ,
2- !50
!
where A and B are constants.
Now we apply boundary conditions.
BC1:
!
BC2: !
or
!
(b) !
(c)
!
P2.67: A 1.0 m long carbon pipe of inner diameter 3.0 cm and outer diameter 5.0 cm is
cut in half lengthwise. Determine the resistance between the inner surface and the outer
surface of one of the half sections of pipe.
One approach is to consider the resistance for the half-section of pipe is twice the resis-
tance for a complete cylindrical section, given by Eqn. (2.84). But we’ll used the
LaPlace equation approach instead.
Laplace: ! ; here we see V only depends on ρ
So: ! ;
Fig. P2.66
2- !51
where A and B are constants.
Now apply boundary conditions.
BC1:
!
BC2:
!
!
J=σE
!
!
P2.68: For a coaxial cable of inner conductor radius a and outer conductor radius b and a
dielectric εr in-between, assume a charge density is added in the dielectric re-
gion. Use Poisson’s equation to derive an expression for V and E. Calculate ρs on each
plate.
!
so
! , where A is a constant.
! , where B is a constant.
Now apply boundary conditions: !
Fig. P2.67
2- !52
Applying the second one gives us:
!
Applying the first one:
!
Therefore,
!
!
where
! ,
so !
!
!
P2.69: For the parallel plate capacitor given in Figure 2.51, suppose a charge density
2- !53
!
is added between the plates. Use Poisson’s equation to derive a new expression for V and
E. Calculate ρs on each plate.
!
!
!
Now apply the boundary conditions:
!
!
!
!
at z = 0, ! !
at z = d, ! !
13. Capacitors
P2.70: A parallel plate capacitor is constructed such that the dielectric can be easily re-
moved. With the dielectric in place, the capacitance is 48 nF. With the dielectric re-
moved, the capacitance drops to 12 nF. Determine the relative permittivity of the dielec-
tric.
!
2- !54
P2.71: A parallel plate capacitor with a 1.0 m2 surface area for each plate, a 2.0 mm plate
separation, and a dielectric with relative permittivity of 1200 has a 12. V potential differ-
ence across the plates. (a) What is the minimum allowed dielectric strength for this ca-
pacitor? Calculate (b) the capacitance, and (c) the magnitude of the charge density on one
of the plates.
(a) !
(b) !
(c) !
P2.72: A conical section of material extends from 2.0 cm ≤ r ≤ 9.0 cm for 0 ≤ θ ≤ 30°
with εr = 9.0 and σ = 0.020 S/m. Conductive plates are placed at each radial end of the
section. Determine the resistance and capacitance of the section.
! , where A and B are constants.
Boundary conditions: r = a, V = 0 and r = b, V = Vb
!
!
!
2- !55
!
!
P2.73: An inhomogeneous dielectric fills a parallel plate capacitor of surface area 50.
cm2 and thickness 1.0 cm. You are given εr = 3(1 + z), where z is measured from the bot-
tom plate in cm. Determine the capacitance.
Place +Q at z = d and –Q at z = 0.
!
!
evaluating the integral:
!
!
Fig. P2.72
2- !56
P2.74: Given E = 5xyax + 3zaz V/m, find the electrostatic potential energy stored in a
volume defined by 0 ≤ x ≤ 2 m, 0 ≤ y ≤ 1 m, and 0 ≤ z ≤ 1 m. Assume ε = εo.
!
P2.75: Suppose a coaxial capacitor with inner radius 1.0 cm, outer radius 2.0 cm and
length 1.0 m is constructed with 2 different dielectrics. When oriented along the z-axis,
εr for 0° ≤ φ ≤ 180° is 9.0, and for 180° ≤ φ ≤ 360° is 4.0. (a) Calculate the capaci-
tance. (b) If 9.0 V is applied across the conductors, determine the electrostatic potential
energy stored in each dielectric for this capacitor.
(a) a coaxial line,
!
But for only half the line,
!
So
!
and
!
So
!
(b) !
Fig. P2.75
Wentworth solutions chapter 3.pdf
3- !1
Solutions for Chapter 3 Problems
1. Magnetic Fields and Cross Products
P3.1: Find AxB for the following:
a. A = 2ax – 3ay + 4az, B = 5ay - 1az
b. A = aρ + 2aφ + 4az, B = 2aρ + 6az
c. A = 2ar + 5aθ + 1aφ, B = ar + 3aφ
(a)
!
(b)
!
(c)
!
P3.2: If a parallelogram has a short side a, a long side b, and an interior angle θ (the
smaller of the two interior angles), the area of the parallelogram is given by
!
Determine how you would use the cross product of a pair of vectors to find the area of a
parallelogram defined by the points O(0,0,0), P(6,0,0), Q(8,12,0) and R(2,12,0).
(Assume dimensions in meters)
!
A = 6ax, B = 2ax + 12ay
A x B = 72az,
Area = 72 m2
( ) ( ) ( )2 3 4 3 20 2 10 17 2 10
0 5 1
x y z
x y z x y z× = − = − − − + = − + +
−
a a a
A B a a a a a a
( ) ( )1 2 4 12 6 8 0 4 12 2 4
2 0 6
z
z z
ρ φ
ρ φ ρ φ× = = − − + − = + −
a a a
A B a a a a a a
( ) ( )2 5 1 15 6 1 0 5 15 5 5
1 0 3
r
r r
θ φ
θ φ θ φ× = = − − + − = − −
a a a
A B a a a a a a
sin .area ab θ=
sin
sin
area ab θ
θ
=
= × =A B A B Fig. P3.2
3- !2
P3.3: Given the vertices of a triangle P(1,2,0), Q(2,5,0) and R(0,4,7), find (a) the interior
angles, (b) a unit vector normal to the surface containing the triangle and (c) the area of
the triangle.
(a) !
!
!
!
!
!
(b) !
(c) !
2. Biot-Savart’s Law
P3.4: A segment of conductor on the z-axis extends from z = 0 to z = h. If this segment
conducts current I in the +az direction, find H(0,y,0). Compare your answer to that of
Example 3.2.
We use Eqn. (3.7) and change the limits:
1 3 ; 3.16x y= + =PQ a a PQ
1 2 7 ; 7.348x y z= − + + =PR a a a PR
21 7 5
22.69
x y z× = − +
× =
PQ PR a a a
PQ PR
( )( )
sin
22.69sin ;
3.16 7.348
78
ab
P
P
θ
θ
θ
× =
=
=
A B A B
�
1 3 ; 3.16
2 1 7 ; 7.348
x y
x y z
= − − =
= − − + =
QP a a QP
QR a a a QR
22.69; 78 ; 180 24Q R P Qθ θ θ θ× = = = − − =QP QR
� � �
0.93 0.31 0.22n x y z
×
= = − +
×
PQ PRa a a a
PQ PR
21 11.4
2
area m= × =PQ PR
Fig. P3.3
3- !3
!
Note that if the line of current is semi-
infinite (goes from z = 0 to z = ∞), we’d
have:
!
P3.5: An infinite length line with 2.0 A current in the +ax direction exists at y = -3.0 m, z
= 4.0 m. A second infinite length line with 3.0 A current in the +az direction exists at x =
0, y = 3.0 m. Find H(0, 0, 0).
This situation is shown in Figure P3.5a.
Ho = H1 + H2
Referring to the figure,
R = 3ay – 4az, R = 5,
aR = 0.6ay – 0.8az
aφ = ax x aR = 0.80ay + 0.60az
!
!
Ho = 159ax + 51ay +38az mA/m
Or with 2 significant digits
( )
2 2 2
0
2 2 2
2 2
4
4
4
h
x
x
I z
z
Iy h
y h y
Ih
y h y
φρ
π ρ ρ
π
π
⎡ ⎤
= ⎢ ⎥
+⎢ ⎥⎣ ⎦
⎡ ⎤−
= ⎢ ⎥
⎢ + ⎥⎣ ⎦
−
=
+
a
H
a
aH
4
xI
yπ
−
=
aH
( )
1
1 2
0.80 0.602
2 5
51 38
y z
y z
I
A
m
mA
m
φπρ
π
=
+
=
= +
H a
a a
a a
2
2 2
3 159
2 3
x
x
I
A mA
mm
φπρ
π
=
= =
H a
a a
Fig. P3.4
Fig. P3.5a
Fig. P3.5b
3- !4
Ho = 160ax + 51ay +38az mA/m
P3.6: A conductive loop in the shape of an equilateral triangle of side 8.0 cm is centered
in the x-y plane. It carries 20.0 mA current clockwise when viewed from the +az
direction. Find H(0, 0, 16cm).
The situation is illustrated in Figure P3.6a. The sketch in Figure P3.6b is used to find the
+x-axis intercept for the triangle. By simple trigonometry we have:
!
Now for one segment we adapt Eqn. (3.7):
!
with ρaρ = RaR,
!
!
H1 = -4.7ax – 0.68az mA/m
Now by symmetry the total H contains only the az component:
Htot = -2.0 mA/m az.
tan 30 ;and since = 4 cm we find = 2.31 cm.b a b
a
= �
1 2 24
a
a
I z
z
φ
πρ ρ
+
−
⎡ ⎤
= ⎢ ⎥
+⎢ ⎥⎣ ⎦
a
H
2.31 16 , 16.17 , 0.99 0.143x z y R x zcm φ= − + = = − × = − −R a a R a a a a a
( )( )
( )
( )3
1 2 2 2
20 10 0.99 0.143 2 4
4 16.17 10 4 16.17
x zx A
x mπ
−
−
− − ⎡ ⎤
= ⎢ ⎥
+⎣ ⎦
a a
H
Fig. P3.6a Fig. P3.6b
3- !5
P3.7: A square conductive loop of side 10.0 cm is centered in the x-y plane. It carries
10.0 mA current clockwise when viewed from the +az direction. Find H(0, 0, 10cm).
We find H for one section of the square
by adopting Eqn. (3.7):
!
!
With ρaρ = RaR, we have
R = -5ax + 10az, |R| = 11.18x10-2 m
aR = -0.447ax + 0.894az,
aφ = -ay x aR = -0.894 ax – 0.447 az
!
Now by symmetry the total H contains only the az component:
HTOT = -10.4az mA/m
P3.8: A conductive loop on the x-y plane is bounded by ρ = 2.0 cm, ρ = 6.0 cm, φ = 0°
and φ = 90°. 1.0 A of current flows in the loop, going in the aφ direction on the ρ = 2.0
cm arm. Determine H at the origin.
By inspection of the figure, we see
that only the arc portions of the
loop contribute to H.
From a ring example we have:
!
For the ρ = a segment of the loop:
!
2 24
a
a
I z
z
φ
πρ ρ
+
−
⎡ ⎤
= ⎢ ⎥
+⎢ ⎥⎣ ⎦
a
H
2 22
Ia
a
φ
πρ ρ
=
+
a
H
( )( )( )
( )
3
1 2 2 2
10 10 5 0.894 0.447
5.2 2.6
2 11.18 10 5 11.18
x z
x z
x mA
mxπ
−
−
− −
= = − −
+
a a
H a a
( )
22
3
2 2 2 04
zIa d
h a
π
φ
π
=
+
∫
aH
/ 22
a 3
04 8
z
z
Ia Id
a a
π
φ
π
= =∫
aH a
Fig. P3.7
Fig. P3.8
3- !6
At ρ = b: ! ;
So !
P3.9: MATLAB: How close do you have to be to the middle of a finite length of current-
carrying line before it appears infinite in length? Consider Hf(0, a, 0) is the field for the
finite line of length 2h centered on the z-axis, and that Hi(0, a, 0) is the field for an
infinite length line of current on the z-axis. In both cases consider current I in the +az
direction. Plot Hf/Hi vs h/a.
Adapting Eqn. (3.7), for the finite length line we have:
!
For the infinite length of line:
!
The ratio we wish to plot is:
!
The MATLAB routine follows.
% M-File: MLP0309
%Consider the field for a finite line of length 2h
%oriented on z-axis with current I in +z direction.
%The field is to be found a distance a away from
%the current on the y axis (point (0,a,0)).
%We want to compare this field with that of an infinite
%length line of current.
%Plot Hf/Hi versus h/a. We expect that as h/a grows large,
02
b 3
/ 24 8
z
z
Ib Id
b bπ
φ
π
−
= =∫
aH a
1 1 1 1 1 4.2
8 8 0.02 0.06TOT z z z
I A
a b m
⎛ ⎞ ⎛ ⎞= − = − =⎜ ⎟ ⎜ ⎟
⎝ ⎠ ⎝ ⎠
H a a a
2 22f
a
I h
h
φ
ρ
πρ ρ
=
⎡ ⎤
= ⎢ ⎥
+⎢ ⎥⎣ ⎦
a
H
2i
I φ
πρ
=
a
H
( )
( )
2 2 2
1
f
i
hH h a
H h a h
a
= =
+ +
3- !7
%the line will appear more 'infinite' to an observation
%point at (0,a,0).
hova=0.01:.01:100;
HfovHi=hova./sqrt(1+(hova).^2);
semilogx(hova,HfovHi)
xlabel('h/a')
ylabel('Hf/Hi')
grid on
P3.10: MATLAB: For the ring of current described in MATLAB 3.2, find H at the
following points (a) (0, 0, 1m), (b) (0, 2m, 0), and (c) (1m, 1m, 0).
%M-File: MLP0310
%Find the magnetic field intensity at any observation point
%resulting from a ring of radius a and current I,
%in the aphi direction centered in the x-y plane.
df=1; %increment in degrees
a=1; %ring radius in m
I=1; %current in A
Ro=input('vector location of observation point: ');
for j=1:df:360;
Fr=j*pi/180;
Rs=[a*cos(Fr) a*sin(Fr) 0];
as=unitvector(Rs);
Fig. P3.9a
Fig. P3.9b
3- !8
dL=a*df*(pi/180)*cross([0 0 1],as);
Rso=Ro-Rs;
aso=unitvector(Rso);
dH=I*cross(dL,aso)/(4*pi*(magvector(Rso))^2);
dHx(j)=dH(1);
dHy(j)=dH(2);
dHz(j)=dH(3);
end
H=[sum(dHx) sum(dHy) sum(dHz)]
Now to run the program:
>> MLP0310
vector location of observation point: [0 0 1]
H = -0.0000 -0.0000 0.1768
>> MLP0310
vector location of observation point: [0 2 0]
H = 0 0 -0.0431
>> MLP0310
vector location of observation point: [1 1 0]
H = 0 0 -0.1907
>> MLP0310
vector location of observation point: [0 1 1]
H = 0.0000 0.0910 0.0768
>>
So we see:
(a)
H = 0.18 az A/m
(b) H = -0.043 az A/m
(c) H = -0.19 az A/m
(extra) H = 9.1ay + 7.7 az mA/m
3- !9
P3.11: A solenoid has 200 turns, is 10.0 cm long, and has a radius of 1.0 cm. Assuming
1.0 A of current, determine the magnetic field intensity at the very center of the solenoid.
How does this compare with your solution if you make the assumption that 10 cm >> 1
cm?
Eqn. (3.10):
!
Or H = 1960 A/m az
The approximate solution, assuming 10cm >> 1cm, is
!
P3.12: MATLAB: For the solenoid of the previous problem, plot the magnitude of the
field versus position along the axis of the solenoid. Include the axis 2 cm beyond each
end of the solenoid.
% M-File: MLP0312
%
% Plot H vs length thru center of a solenoid
%
clc
clear
% initialize variables
N=200; %number of turns
h=0.10; %height of solenoid
a=0.01; %radius of solenoid
I=1; %current
dz=0.001; %step change in z
z=-.02:dz:h+.02;
zcm=z.*100;
( )
( )
( ) ( )
2 2 22
2 2 22
2
200 1 0.1 0.05 0.05 1961
2 0.1 0.05 0.010.1 0.05 0.01
z
z z
NI h z z
h z ah z a
A A
m m
⎡ ⎤
−⎢ ⎥= +
⎢ ⎥+− +⎣ ⎦
⎡ ⎤
−⎢ ⎥= + =
⎢ ⎥+− +⎣ ⎦
H a
H a a
( )200 1 2000
0.1z z z
ANI A
h m m
= = =H a a a
3- !10
A1=(h-z)./sqrt((h-z).^2+a^2);
A2=z./sqrt(z.^2+a^2);
Atot=A1+A2;
H=N*I.*Atot/(2*h);
% generate plot
plot(zcm,H)
xlabel('z(cm)')
ylabel('H (A/m)')
grid on
P3.13: A 4.0 cm wide ribbon of current is centered about the y-axis on the x-y plane and
has a surface current density K = 2π ay A/m. Determine the magnetic field intensity at
the point (a) P(0, 0, 2cm), (b) Q(2cm, 2cm, 2cm).
(a) Because of the symmetry (Figure P3.13a),
we can use a modified Eqn. (3.14):
!
(b) Referring to Figure P3.13b;
!
!
!
1
1
tan
2 2tan 1.57
2
y
x
x x
K d
a
A
m
π
π
π
−
−
⎛ ⎞= ⎜ ⎟
⎝ ⎠
⎛ ⎞= =⎜ ⎟
⎝ ⎠
H a
a a
( )
;
2
; r x z
Id
R where d x a
φ
ρ
πρ
ρ
=
= = − +
H a
a a R a a
( )
( )
( )
( )
( )
2 2
2 2
2 2
, and
. Sox zR
x z
y R
so d x a
d x a
d x a
a x d
d x a
φ
ρ = − +
− +
=
− +
+ −
= × =
− +
a a
a
a a
a a a
( )
( )2 22
y x z zK a x d dx
d x aπ
+ −
=
− +∫
a a a
H
Fig. P3.12
Fig. P3.13a
Fig. P3.13b
3- !11
This is separated into 3 integrals, each
one solved via numerical integration,
resulting in:
H=1.1083ax +0.3032az – 1.1083az; or
H = 1.1ax – 0.80az A/m
3. Ampere’s Circuit Law
P3.14: A pair of infinite extent current sheets exists at z = -2.0 m and at z = +2.0 m. The
top sheet has a uniform current density K = 3.0 ay A/m and the bottom one has K = -3.0
ay A/m. Find H at (a) (0,0,4m), (b) (0,0,0) and (c) (0,0,-4m).
We apply !
(a) !
(b)
!
(c) H = 0
P3.15: An infinite extent current sheet with K = 6.0 ay A/m exists at z = 0. A conductive
loop of radius 1.0 m, in the y-z plane centered at z = 2.0 m, has zero magnetic field
intensity measured at its center. Determine the magnitude of the current in the loop and
show its direction with a sketch.
Htot = HS + HL
!
For the loop, we use Eqn. (3.10):
!
where here
1 ,
2 N
= ×H K a
( ) ( )1 13 -3 0
2 2y z y z
= × + × =H a a a a
( ) ( )1 13 -3
2 2
3
y z y z
x
A
m
= ×− + ×
= −
H a a a a
a
( )1 1 6 3
2 2S N y z x
A
m
= × = × =H K a a a a
2 z
I
a
=H a
Fig. P3.14
Fig. P3.15
3- !12
!
(sign is chosen opposite HS).
So, I/2 = 3 and I = 6A.
P3.16: Given the field H = 3y2 ax, find the current passing through a square in the x-y
plane that has one corner at the origin and the opposite corner at (2, 2, 0).
Referring to Figure P3.6, we evaluate the circulation of H around the square path.
!
!
!
!
!
So we have Ienc = 24 A. The negative
Sign indicates current is going in the
-az direction.
P3.17: Given a 3.0 mm radius solid wire centered on the z-axis with an evenly distributed
2.0 amps of current in the +az direction, plot the magnetic field intensity H versus radial
distance from the z-axis over the range 0 ≤ ρ ≤ 9 mm.
Figure P3.17 shows the situation along with the Amperian Paths. We have:
!
This will be true for each Amperian path.
( )
2 2L x x
I I
a
−
= − =H a a
b c d a
enc
a b c d
d I= = + + +∫ ∫ ∫ ∫ ∫H L��
( )23 0 0
b
x x
a
dx =∫ a a�
23 0
c
x y
b
y dy =∫ a a�
( )
0
2
2
3 2 12 24
d
x x
c
dx dx= = −∫ ∫a a�
0
a
d
=∫
, ; 2enc encd I where H and d d H Iφ φ φ φρ φ πρ= = = =∫ H L H a L a� ��
Fig. P3.16
3- !13
AP1: !
So: !
AP2: Ienc = I,
% MLP0317
% generate plot for ACL problem
a=3e-3; %radius of solid wire (m)
I=2; %current (A)
N=30; %number of data points to plot
rmax=9e-3; %max radius for plot (m)
dr=rmax/N;
for i=1:round(a/dr)
r(i)=i*dr;
H(i)=(I/(2*pi*a^2))*r(i);
end
for i=round(a/dr)+1:N
r(i)=i*dr;
H(i)=I/(2*pi*r(i));
end
plot(r,H)
xlabel('rho(m)')
ylabel('H (A/m)')
grid on
2 2
2 2 2
0 0
I I I, = ,
a a aenc z enc
I d I d d
ρ π ρ
ρ ρ φ
π π
= = =∫ ∫ ∫J S J a�
2 for 2
I a
a φ
ρ ρ
π
= ≤H a
for
2
I aφ ρπρ
= ≥H a
Fig. P3.17a Fig. P3.17b
3- !14
P3.18: Given a 2.0 cm radius solid wire centered on the z-axis with a current density J =
3ρ A/cm2 az (for ρ in cm) plot the magnetic field intensity H versus radial distance from
the z-axis over the range 0 ≤ ρ ≤ 8 cm.
We’ll let a = 2 cm.
!
AP1 (ρ < a): !
and !
AP2 (ρ > a): Ienc = 2πa3, so
The MATLAB plotting routine is as follows:
% MLP0318
% generate plot for ACL problem
a=2; %radius of solid wire (cm)
N=40; %number of data points to plot
rmax=8; %max radius for plot (cm)
dr=rmax/N;
for i=1:round(a/dr)
r(i)=i*dr;
H(i)=r(i)^2;
end
for i=round(a/dr)+1:N
r(i)=i*dr;
H(i)=a^3/r(i);
end
plot(r,H)
xlabel('rho(cm)')
ylabel('H (A/cm)')
grid on
P3.19: An infinitesimally thin metallic cylindrical shell of radius 4.0 cm is centered on
the z-axis and carries an evenly distributed current of 10.0 mA in the +az direction. (a)
, ; 2enc encd I where H and d d H Iφ φ φ φρ φ πρ= = = =∫ H L H a L a� ��
33 2enc zI d d dρρ ρ ρ φ πρ= = =∫ ∫J S a a� �
2 for aφρ ρ= ≤H a
3
for a aφ ρρ
= ≥H a
Fig. P3.18
3- !15
Determine the value of the surface current density on the conductive shell and (b) plot H
as a function of radial distance from the z-axis over the range 0 ≤ ρ ≤ 12 cm.
(a) !
(b) for ρ < a, H = 0. For ρ > a we have:
!
The MATLAB routine to generate the plot is as follows:
% MLP0319
% generate plot for ACL problem
a=4; %radius of solid wire (cm)
N=120; %number of data points to plot
I=10e-3; %current (A)
rmax=12; %max plot radius(cm)
dr=rmax/N;
for i=1:round(a/dr)
r(i)=i*dr;
H(i)=0;
end
for i=round(a/dr)+1:N
r(i)=i*dr;
H(i)=100*I/(2*pi*r(i));
end
plot(r,H)
xlabel('rho(cm)')
ylabel('H (A/m)')
grid on
( )
10 39.8 ; so 40
2 2 0.04s z
I mA mA mAK
a m m mπ π
= = = =K a
2
I
φπρ
=H a
Fig. P3.19a
Fig. P3.19b
3- !16
P3.20: A cylindrical pipe with a 1.0 cm
wall thickness and an inner radius of 4.0 cm is
centered on the z-axis and has an evenly distributed 3.0 amps of current in the +az
direction. Plot the magnetic field intensity H versus radial distance from the z-axis over
the range 0 ≤ ρ ≤ 10 cm.
For each Amperian Path:
!
Now, for ρ < a, Ienc = 0 so H = 0.
For a < ρ < b, !
!
% MLP0320
% generate plot for ACL problem
a=4; %inner radius of pipe (cm)
b=5; %outer radius of pipe(cm)
N=120; %number of data points to plot
I=3; %current (A)
rmax=10; %max radius for plot (cm)
dr=rmax/N;
aoverdr=a/dr
boverdr=b/dr
, ; 2enc encd I where H and d d H Iφ φ φ φρ φ πρ= = = =∫ H L H a L a� ��
( )2 2
, where and =enc z z
II d d d d
b a
ρ ρ φ
π
= =
−∫ J S J a S a�
( ) ( )
( )2 22 2 2
2 22 2 2 2
0
,
2enc a
aI a II d d I
b ab a b a
ρ π
φ
ρρ
ρ ρ φ
ρπ π
−−
= = =
−− −∫ ∫ H a
Fig. P3.20bFig. P3.20a
3- !17
for i=1:round(a/dr)
r(i)=i*dr;
H(i)=0;
end
for i=round(a/dr)+1:round(b/dr)
r(i)=i*dr;
num(i)=I*(r(i)^2-a^2);
den(i)=2*pi*(b^2-a^2)*r(i);
H(i)=100*num(i)/den(i);
end
for i=round(b/dr)+1:N
r(i)=i*dr;
H(i)=100*I/(2*pi*r(i));
end
plot(r,H)
xlabel('rho(cm)')
ylabel('H (A/m)')
grid on
P3.21: An infinite length line carries current I in the +az direction on the z-axis, and this
is surrounded by an infinite length cylindrical shell (centered about the z-axis) of radius a
carrying the return current I in the –az direction as a surface current. Find expressions for
the magnetic field intensity everywhere. If the current is 1.0 A and the radius a is 2.0 cm,
plot the magnitude of H versus radial distance from the z-axis from 0.1 cm to 4 cm.
Fig. P3.21a
3- !18
!
The MATLAB routine used to generate
Figure P3.21b is as follows:
% MLP0321
% generate plot for ACL
problem
clc
clear
a=2; %inner radius of
cylinder(cm)
N=80; %number of data
points to plot
I=1; %current (A)
rmax=4; %max radius for plot (cm)
dr=rmax/N;
for i=1:40
r(i)=.1+(i-1)*dr;
H(i)=100*I/(2*pi*r(i));
end
for i=40:N
r(i)=i*dr;
H(i)=0;
end
plot(r,H)
xlabel('rho(cm)')
ylabel('H (A/m)')
grid on
P3.22: Consider a pair of collinear cylindrical shells centered on the z-axis. The inner
shell has radius a and carries a sheet current totaling I amps in the +az direction while the
outer shell of radius b carries the return current I in the –az direction. Find expressions
for the magnetic field intensity everywhere. If a = 2cm, b = 4cm and I = 4A, plot the
magnitude of H versus radial distance from the z-axis from 0 to 8 cm.
; for 0< , and for , 0.
2enc
Id I a aφρ ρπρ
= < = > =∫H L H a H�
Fig. P3.21b
3- !19
!
The MATLAB routine used to generate Figure P3.22b is as follows:
% MLP0322
% generate plot for ACL problem
a=2; %inner radius of coax (cm)
b=4; %outer radius of coax(cm)
N=160; %number of data points to plot
I=4; %current (A)
rmax=8; %max radius for plot (cm)
dr=rmax/N;
aoverdr=a/dr
boverdr=b/dr
for i=1:round(a/dr)
r(i)=i*dr;
H(i)=0;
end
for i=round(a/dr)+1:round(b/dr)
r(i)=i*dr;
H(i)=100*I/(2*pi*r(i));
end
for i=round(b/dr)+1:N
; for 0< , = 0;
for a< , ;
2
and for , 0.
encd I a
Ib
b
φ
ρ
ρ
πρ
ρ
= <
< =
> =
∫H L H
H a
H
�
Fig. P3.22a Fig. P3.22b
3- !20
r(i)=i*dr;
H(i)=0;
end
plot(r,H)
xlabel('rho(cm)')
ylabel('H (A/m)')
grid on
P3.23: Consider the toroid in Figure 3.55 that is tightly wrapped with N turns of
conductive wire. For an Amperian path with radius less than a, no current is enclosed
and therefore the field is zero. Likewise, for radius greater than c, the net current
enclosed is zero and again the field is zero. Use Ampere’s Circuital Law to find an
expression for the magnetic field at radius b, the center of the toroid.
!
Within the toroid, H = Hφ aφ, so
!
Then, Ienc by the Amperian path is: Ienc = NI.
!
4. Curl and the Point Form of Ampere’s Circuital Law
P3.24: Find ! for the following fields:
a. A = 3xy2/z ax
b. A = ρsin2φ aρ – ρ2 z cosφ aφ
c. A = r2sinθ ar + r/cos φ aθ
(a) !
(b) !
!
;encd I=∫H L�
2 .
b
d H d bHφ φ φ φρρ φ π== ⋅ =∫ ∫H L a a�
.
2
NI
b φπ
∴ =H a
∇×A
( )
2
2
2
3 63 x xx y z y z
A A xy xyxy z
z y z z
∂ ∂ −
∇× = − = −
∂ ∂
a a a a a
( ) ( )2 2 1sin cos z
A A A
z A
z z
φ ρ ρ
ρ φ ρ φ φρ φ ρ φ ρρ ρ φ
∂ ∂ ∂⎡ ⎤∂
∇× − = − + + −⎢ ⎥∂ ∂ ∂ ∂⎣ ⎦
a a a a a
2
2 3 cos 1cos 0 2sin cosz z
z
ρ φ
ρ φ
ρ φ ρ φ φ
ρ ρ
= + − −a a a a
3- !21
!
(c)
!
!
P3.25: Find J at (3m, 60°, 4m) for H = (z/sinφ) aρ – (ρ2/cosφ) az A/m.
!
!
Now find J by evaluating ! at the given point:
!
P3.26: Suppose H = y2ax + x2ay A/m.
a. Calculate ! around the path ! , where A(2m,0,0),
B(2m,4m,0), C(0,4m,0) and D(0,0,0).
b. Divide this by the area S (2m*4m = 8m2).
c. Evaluate ! at the center point.
d. Comment on your results for (b) & (c).
(a) Referring to the figure, we evaluate !
[ ]2 cos 3 cos 2sin cos zzρρ φ ρ φ φ φ= − +a a
( )
2 sin
cos
1 1 1 1
sin sin
r
r r
r
rr
rAA A A
r r r r
θ
θθ
θ φ
θ
φ
θ φ θ φ θ
⎛ ⎞
∇× +⎜ ⎟
⎝ ⎠
∂⎡ ⎤⎡∂ ⎤ ⎡ ⎤∂ ∂−
= + + −⎢ ⎥⎢ ⎥ ⎢ ⎥∂ ∂ ∂ ∂⎣ ⎦ ⎣ ⎦ ⎣ ⎦
a a
a a a
( ) ( )
1 2
2
2
cos 1 1 sin
sin cos
sin 2 cos
sin cos cos
r
r
r r r
r r r r
r
φ φ
φ
φ
θ
θ φ φ θ
φ
θ
θ φ φ
−∂ ⎛ ⎞− ∂ ∂
= + −⎜ ⎟∂ ∂ ∂⎝ ⎠
⎡ ⎤−
= + −⎢ ⎥
⎣ ⎦
a a a
a a
1 1z z
z
A AA A
z
ρ ρ
ρ φρ φ ρ ρ φ
∂ ∂⎡ ⎤∂ ∂ −
∇× = + − +⎢ ⎥∂ ∂ ∂ ∂⎣ ⎦
H a a a
2 2
sin 1 2 cos
cos sin cos sin z
z
ρ φ
ρ φ ρ φ
φ φ φ ρ φ
⎛ ⎞−
= + + +⎜ ⎟
⎝ ⎠
a a a
∇×H
210 13 0.89 z
A
mρ φ
= − + +J a a a
d∫ H L�� A B C D A→ → → →
d∫ H L��
∇×H
B C D A
enc
A B C D
d I= = + + +∫ ∫ ∫ ∫ ∫H L�
3- !22
!
!
!
!
So we have !
(b) dividing by S = 8m2, we have -2 C/m2
(c) Evaluating the curl of H:
! , and at the center point (x = 1 and y = 2) we have
!
(d) In this particular case, ! even though S is of appreciable size.
P3.27: For the coaxial cable example 3.8, we found:
!
a. Evaluate the curl in all 4 regions.
b. Calculate the current density in the conductive regions by dividing the current by
the area. Are these results the same as what you found in (a)?
(a) !
!
( )
4
2
0 2
4 4 16
B
A x
x dy C
=
= = =∫ ∫
0
2
2 4
32
C
B y
y dx C
=
= = −∫ ∫
0
2
4 0
0
D
C x
x dy
=
= =∫ ∫
2
2
0 0
0
A
D y
y dy
=
= =∫ ∫
16d C= −∫H L�
( )2 2y x z z
A A x y
x y
∂⎛ ⎞∂
∇× = − = −⎜ ⎟∂ ∂⎝ ⎠
H a a
22 zcenter
C
m
∇× = −H a
,d S∇× = ∫H H L��
( )
2
2 2
2 2
for a, ,
2
for a< b, ,
2
Ifor b< c, = ,
2 c
and for c< , =0.
I
a
I
c
b
φ
φ
φ
ρ
ρ
π
ρ
πρ
ρ
ρ
ρπ
ρ
≤ =
≤ =
⎛ ⎞−
≤ ⎜ ⎟
− ⎝ ⎠
H a
H a
H a
H
2
2 2
1 for
2 z z
I I a
a a
ρ
ρ
ρ ρ π π
⎛ ⎞∂
∇× = = ≤⎜ ⎟∂ ⎝ ⎠
H a a
1 0 for a
2 z
I bρ
ρ ρ π
∂ ⎛ ⎞∇× = = < <⎜ ⎟∂ ⎝ ⎠
H a
Fig. P3.26
3- !23
!
!
(b) !
!
Comment: ! is confirmed.
P3.28: Suppose you have the field H = r cos θ aφ A/m. Now consider the cone specified
by θ = π/4, with a height a as shown in Figure 3.56. The circular top of the cone has a
radius a.
a. Evaluate the right side of Stoke’s theorem through the dS = dSaθ surface.
b. Evaluate the left side of Stoke’s theorem by integrating around the loop.
(a) !
ar derivative:
!
aθ derivative:
( )
( ) ( )
2 2
2 2 2 2
1 for b<
2 z z
I c I c
c b c b
ρ
ρ
ρ ρ π π
⎛ ⎞−∂ −
⎜ ⎟∇× = = ≤
⎜ ⎟∂ − −⎝ ⎠
H a a
0 for cρ∇× >H =
( )2 2for , , z zIIa S a S aρ π π≤ = = =J a a
( ) ( )
2 2
2 2
for b< , , z
Ic S c b
c b
ρ π
π
−
≤ = − =
−
J a
∇×H = J
( ) ( )1 1sinsin r
rH
H
r r r
φ
φ θθθ θ
∂∂⎡ ⎤∇× = −⎢ ⎥∂ ∂⎣ ⎦
H a a
( ) ( ) ( ) ( )
2 2
2 2
cos sin1sin cos cos sin ; sin
sin sinr r
r r H
r φ
θ θ
θ θ θ θ θ
θ θ θ θ
−∂ ∂⎡ ⎤= − =⎢ ⎥∂ ∂⎣ ⎦
a a
Fig. P3.28
3- !24
!
So, !
Now we must integrate this over the aθ surface:
!
(b) !
Clearly in this case the circulation of H is the easiest approach.
5. Magnetic Flux Density
P3.29: An infinite length line of 3.0 A current in the +ay direction lies on the y-axis. Find
the magnetic flux density at P(7.0m,0,0) in (a) Teslas, (b) Wb/m2, and (c) Gauss.
!
!
!
P3.30: Suppose an infinite extent sheet of current with K = 12ax A/m lies on the x-y
plane at z = 0. Find B for any point above the sheet. Find the magnetic flux passing
through a 2m2 area in the x-z plane for z > 0.
!
This is valid at any point above the sheet.
( )2 cos1 2cosr
r r θ θ
θ
θ
∂−
= −
∂
a a
( )2 2cos sin
2cos
sin r θ
θ θ
θ
θ
−
∇× = −H a a
( )
( )
( )
2 2
2 2
2
4 0 0
cos sin
2cos sin
sin
2 sin cos 2 sin cos 2
r
a
d r drd
r drd rdr d a
θ θ
π
π
θ
θ θ
θ θ φ
θ
θ θ φ θ θ φ π
=
⎡ ⎤−
⎢ ⎥∇× = − −
⎢ ⎥⎣ ⎦
= = =
∫ ∫
∫ ∫ ∫
H S a a a� �
2
2 2
2
4 0
cos cos 2
r a
d r ad a d a
π
πφ φ θ
θ φ θ φ π
= =
= = =∫ ∫ ∫H L a a� ��
( )
( )3 68
2 2 7 z z
I A mA
m mφπρ π
= = − = −H a a a
7 9 9
24 10 68 86 10 86 10o z z z
H A Wb Wbx x x T
m m HA m
µ π − − −⎛ ⎞⎛ ⎞ ⎛ ⎞= = − = − = −⎜ ⎟⎜ ⎟ ⎜ ⎟
⎝ ⎠⎝ ⎠ ⎝ ⎠
B H a a a
( )9 610,00086 10 860 10z zGx T x GT
− −= − = −B a a
( )7 6
2
1 ;
2
4 10
12 7.54 10
2 2
N
o
N x z y
x H m A Wbx
m m
πµ
−
−
= ×
⎛ ⎞= × = × = −⎜ ⎟
⎝ ⎠
H K a
B K a a a a
3- !25
Now, !
P3.31: An infinite length coaxial cable exists along the z-axis, with an inner shell of
radius a carrying current I in the +az direction and outer shell of radius b carrying the
return current. Find the magnetic flux passing through an area of length h along the z-
axis bounded by radius between a and b.
For a < ρ < b, !
6. Magnetic Forces
P3.32: A 1.0 nC charge with velocity 100. m/sec in the y direction enters a region where
the electric field intensity is 100. V/m az and the magnetic flux density is 5.0 Wb/m2 ax.
Determine the force vector acting on the charge.
!
!
P3.33: A 10. nC charge with velocity 100. m/sec in the z direction enters a region where
the electric field intensity is 800. V/m ax and the magnetic flux density 12.0 Wb/m2 ay.
Determine the force vector acting on the charge.
!
( )( )-6 22-7.54x10 2 15y yWbd m Wbmφ µ
⎛ ⎞= = − =⎜ ⎟
⎝ ⎠∫
B S B S = a a� �
( )
, ,
2 2
ln
2 2
ln Wb
2
o
bo o
a
o
II
I Id d dz h
Ih b
a
φ φ
φ
φ
µ
πρ πρ
µ µ
φ ρ ρ
π ρ π
µ
φ
π
= =
= = =
⎛ ⎞= ⎜ ⎟
⎝ ⎠
∫ ∫
H a B a
a
B S a� �
( ) 2; 100 5 500y z
x
m Wb Wbq
s m sm
= + × × = × = −F E u B u B a a
a
910 100 500 400z z z
V Wb Vs mNC nN
m sm Wb VC
− ⎛ ⎞= + − = −⎜ ⎟
⎝ ⎠
F a a a
( ) 9 210 10 800 100 12 4x z x
y
V m Wbq x C N
m s m
µ−
⎛ ⎞
= + × = + × = −⎜ ⎟⎜ ⎟
⎝ ⎠
F E u B a a a
a
3- !26
P3.34: A 10. nC charged particle has a velocity v = 3.0ax + 4.0ay + 5.0az m/sec as it
enters a magnetic field B = 1000. T ay (recall that a tesla T = Wb/m2). Calculate the force
vector on the charge.
!
The cross-product:
!
Evaluating we find: F = -50ax + 30az µN
P3.35: What electric field is required so that the velocity of the charged particle in the
previous problem remains constant?
!
!
P3.36: An electron (with rest mass Me= 9.11x10-31kg and charge q = -1.6 x 10-19 C) has a
velocity of 1.0 km/sec as it enters a 1.0 nT magnetic field. The field is oriented normal to
the velocity of the electron. Determine the magnitude of the acceleration on the electron
caused by its encounter with the magnetic field.
!
!
!
( ) ( )9 210 10 3 4 5 1000x y z y
m Wbq x C
s m
− ⎡ ⎤= × = + + ×⎢ ⎥⎣ ⎦
F u B a a a a
3 4 5 5000 3000
0 1000 0
x y z
x z
⎡ ⎤
⎢ ⎥ = − +⎢ ⎥
⎢ ⎥⎣ ⎦
a a a
a a
0 (constant velocity)dm m
dt
= = =F va
( )
( ) 2
0;
4 5 1000 3000 5000
5 3
x y z y z x
x z
q
m Wb V V
s m m m
kV
m
= + × =
= − × + + × = − +
= −
F E u B
E u B = - 3a a a a a a
E a a
( );m q= = ×F u Ba
( );q
m
= ×u Ba
( )( )( )
( )
19 9
2
3
231
1.6 10 1000 10
175 10
9.11 10
m Wbx CquB s m ma x sm x kg
− −
−
−
= = =
3- !27
P3.37: Suppose you have a surface current K = 20. ax A/m along the z = 0 plane. About a
meter or so above this plane, a 5.0 nC charged particle is moving along with velocity v =
-10.ax m/sec. Determine the force vector on this particle.
!
!
!
P3.38: A meter or so above the surface current of the previous problem there is an infinite
length line conducting 1.0 A of current in the –ax direction. Determine the force per unit
length acting on this line of current.
!
!
P3.39: Recall that the gravitational force on a mass m is ! where, at the earth’s
surface, g = 9.8 m/s2 (-az). A line of 2.0 A current with 100. g mass per meter length is
horizontal with the earth’s surface and is directed from west to east. What magnitude and
direction of uniform magnetic flux density would be required to levitate this line?
!
!
By inspection, B = Bo(-ax)
1 1 20 10
2 2N x z y
A
m
= × = × = −H K a a a a
( )7 72 210 4 10 40 10o y yWb Wbx xm mµ π π
− −= = − = −B H a a
( )9 7 25 10 10 40 10 0.63x y z
m Wbq x C x pN
s m
π− −⎛ ⎞= × = − ×− =⎜ ⎟
⎝ ⎠
F a a au b
0
7
12 2 2 1 2 240 10x y
L
WbI d I dx x
m
π −= × = ×−∫ ∫ a aF L B
( )( )712 2
712
40 10 ;
40 10 12.6
z
z z
I L x
Nx
L m
π
µ
π
−
−
= − −
= =
a
a a
F
F
,m=F g
( ) ( )
2
2; for 1 m 100 9.8 0.98 1000g g z z
m Ns kgm g N
s kg m g
⎛ ⎞⎛ ⎞= = − = −⎜ ⎟⎜ ⎟
⎝ ⎠ ⎝ ⎠
a aF g F
1
0
0.98
m
g zId N= × = − = +∫ aF L B F
3- !28
!
The unit conversion to arrive at
Newtons is as follows:
!
So we have Bo = 0.490 Wb/m2, and
B = 0.490 Wb/m2 (-ax) (directed north)
P3.40: Suppose you have a pair of parallel lines each with a mass per unit length of 0.10
kg/m. One line
sits on the ground and conducts 200. A in the +ax direction, and the other
one, 1.0 cm above the first (and parallel), has sufficient current to levitate. Determine the
current and its direction for line 2.
Here we will use !
!
So solving for I2:
!
P3.41: In Figure 3.57, a 2.0 A line of current is shown on the z-axis with the current in the
+az direction. A current loop exists on the x-y plane (z = 0) that has 4 wires (labeled 1
through 4) and carries 1.0 mA as shown. Find the force on each arm and the total force
acting on the loop from the field of the 2.0 A line.
!
!
( )
( )( ) ( )22 1 2
L
y o x o z
o
o o
Idy B ILB
WbA m B B N
m
× − =
⎛ ⎞= =⎜ ⎟
⎝ ⎠
∫ a a a
2
Am Wb Vs W J Nm N
m Wb VAWs J
=
7
31 212
2 2
4 10 200 4 10
2 2 0.01
o
y z z
I I x H m A I x I
L y m
µ π
π π
−
−= = =a a aF
( )( )
2
20.10 9.8 0.98F mg Ns Nkg m m s
L L kgm m
= = =
2 3
0.98 245 in the - direction.
4 10 x
I A
x −
= = a
1
12 2 2 1 1; = 2
oII d φ
µ
πρ
= ×∫ aF L B B
2: 5 , 0
: 0
A B d d
C D
φ φ φ
φ φ
φ→ = × =
→ × =
a a a
a a
L
Fig. P3.39
3- !29
!
!
So for B to C: F12 = -0.20 nN az
Likewise, from D to A: F12 = +0.20 nN az
P3.42: MATLAB: Modify MATLAB 3.4 to find the differential force acting from each
individual differential segment on the loop. Plot this force against the phi location of the
segment.
%MLP0342
%modify ML0304 to find dF acting from the field
% of each segment of current; plot vs phi
clear
clc
I=1; %current in A
a=1; %loop radius, in m
mu=pi*(4e-7); %free space permeability
az=[0 0 1]; %unit vector in z direction
DL1=a*2*(pi/180)*[0 1 0]; %Assume 2 degree increments
%DL1 is the test element vector
%F is the angle phi in radians
%xi,yi is location of ith element on the loop
%Ai & ai = vector and unit vector from origin
% to xi,yi
%DLi is the ith element vector
3
1 1 2
12 2
5
3: ln
2 2 5
o o
z
I I IB C I d ρ φ
µ µ
ρ
πρ π
⎛ ⎞→ = × = ⎜ ⎟
⎝ ⎠∫
a a aF
( )( )
7
3
12
4 10 32 10 ln 204
2 5 z z
x H m A A pNπ
π
−
− ⎛ ⎞= = −⎜ ⎟
⎝ ⎠
a aF
Fig. P3.41
3- !30
%Ri1 & ri1 = vector and unit vector from ith
% point to test point
for i=1:179
phi(i)=i*2;
F=2*i*pi/180;
xi=a*cos(F);
yi=a*sin(F);
Ai=[xi yi 0];
ai=unitvector(Ai);
DLi=(pi*a/90)*cross(az,ai);
Ri1=[a-xi -yi 0];
ri1=unitvector(Ri1);
num=mu*I*cross(DLi,ri1);
den=4*pi*(magvector(Ri1)^2);
B=num/den;
dFvect=I*cross(DL1,B);
dF(i)=dFvect(1);
end
plot(phi,dF)
xlabel('angle in degrees')
ylabel('the differential force, N')
P3.43: MATLAB: Consider a circular conducting loop of radius 4.0 cm in the y-z plane
centered at (0,6cm,0). The loop conducts 1.0 mA current clockwise as viewed from the
Fig. P3.42
3- !31
+x-axis. An infinite length line on the z-axis conducts 10. A current in the +az direction.
Find the net force on the loop.
The following MATLAB routine shows the force as a function of radial position around
the loop. Notice that while there is a net force in the -y direction, the forces in the z-
direction cancel.
% MLP0343
% find total force and torque on a loop of
% current next to a line of current
% variables
% I1,I2 current in the line and loop (A)
% yo center of loop on y axis (m)
% uo free space permeability (H/m)
% N number of segements on loop
% a loop radius (m)
% dalpha differential loop element
% dL length of differential section
% DL diff section vector
% B1 I1's mag flux vector (Wb/m^2)
% Rv vector from center of loop
% to the diff segment
% ar unit vector for Rv
% y,z the location of the diff segment
clc
clear
% initialize variables
I1=10;
I2=1e-3;
yo=.06;
uo=pi*4e-7;
N=180;
a=0.04;
dalpha=360/N;
dL=a*dalpha*pi/180;
ax=[1 0 0];
Fig. P3.43
3- !32
% perform calculations
for i=1:N
dalpha=360/N;
alpha=(i-1)*dalpha;
phi(i)=alpha;
z=a*sin(alpha*pi/180);
y=yo+a*cos(alpha*pi/180);
B1=-(uo*I1/(2*pi*y))*[1 0 0];
Rv=[0 y-.06 z];
ar=unitvector(Rv);
aL=cross(ar,ax);
DL=dL*aL;
dF=cross(I2*DL,B1);
dFx(i)=dF(1);
dFy(i)=dF(2);
dFz(i)=dF(3);
end
plot(phi,dFy,phi,dFz,'--k')
legend('dFy','dFz')
Fnet=sum(dFy)
Running the program we get:
Fnet = -4.2932e-009
>>
So Fnet = -4.3 nN ay
P3.44: MATLAB: A square loop of 1.0 A current of side 4.0 cm is centered on the x-y
plane. Assume 1 mm diameter wire, and estimate the force vector on one arm resulting
from the field of the other 3 arms.
% MLP0344 V2
%
% Square loop of current is centered on x-y plane.
Viewed
% from the +z axis, let current go clockwise. We want to
% find the force on the arem at x = +2 cm resulting from
the
% current in arms at y = -2 cm, x = -2 cm and y = +2 cm.
3- !33
% Wentworth, 12/3/03
% Variables
% a side length (m)
% b wire radius (m)
% I current in loop (A)
% uo free space permeability (H/m)
% N number of segments for each arm
% xi,yi location of test arm segment (at x = +2 cm)
% xj,yj location of source arm segment (at y = -2 cm)
% xk,yk location of source arm segment (at x = -2 cm)
% xL,yL location of source arm segment (at y = +2 cm)
% dLi differential test segment vector
% dLj, dLk, dLL diff vectors on sources
% Rji vector from source point j to test point i
% aji unit vector of Rji
clc;clear;
a=0.04;
b=.0005;
I=1;
uo=pi*4e-7;
N=80;
for i=1:N
xi=(a/2)+b;
yi=-(a/2)+(i-0.5)*a/N;
ypos(i)=yi;
dLi=(a/N)*[0 -1 0];
for j=1:N
xj=-(a/2)+(j-0.5)*a/N;
yj=-a/2-b;
dLj=(a/N)*[-1 0 0];
Rji=[xi-xj yi-yj 0];
aji=unitvector(Rji);
num=I*CROSS(dLj,aji);
den=4*pi*(magvector(Rji))^2;
H=num/den;
dHj(j)=H(3);
end
for k=1:N
3- !34
yk=(-a/2)+(k-0.5)*a/N;
xk=-(a/2)-b;
dLk=(a/N)*[0 1 0];
Rki=[xi-xk yi-yk 0];
aki=unitvector(Rki);
num=I*CROSS(dLk,aki);
den=4*pi*(magvector(Rki))^2;
H=num/den;
dHk(k)=H(3);
end
for L=1:N
xL=(-a/2)+(L-0.5)*a/N;
yL=(a/2)+b;
dLL=(a/N)*[1 0 0];
RLi=[xi-xL yi-yL 0];
aLi=unitvector(RLi);
num=I*CROSS(dLL,aLi);
den=4*pi*(magvector(RLi))^2;
H=num/den;
dHL(L)=H(3);
end
H=sum(dHj)+sum(dHk)+sum(dHL);
B=uo*H*[0 0 1];
F=I*CROSS(dLi,B);
dF(i)=F(1);
end
Ftot=sum(dF)
plot(ypos,dF)
Running the program:
Ftot =
7.4448e-007
So Ftot = 740 nN
Fig. P3.44
3- !35
P3.45: A current sheet K = 100ax A/m exists at z = 2.0 cm. A 2.0 cm diameter loop
centered in the x-y plane at z = 0 conducts 1.0 mA current in the +aφ direction. Find the
torque on this loop.
!
P3.46: 10 turns of insulated wire in a 4.0 cm diameter coil are centered in the x-y plane.
Each strand of the coil conducts 2.0 A of current in the aφ direction. (a) What is the
magnetic dipole moment of this coil? Now suppose this coil is in a uniform magnetic
field B = 6.0ax + 3.0ay + 6.0az Wb/m2, (b) what is the torque on the coil?
(a) !
(b) !
!
P3.47: A square conducting loop of
side 2.0 cm is free to rotate about one side that is
fixed on the z-axis. There is 1.0 A current in the loop, flowing in the –az direction on the
fixed side. A uniform B-field exists such that when the loop is positioned at φ = 90°, no
torque acts on the loop, and when the loop is positioned at φ = 180° a maximum torque of
8.0 µN-m az occurs. Determine the magnetic flux density.
At φ = 90°, ! . Also, since! B is in direction of m,
and therefore B = ±Boax.
At φ = 180°, ! , and !
Therefore, B = -Boax and mBo = 8x10-6, so
!
( ) ( )( )
( )
23 9 2
2
;
10 0.01 314 10
1 1; 100 50 ;
2 2
50 ;
20
N z z
o N x z y
o y
x
IS A m x Am
A A
m m
Wb
m
pNm
π
µ
µ
− −
= ×
= = =
⎛ ⎞= = × = × − =⎜ ⎟
⎝ ⎠
=
= ×
B
a a a
B H H a a a a
B a
B = - a
τ
τ
m
m
K
m
( )( ) ( )( )2 210 2 0.02 25.1x zNIS A m mAmπ= = =a am
( ) 225.1 6 3 6 0.151 0.075 Nmz x y z y x
WbmA
m
= × = × + + = −B a a a a a aτ m
75 151 x y mNm∴ = − +a aτ
20.0004N xIS Am= =a am 0,= × =Bmτ
20.0004N yIS Am= =a am 68 10 .zx Nm
−= × =B amτ
6
2 2
8 10 20 , and 20 .
0.0004o x
x mWb mWbB
m m
−
= = = −B a
3- !36
7. Magnetic Materials
P3.48: A solid nickel wire of diameter 2.0 mm evenly conducts 1.0 amp of current.
Determine the magnitude of the magnetic flux density B as a function of radial distance
from the center of the wire. Plot to a radius of 2 mm.
!
!
!
!
!
% MLP0348
% generate plot for ACL problem
a=2e-3; %radius of solid wire (m)
I=1; %current (A)
N=30; %number of data points to plot
rmax=4e-3; %max radius for plot (m)
dr=rmax/N;
uo=pi*4e-7;
ur=600;
for i=1:round(a/dr)
r(i)=i*dr;
B(i)=(ur*uo*I/(2*pi*a^2))*r(i);
end
for i=round(a/dr)+1:N
r(i)=i*dr;
B(i)=uo*I/(2*pi*r(i));
End
rmm=r*1000;
plot(rmm,B)
( )22 23
1 31.8
1 10
z z z
I A kA
a mx mπ π −
= = =J a a a
encd I d= =∫ ∫H L J S� �
2
2 22
I IH d d
a aφ
πρ ρ ρ φ ρ
π
= =∫
2 2for ; 2 2
r o IIa
a aφ φ
µ µ ρρ
ρ
π π
≤ = =H a B a
for ;
2 2
oIIa φ φ
µ
ρ
πρ πρ
≥ = =H a B a
Fig. P3.48
3- !37
xlabel('rho(cm)')
ylabel('B (Wb/m^2)')
grid on
8. Boundary Conditions
P3.49: A planar interface separates two magnetic media. The magnetic field in media 1
(with µr1) makes an angle α1 with a normal to the interface. (a) Find an equation for α2,
the angle the field in media 2 (that has µr2) makes with a normal to the interface, in terms
of α1 and the relative permeabilities in the two media. (b) Suppose media 1 is nickel and
media 2 is air, and that the magnetic field in the nickel makes an 89° angle with a normal
to the surface. Find α2.
!
!
!
P3.50: MATLAB: Suppose the z = 0 plane separates two magnetic media, and that no
surface current exists at the interface. Construct a program that prompts the user for µr1
(for z < 0), µr2 (for z > 0), and one of the fields, either H1 or H2. The program is to
calculate the unknown H. Verify the program using Example 3.11.
% M-File: MLP0350
%
% Given H1 at boundary between a pair of
1 1 1 2 1 1 1 1
1
2 1 1 1 2 2 2 1
2
; ;
;
N N T T T T T T N N r o N N
r
N N N N r o N N r o N N N N
r
H H H H B H
B B H H H H
µ µ
µ
µ µ µ µ
µ
= + = =
= = = ∴ =
H a a a a a a
a a a a
2 1
2
2 1
1
2
12 2
2 1
1 1 1
tan ;
tan tan
T T
N r
N
r
Tr r
r N r
α
µ
µ
µ µ
α α
µ µ
= =
= =
H H
H
H
H
H
1 2
2 1
1
1
2
tan tan
1tan tan89 5.5
600
r
r
µ
α α
µ
α
−
−
⎡ ⎤
= ⎢ ⎥
⎣ ⎦
⎡ ⎤= =⎢ ⎥⎣ ⎦
� �
Fig. P3.49
3- !38
% materials with no surface current at boundary,
% calculate H2.
%
clc
clear
% enter variables
disp('enter vectors quantities in brackets,')
disp('for example: [1 2 3]')
ur1=input('relative permeability in material 1: ');
ur2=input('relative permeability in material 2: ');
a12=input('unit vector from mtrl 1 to mtrl 2: ');
F=input('material where field is known (1 or 2): ');
Ha=input('known magnetic field intensity vector: ');
if F==1
ura=ur1;
urb=ur2;
a=a12;
else
ura=ur2;
urb=ur1;
a=-a12;
end
% perform calculations
Hna=dot(Ha,a)*a;
Hta=Ha-Hna;
Htb=Hta;
Bna=ura*Hna; %ignores uo since it will factor out
Bnb=Bna;
Hnb=Bnb/urb;
display('The magnetic field in the other medium is: ')
Hb=Htb+Hnb
Now run the program (for Example 3.11):
enter vectors quantities in brackets,
for example: [1 2 3]
relative permeability in material 1: 6000
relative permeability in material 2: 3000
unit vector from mtrl 1 to mtrl 2: [0 0 1]
material where field is known (1 or 2): 1
known magnetic field intensity vector: [6 2 3]
3- !39
ans =
The magnetic field in the other medium is:
Hb = 6 2 6
For a second test, run the program for problem P3.52(a).
enter vectors quantities in brackets,
for example: [1 2 3]
relative permeability in material 1: 4
relative permeability in material 2: 1
unit vector from mtrl 1 to mtrl 2: [0 0 -1]
material where field is known (1 or 2): 1
known magnetic field intensity vector: [3 0 4]
ans =
The magnetic field in the other medium is:
Hb = 3 0 16
P3.51: The plane y = 0 separates two magnetic media. Media 1 (y < 0) has µr1 = 3.0 and
media 2 (y > 0) has µr2 = 9.0. A sheet current K = (1/µo) ax A/m exists at the interface,
and B1 = 4.0ay + 6.0az Wb/m2. (a) Find B2. (b) What angles do B1 and B2 make with a
normal to the surface?
(a) !
(b) !
(c) !
(d) !
(e) !
(f) !
(g) !
Now for step (e):
1 4N y=B a
2 1 4N N y= =B B a
1 6T z=B a
1
1
1
2T
T z
r o oµ µ µ
= =
BH a
2
3 (see below)T z
oµ
=H a
2 2 27T r o T zµ µ= =B H a
2 24 27y z
Wb
m
= +B a a
Fig. P3.51
3- !40
!
!
Angles: !
P3.52: Above the x-y plane (z > 0), there exists a magnetic material with µr1 = 4.0 and a
field H1 = 3.0ax + 4.0az A/m. Below the plane (z < 0) is free space. (a) Find H2,
assuming the boundary is free of surface current. What angle does H2 make with a
normal to the surface? (b) Find H2, assuming the boundary has a surface current K = 5.0
ax A/m.
(a)
(1) ! , (2) ! , (3) ! ,(4) !
(5) ! , (6) ! , (7) !
!
(b)
Now step (6) becomes , shere a21 = az.
Let’s let ! , then !
Solve for A and B:
! ; so A = 3 and B = 5
Finally, ! .
( ) ( )21 1 2 1 2
1; y T z T z x
o
H H
µ
× − = − × − =a H H K a a a a
( )1 2 2 1 2
1 1 1 2 3; ; T T x x T T T
o o o o o
H H H H H
µ µ µ µ µ
− − = − = = + =a a
1 21 1
1 2
1 2
tan 56 ; tan 82T T
N N
B B
B B
α α− −
⎛ ⎞ ⎛ ⎞
= = = =⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟
⎝ ⎠ ⎝ ⎠
� �
1 4N z=H a 1 16N o zµ=B a 2 1 16N N o zµ= =B B a
2
2 16NN z
oµ
= =
BH a
1 3T x=H a 2 1 3T T x= =H H a
2 3 16x z
A
m
= +H a a
21
2
2
tan 10.6T
N
H
H
α −
⎛ ⎞
= =⎜ ⎟⎜ ⎟
⎝ ⎠
�
( )21 1 2× − =a H H K
2 16x y zA B= + +H a a a ( )21 3 4 16 5x z x y z xA B× + − − − =a a a a a a a
( )0 0 1 3 5
3 12
x y z
x y xB A
A B
⎡ ⎤
⎢ ⎥ = + − =⎢ ⎥
⎢ ⎥− − −⎣ ⎦
a a a
a a a
2 3 5 16x y z
A
m
= + +H a a a
3- !41
P3.53: The x-z plane separates magnetic material with µr1 = 2.0 (for y < 0) from magnetic
material with µr2 = 4.0 (for y > 0). In medium 1, there is a field H1 = 2.0ax + 4.0ay +
6.0az A/m. Find H2 assuming the boundary has a surface current K = 2.0ax – 2.0az A/m.
!
!
!
!
!
P3.54: An infinite length line of 2π A current in the +az direction exists on the z-axis.
This is surrounded by air for ρ ≤ 50 cm, at which point the magnetic media has µr2 = 9.0
for ρ > 50 cm. If the field in media 2 at ρ = 1.0 m is H = 5.0aφ A/m, find the sheet
current density vector at ρ = 50. cm, if any.
Method 1:
From just the line of current we would have !
Now, since ! is the contribution from the sheet
current.
!
!
Method 2:
From I1 at boundary we have
1 1 2 1 2
8(1) 4 ,(2) 8 ,(3) 8 ,(4) 2
4
o
N y N o y N N o y N y y
o
µ
µ µ
µ
= = = = = =H a B a B B a H a a
( )1 21 1 2 2(5) 2 6 ,(6) , let ,T x z T x x z zH H= + × − = = +H a a a H H K H a a
( ) ( )( )so 2 6 2 2 ,y x x z z x zH H− × − + − = −a a a a a
( ) ( )0 1 0 6 2 2 2 , so 8, 4
2 0 6
x y z
z x x z x z z x
x z
H H H H
H H
− = − + − = − = =
− −
a a a
a a a a
2 2(7) 4 8 , 4 2 8 T x z x y z
A
m
= + ∴ = + +H a a H a a a
( )
1
1 1 .2 1
I
φ φπ
= =H a a
1 2 25 , then 4TOT φ φ= = + =H a H H H a
( ) ( )
2 2
2 2
84 , so 8 , then 8
2 1 2 2 0.5z z
I I AI A
a mφ φ
π
π
π π π
= = = = = =H a a K a a
8 z
A
m
∴ =K a
3- !42
! since H varies
as 1/ρ. So
!
!
9. Inductance and Magnetic Energy
P3.55: Consider a long pair of straight parallel wires, each of radius a, with their centers
separated by a distance d. Assuming d >> a, find the inductance per unit length for this
pair of wires.
Each wire can be solved using the
energy approach:
!
The fields are not confined to a
volume, so we must use the flux
linkage approach to find inductance
outside the wires.
!
!
Finally, with ! .
P3.56: In problem P3.23 the task was to find the field at the center (radius b) of an N-turn
toroid. If the radius of the toroid is large compared to the diameter of the coil (that is, if b
>> c-a), then the field is approximately constant from radius a to radius c. (a) Obtain an
expression for the toroid’s inductance. (b) Find L if there are 600 turns around a 99.8%
iron core with a = 8.0 cm and c = 9.0 cm.
( )1
2
2 , but 5 at = 1.0m corresponds to 10 at =0.5m
2 0.5
φ
φ φ φ
π
ρ ρ
π
= =
a
H a a a
( ) ( )2 10 , 8 8 zρ φ φ ρ φ− × − = − × − = =a a a K a a a K
8 z
A
m
∴ =K a
2
8 4wires
L
h
µ µ
π π
= =
( )1 2
1
, ,
2 2
, 2 ,TOT
I I
d
d d
φ φπρ π ρ
φ φ
= =
−
= =∫ ∫
H a H a
B S B S� �
0
22 ln
2 2
d a h
TOT
a
I I d I d ad dz dz h
aφ φ
µ µ ρ µ
φ ρ
πρ π ρ π
− −⎛ ⎞= = = ⎜ ⎟
⎝ ⎠∫ ∫ ∫
a a�
1, we arrive at ln
4
TOT
TOT
L d aL
I h a
φ µ
π
⎡ − ⎤⎛ ⎞= = + ⎜ ⎟⎢ ⎥
⎝ ⎠⎣ ⎦
Fig. P3.55
3- !43
From P3.23 we found: ! at radius b for the toroid.
!
and with area ! we have !
!
!
Plugging in our numbers we have: !
P3.57: MATLAB: Consider that a solid wire of radius a = 1.0 mm is bent into a circular
loop of radius 10. cm. Neglecting internal inductance of the wire, write a program to find
the inductance for this loop.
% M-File: MLP0357
%
% Inductance inside a conductive loop
%
% This modifies ML0302 to calculate inductance of
% a conductive loop. It does this by calculating the
% mag field at discrete points along a pie wedge,
% then calculates flux through each portion of the
% wedge. Then it multiplies by the number of wedges
% in the 'pie'.
%
% Wentworth, 1/19/03
%
% Variables:
% I current(A) in +phi direction on ring
% a ring radius (m)
% b wire radius (m)
% Ndeg number of increments for phi
% f angle of phi in radians
2
NI
b φπ
=H a
2
NId dS
b φ φ
µ
φ
π
= =∫ ∫B S a a� �
( )2 ,b aπ − ( )
2 .
2
NI b a
b
µ
φ = −
( )
2
2
2
N IN b a
b
µ
λ φ= = −
( )
2
2
2
NL b a
I b
λ µ
= = −
( )( )( ) ( )
( )
2275000 4 10 600 .085 .080
0.33
2 0.085
x
L H
π − −
= =
3- !44
% df differential change in phi
% dL differential length vector on the ring
% dLmag magnitude of dL
% dLuv unit vector in direction of dL
% [xL,yL,0] location of source point
% Ntest number of test points
% Rsuv unit vector from origin to source point
% R vector from the source to test point
% Ruv unit vector for R
% Rmag magnitude of R
% dH differential contribution to H
% dHmag magnitude of dH
% radius radial distance from origin
% Hz total magnetic field at test point
% Bz total mag flux density at test point
% flux flux through each differential segment
clc %clears the command window
clear %clears variables
% Initialize Variables
a=0.1;
b=1e-3;
I=1;
Ndeg=180;
Ntest=60;
uo=pi*4e-7;
df=360/Ndeg;
dLmag=(df*pi/180)*a;
dr=(a-b)/Ntest;
% Calculate flux thru each segment of pie wedge
for j=1:Ntest
x=(j-0.5)*dr;
for i=(df/2):df:360
f=i*pi/180;
xL=a*cos(f);
yL=a*sin(f);
Rsuv=[xL yL 0]/a;
dLuv=cross([0 0 1],Rsuv);
dL=dLmag*dLuv;
R=[x-xL -yL 0];
3- !45
Rmag=magvector(R);
Ruv=R/Rmag;
dH=I*cross(dL,Ruv)/(4*pi*Rmag^2);
dHmag(i)=magvector(dH);
end
Hz(j)=sum(dHmag);
Bz(j)=uo*Hz(j);
dSz(j)=x*df*(pi/180)*dr;
flux(j)=Bz(j)*dSz(j);
end
fluxwedge=sum(flux);
Inductance=Ndeg*fluxwedge
Now run the program:
Inductance = 5.5410e-007
or
L = 550 nH
P3.58: Find the mutual inductance between an infinitely long wire and a rectangular wire
with dimensions shown in Figure 3.58.
!
!
!
P3.59: Consider a pair of concentric conductive loops, centered in the same plane, with
radii a and b. Determine the mutual inductance between these loops if b >> a.
In this case will drive the b-radius loop with current I. Here, at the center of the b-radius
loop we have from Eqn. (3.10):
1 1
1 1, 2 2
oI I
φ φ
µ
πρ πρ
= =H a B a
1 1
12 1 2
0
ln
2 2
o
o
a b
o o o
o
I d I b ad dz
ρ
ρ
µ ρ µ ρ
φ
π ρ π ρ
+ ⎛ ⎞+
= = = ⎜ ⎟
⎝ ⎠
∫ ∫ ∫B S�
12
12
1
ln
2
o o
o
b aM
I
φ µ ρ
π ρ
⎛ ⎞+
= = ⎜ ⎟
⎝ ⎠
3- !46
!
Then, !
So !
P3.60: A 4.0 cm diameter solid nickel wire, centered on the z-axis, conducts current with
a density J = 4ρ A/cm2 az (where ρ is in cm). Find the internal inductance per unit length
for the wire with this current distribution.
!
!
!
!
Now we solve for I: ! and then solve for L:
!
!
10. Magnetic Circuits
P3.61: Referring to Figure 3.48(a), suppose 2.0 Amps flows through 80 turns of a toroid
that has a core cross sectional area of 2.0 cm2 and a mean radius of 80. cm. The core is
99.8% pure iron. (a) How much magnetic flux exists in the toroid? (b) How much energy
is stored in the magnetic field contained by the toroid?
1 2
o
z
I
b
µ
=B a
2 2
1 1 1
12 1 2
0
2
2 2 2 2
a
o o oI I a Id d d
b b b
ρ
µ πµ ρ πµ
φ ρ ρ φ
=
= = = =∫ ∫B S�
2
12
12
1 2
oaM
I b
φ πµ
= =
2
2 3
0 0
8, 2 4
3enc
d I H d d
ρ π
φ
π
πρ ρ ρ φ ρ= = =∫ ∫ ∫H L��
3
2 28 4 4, B
6 3 3
Hφ φ
πρ µ
ρ ρ
πρ
= = =
22 2
5 6
0 0 0
1 4 4 1 16 8
2 3 3 2 9 27
a h
mW d d dz d d dz a h
π
φ φ
µρ ρ
ρ ρ φ µ ρ ρ φ µ π= = =∫ ∫ ∫ ∫a a�
21 ;
2m
W LI=
38 ,
3
aI d π= =∫ J S�
1
12
hL µ
π
=
71 1 4 10600 20 .
12 12
L x HL
h m
µ π µ
π π
−
= = = =
3- !47
(a) !
(b) !
P3.62: In Figure 3.59, a 2.0 cm diameter toroidal core with µr1 = 10,000 is wrapped with
a 1.0 cm thick layer of µr2 = 3000. The toroid has a 1.0 m mean radius. For 20. A of
current driven through 50 loops of wire, find the magnetic field intensity in each material
of the toroid.
This toroid can be modeled with the circuit of Figure P3.62.
Vm = NI = (50)(20A)=1000 A-turns
!
!
!
!
( )( )( )( )
( )
7
2
5000 4 10 80 2
; 0.200 , 40
2 2 0.8o
xNI WbBA A B Wbm
πµ
φ φ µ
πρ π
−
= = = = =
21 1 2 3.2
2 2m o
BW BH dv A mJπρ
µ
= = =∫
( )
6 6
1 22 2 2
1 2
2 2; 1.592 10 ; 1.768 10 ;o o
r o r o
x x
A a b a
πρ πρ
µ µ µ π µ µ π
= =
−
lR = R = R =
( )
6
6
1 12
1 1
628 10628 10 ; 2; 159
0.01
m
r o
V x B Ax BA B H
m
φ
µ µπ
−
−= = = = = = =R
( ) ( )( )
6
6
2 2 2
2
566 10566 10 ; 0.6001
0.02 0.01
mV xx BA Bφ
π
−
−= = = = =
−R
2
2
159
r o
B AH
mµ µ
= =
Fig. P3.62
3- !48
P3.63: Referring to Figure 3.49(a), the 2.0 cm diameter core is characterized by the
magnetization curve of Figure 3.60. The toroid has a mean radius of 60. cm. For 10. A
of current driven through 100 loops of wire, find the magnetic field intensity in the 1.0
mm gap.
Referring to the model of Figure 3.49(b),
Vm = NI = (100)(10A)=1000 A-turns; A = πa2 = 314x10-6
! ;
Approach:
(1) Assume B: B = 0.4
(2) Read H from chart: H = 200
(3) Evaluate : Rc =6x106
(4) RTOT = Rc + 2.53x106: RTOT =8.53x106
(5) φ = Vm/ RTOT: φ =117x10-6
(6) B = φ/A: B = 0.373
Insert this value of B into step (1) and repeat. After 4 or 5 iterations the routine
converges to a solution B = 0.382 Wb/m2 and H = 190 kA/m. Then, in the gap, we have
!
P3.64: Referring to Figure 3.52, suppose the cross sectional area of the bar is 3.0 cm2,
while that of the electromagnet core is 2.0 cm2. Also, the bar has a relative permittivity
of 3000, while that of the magnetic core is 10,000. The dimensions for h and w are 12.
cm and 16. cm, respectively. If the mass of the bar is 20. kg, how much current must be
driven through 24 loops to hold up the bar against gravity?
We will follow an approach similar to Example 3.18, but realize that Ac ≠ Ab.
!
62.53 10GG
o
x
Aµ
lR = =
32 12 10o
C
x H
A B
πρ
µ
R = =
300 .
o
B kAH
mµ
= =
2
3
2
2(.12) 0.16 100 159 10
10,000 (2 )c c o
cm x
A cm mµ µ
+ ⎛ ⎞
⎜ ⎟
⎝ ⎠
lR = = =
3- !49
!
!
!
Solve for I: !
P3.65: Consider a 1.0 mm air gap in Figure 3.49(a). The toroid mean radius and cross
sectional area are 50. cm and 2.0 cm2, respectively. If the magnetic core has a µr = 6000,
and 4.0 A is being driven through 30 loops, determine the force pulling the gap closed.
!
!
In the gap, B = 19.79x10-6/A = 98.97x10-3 = µoH, H = 78.76x103A/m
!
F = 1.6N
2
3
2
0.16 100 141 10
3,000 (3 )b b o
cm x
A cm mµ µ
⎛ ⎞
⎜ ⎟
⎝ ⎠
lR = = =
NI, ; H=mm o c
TOT TOT TOT o c
V NIV NI HA
A
φ µ
µ
= = = =R R R
2 2
2 NI NI 1
o c o c
TOT o c TOT o c
F H A A mg
A A
µ µ
µ µ
⎛ ⎞ ⎛ ⎞
= = = =⎜ ⎟ ⎜ ⎟
⎝ ⎠ ⎝ ⎠R R
2.8TOTo cI mg A AN
µ= =
R
( )( )2 62, 6.06 10oo TOT
r o o
A turns
F H A x
A A Wb
πρ
µ
µ µ µ
= + =
lR =
( )( ) 6
6
30 4
19.79 10
6.06 10
m
TOT
V x BA
x
φ −= = = =R
( )( ) ( )
2
22 7 3 14 10 78.76 10 2 1.6
100o
F H A x xµ π − ⎛ ⎞= = =⎜ ⎟
⎝ ⎠
Wentworth solutions chapter 4.pdf
4- !1
Solutions for Chapter 4 Problems
1. Current Continuity and Relaxation Time
P4.1: How long does it take for charge density to drop to 1% of its initial value in
polystyrene?
Polystyrene has εr = 2.56 and σ = 10-17 S/m.
!
!
Solving for t we get
t = 10.4x106 sec = 120 days
P4.2: At a particular point in a slab of silver, a charge density of 109 C/m3 is introduced.
Plot ρv versus time for a duration of 10 relaxation times.
For silver we have εr = 1 and σ = 6.2x107 S/m
!
The relaxation time is: !
The plot is obtained with the following MATLAB routine.
% MLP0402
%
% Plot rhov vs time
% for a charge placed in silver
rhovo=1e9; %initial charge density, C/m^3
er=1; %relative permittivity of silver
sig=6.2e7; %conductivity, S/m
eo=8.854e-12;
tau=er*eo/sig;
t=tau/10:tau/10:10*tau;
rho=rhovo*exp(-t./tau);
subplot(2,1,1)
plot(t,rho)
( )/ t
v oe
σ ερ ρ −=
( )( )
9
17
441 10
12
100.01 exp
2.56 8.854 10
x t
o o ot ex
ρ ρ ρ
−
−
−
−
⎛ ⎞−
= =⎜ ⎟
⎜ ⎟
⎝ ⎠
( ) 18/ 9 7 10
3
C10
m
t x t
v oe e
σ ερ ρ − −= =
21142 10 secxετ
σ
−= =
4- !2
xlabel('t (s)')
ylabel('rhov (C/m^3)')
grid on
subplot(2,1,2)
loglog(t,rho)
xlabel('t (s)')
ylabel('rhov (C/m^3)')
grid on
P4.3: A current density is given by J = ρe-.01taρ A/m2. Find the charge density after 10
seconds if it has an initial value of zero.
!
!
At t = 0, ρv = 0 = 200 + C, C = -200.
So we have !
P4.4: At t = 0 seconds, 60.0 µC is evenly distributed throughout a 2.00 cm diameter pure
silicon sphere. (a) Find the initial charge density. (b) How long does it take to drop to
10% of its initial value? (c) What will be the final surface charge density?
( )2 0.01 0.011 2t tv e et t
ρ
ρ
ρ
− −∂ ∂∇ = − = =
∂ ∂
J�
0.01 0.012 ; 200t tv vd e dt e Cρ ρ
− −= − = +
( )0.01(10) 3200 1 19v
Ce
m
ρ −= − = −
Fig. P4.02
4- !3
(a) Volume !
(b) !
For pure silicon, !
!
(c) !
2. Wave Fundamentals
P4.5: A propagating electric field is given by
!
(a) Determine the attenuation constant, the wave frequency, the wavelength, the
propagation velocity and the phase shift. (b) How far must the wave travel before its
amplitude is reduced to 1.0 V/m?
(a) !
(b) !
P4.6: A 10.0 MHz magnetic field travels in a fluid for which the propagation velocity is
1.0x108 m/sec. Initially, we have H(0,0)=2.0 ax A/m. The amplitude drops to 1.0 A/m
after the wave travels 5.0 meters in the y direction. Find the general expression for this
wave.
The general expression for the wave is: !
!
!
( )3 6 3 3
4 0.01 4.19 10 ; 14.3
3 vo
Q Cv m x m
v m
π ρ−= = = =
/0.10 tv vo voe
τρ ρ ρ −= =
( )( )
( )
12
9
4
11.8 8.854 10
237.4 10 sec
4.4 10
x
x
x
ε
τ
σ
−
−
−
= = =
( )ln 0.10 2.303; 2.303 547t t nsτ
τ
−
= = − = =
2 3 2
23 2
60; area=4 r 1.257 10 ; 47.8
1.257 10s s
Q C mCx m marea x m
µ
ρ π ρ−
−
= = = =
( ) .01 7, 100. cos 10 .
4
z VE
z t e t z
m
π
π π− ⎛ ⎞= × + −⎜ ⎟
⎝ ⎠
70.01 ; 5 ; 2 ; 1 10 ; 45
4p
mNp m f MHz m u x
s
π
α λ φ
−
= = = = = = − �
0.01 1( ) 100 1; 0.01 ln ; 460
100
zE z e z z m− ⎛ ⎞= = − = =⎜ ⎟
⎝ ⎠
( )( , ) cosyo x
Ay t H e t y
m
α ω β φ−= − +H a
( )6 62 2 10 10 20 10 radf x x sω π π π= = =
8 21 10 ; 10 ; 0.2pp
u radu x m s m
f m
π
λ β π
λ
= = = = =
4- !4
!
!
Finally,
!
P4.7: MATLAB: Modify the simple wave program in MATLAB 4.1 to include
attenuation. Generate a plot for the case where the amplitude is 4 V/m, the attenuation
constant is 0.001 Np/m, and the frequency is 1 MHz. Take your snapshot in time at 0
seconds, and let your phase shift be 0°.
% M-File: MLP0407
%
% This is a modification of ML0401 that
% includes attenuation. It plots a wave (in vac)
% versus position (in z direction) for a fixed time.
%
% Wentworth, 1/19/03
%
% Variables:
% atten attenuation constant
% Eo wave amplitude (V/m)
% f frequency (Hz)
% omega angular frequency (rad/sec)
% t time snapshot (sec)
% phi phase constant (degrees)
% phir phase constant (radians)
% c speed of light in vacuum (m/s)
% lambda wavelength (m)
% B phase constant (1/m)
% E electric field intensity
% z position
clc %clears the command window
clear %clears variables
% Initialize Variables
atten=0.001;
Eo=4;
(0,0) 2.0 ; 2.0, 0x o
A H
m
φ= ∴ = =H a
(5)( 5) 1 2.0 ; solving we get 0.14yoH y H e e
α α α− −= = = = =
( )0.14 6( , ) 2.0 cos 20 10 0.2y x Ay t e x t y mπ π
−= −H a
4- !5
f=1e6;
t=0;
phi=0;
phir=phi*pi/180;
c=2.998e8;
lambda=c/f;
B=2*pi/lambda;
omega=2*pi*f;
% Perform Calculation
z=0:4*lambda/100:4*lambda;
E=Eo*exp(-atten.*z).*cos(omega*t-B*z+phir);
% Generate the Plot
plot(z,E)
axis('tight') %sets axes min & max data values
grid
xlabel('z(m)')
ylabel('E(V/m)')
P4.8: MATLAB: Modify the traveling wave program in MATLAB 4.2 to include
attenuation. Use the parameters from problem P4.7, except for the fixed time of course.
% M-File: MLP0408
%
% This program illustrates a traveling wave including
Fig. P4.7
4- !6
% attenuation. It modifies ML0402.
%
% Wentworth, 1/19/03
%
% Variables:
% atten attenuation (Np/m)
% Eo wave amplitude (V/m)
% f frequency (Hz)
% omega angular frequency (rad/sec)
% t time snapshot (sec)
% phi phase constant (degrees)
% phir phase constant (radians)
% c speed of light in vacuum (m/s)
% lambda wavelength (m)
% B phase constant (1/m)
% E electric field intensity
% z position
clc %clears the command window
clear %clears variables
% Initialize Variables
atten=0.001
Eo=4;
f=1e6;
t=1;
phi=0;
phir=phi*pi/180;
c=2.998e8;
lambda=c/f;
B=2*pi/lambda;
omega=2*pi*f;
% Perform Calculation
z=0:4*lambda/100:4*lambda;
E=Eo.*exp(-atten.*z).*cos(omega*t-B*z+phir);
% Generate a Reference Frame
plot(z,E)
axis([0 4*lambda -2*Eo 2*Eo])
grid
xlabel('z(m)')
4- !7
ylabel('E(V/m)')
pause
% Make the Movie
t=0:1/(40*f):1/f;
for n=1:40;
E=Eo.*exp(-atten.*z).*cos(omega*t(n)-B*z+phir);
plot(z,E)
axis([0 4*lambda -2*Eo 2*Eo])
grid
title('General Wave Equation');
xlabel('z(m)');
ylabel('E(V/m)');
M(:,1)=getframe;
end
3. Faraday’s Law and Transformer EMF
P4.9: The magnetic flux density increases at the rate of 10 (Wb/m2)/sec in the z direction.
A 10 cm x 10 cm square conducting loop, centered at the origin in the x-y plane, has 10
ohms of distributed resistance. Determine the direction (with a sketch) and magnitude of
the induced current in the conducting loop.
!
210 z
d Wb
dt m s
=
B a
Fig. P4.8 (at end of animation)
4- !8
!
!
I=!
I=10mA clockwise (when viewed
From +z)
P4.10: A bar magnet is dropped through a
conductive ring. Indicate in a sketch the direction of the induced current when the falling
magnet is just above the plane of the ring and when it is just below the plane of the ring,
as shown in Figure 4.22.
Refer to Figure P4.10.
When the north pole first goes through the loop, flux is increasing and the current
induced to oppose this change in flux is as shown.
When the south pole is exiting the loop, flux is decreasing and the current induced acts to
oppose this change in flux.
P4.11: Considering Figure 4.7, suppose the area of a single loop of the pair is 100 cm2,
and the magnetic flux density is constant over the area of the loops but changes with time
210emf z z
d WbV d dxdy
dt m s
= − = −∫ ∫
B S a a� �
0.1 0.1emf
Wb VsV V
s Wb
= − = −
0.1 10
10
VI mA= =
Ω
Fig. P4.9
Fig. P4.10
4- !9
as where Bo = 4.0 mWb/m2 and χ = 0.30 Np/sec. Determine VR at 1, 10,
and 100 seconds.
!
!
at t = 1 sec, VR = 17.8 µV
at t = 10 sec, VR = 1.20 µV
at t = 100 sec, VR = 2.25x10-18 V
P4.12: Sometimes a transformer is used as an impedance converter, where impedance is
given by v/i. Find an expression for the impedance Z1 seen by the primary side of the
transformer in Figure 4.11 that has a load impedance Z2 terminating the secondary.
We have ! and !
!
!
P4.13: A 1.0 mm diameter copper wire is shaped into a square loop of side 4.0 cm. It is
placed in a plane normal to a magnetic field increasing with time as B = 1.0 t Wb/m2 az,
where t is in seconds. (a) Find the magnitude of the induced current and indicate its
direction in a sketch. (b) Calculate the magnetic flux density at the center of the loop
resulting from the induced current, and compare this with the original magnetic flux
density that generated the induced current at t = 1.0 sec.
We find the distributed resistance of the loop and work the problem assuming this
resistance is lumped in one spot as shown in the figure.
(a) The induced current is Vemf divided by the distributed resistance of the wire loop.
!
,to zB e
χ−=B a
d; ; 2
dt
t t
emf o z emf oV N d B e V B e St
χ χχ χ− −
∂
= − = − =
∂∫
B BS a�
( )( )( )
2
3 2 0.30 6 0.3012 2 0.30 4 10 100 24 10
100
t t t
R o
mV B Se x cm e x e
cm
χχ − − − − −⎛ ⎞= = =⎜ ⎟
⎝ ⎠
1
2 1
2
Ni i
N
= 22 1
1
Nv v
N
=
2
21
11 2 2
1 2 1
1 2 11
1
2
,
N v
Nv v NZ Z Z
i i NN i
N
⎛ ⎞
⎜ ⎟ ⎛ ⎞⎝ ⎠= = = = ⎜ ⎟⎛ ⎞ ⎝ ⎠
⎜ ⎟
⎝ ⎠
2
1
1 2
2
NZ Z
N
⎛ ⎞
= ⎜ ⎟
⎝ ⎠
( )
( )27
4 0.04 )1 1 3.5
5.8 10 0.0005
ml mR m
A x mσ π
Ω
= = = Ω
4- !10
!
! (note that this answer has no time dependence)
(b) The field at the center of the loop from a single arm of the loop is found from Eqn.
(3.7):
!
So !
P4.14: The mean length around a nickel core of a transformer like the one shown in
Figure 4.11 is 16 cm, and its cross sectional area is 1 cm2. There are 30 turns on the
primary side and 45 on the secondary side. If the current on the primary side is 1.0
sin20πx106t mA, (a) calculate the amplitude of the magnetic flux in the core in the
absence of the output windings. (b) With the output windings in place, calculate i2.
(a) !
!
(b) !
0.04 0.04
2 2
0 0
1 ; 1 1.6z emf
d Wb d WbV d dx dy mV
dt m s dt m s
= = − = − = −∫ ∫ ∫
B Ba S�
1.6
0.46
3.5ind
mVI A
m
= =
Ω
( ) ( ) ( )
( )
( )
2 2 2
0.461 1 2.59 ;
4 2 22 2 0.02
a
z z z z
a
I z I A
maz
ρ
π π πρ ρ
+
−
⎡ ⎤
= − = = − = −⎢ ⎥
+⎢ ⎥⎣ ⎦
H a -a a a
24 13 .o z
Wb
m
µ
µ= = −B H a
1 1; ; m m
r o
V V N i
A
φ
µ µ
= = =
lRR
( )( )( )( )( )
( ) ( )
7 2 2
91 1
2
30 1 600 4 10 1
14 10
0.16 100
r o
mA x cmN i A m x Wb
m cm
πµ µ
φ
−
−= = =l
( )( ) ( )6 612 1
2
30 21sin 20 10 sin 20 10
45 3
Ni i x t mA x t mA
N
π π= = =
Fig. P4.13
4- !11
P4.15: A triangular wire loop has its vertices at the points (2, 0, 0), (0, 3, 0) and (0, 0, 4),
with dimensions in meters. A time-varying magnetic field is given by B = 4t ay Wb/m2 (t
in seconds). If the wire has a total distributed resistance of 2 Ω, calculate the induced
current and indicate its direction in a carefully drawn sketch.
!
!
!
Vemf = -(4)(4)=-16V
!
4. Faraday’s Law and Motional EMF
P4.16: Referring to Figure 4.23, suppose a conductive bar of length h = 2.0 cm moves
with velocity u = -1.0 m/s aρ towards an infinite length line of current I = 4.0 A. Find an
expression for the voltage from one end of the bar to the other when ρ reaches 10 cm and
indicate which end is positive.
In Figure P4.16, an imaginary circuit has been chosen. For the chosen circulation
direction, we have the sign for Vemf as shown. Then,
! ,
! Therefore, the bottom of the bar is positive.
2
,
4
1
2
emf
y
V d
t
Wb
t m s
∂
= −
∂
∂
=
∂
×
= ×
×
∫
B S
B a
M NS M N
M N
�
2 3 , 2 4x y x z= − + = − +M a a N a a
26 4 3 x y z m= + +S a a a
16 8
2
emfV VI A
R
= = =
Ω
( )
0
1 1
2 2
h
o o
emf z
I ImV d dz dz
sρ φ
µ µ
πρ πρ
⎛ ⎞ ⎛ ⎞= × = × = −⎜ ⎟⎜ ⎟
⎝ ⎠⎝ ⎠
∫ ∫ ∫u B L - a a a� �� �
160 .emfV nV= −
Fig. P4.15
Fig. P4.16
4- !12
P4.17: Suppose we have a conductive bar moving along a pair of conductive rails as in
Figure 4.12, only now the magnetic flux density is B = 4.0ax + 3.0az Wb/m2. If R = 10.
Ω, w = 20. cm, and uy = 3.0 m/s, calculate the current induced and indicate its direction.
!
! (clockwise when viewed from the +z axis)
P4.18: The radius r of a perfectly conducting metal loop in free space, situated in the x-y
plane, increases at the rate of (πr)-1 m/sec. A break in the loop has a small 2.0 ohm
resistor across it. Meanwhile, there exists a magnetic field B = 1.0 az T. Determine the
current induced in the loop, and show in a sketch the direction of flow.
Here we’ve assumed dS = -dSaz to get iind and Vemf as shown. Our approach will be to
find φ, then Vemf = -dφ/dt.
!
!
P4.19: Rederive Vemf for the rectangular loop of Figure 4.16 if the magnetic field is now
B = Boaz.
We see in Figure P4.19a that ! for the ! line sections.
For the ! section we have:
( )23 3 0.2 1.8emf y z
Wb mV u dx m V
m s
⎛ ⎞⎛ ⎞= − = − = −⎜ ⎟⎜ ⎟
⎝ ⎠⎝ ⎠∫
B a�
1.8 0.18
10
VI A= =
Ω
( )
2
0
1
2
z z
r
d d d
d r
φ ρ ρ φ
π ρ ρ π
= = −
= − = −
∫ ∫
∫
B S a a� �
12 2 2
2
2 1 , clockwise as shown
2
emf
d drr r V
dt dt r
V V
VI A
φ
π π
π
⎛ ⎞= − = − = −⎜ ⎟
⎝ ⎠
= +
= =
Ω
( ) 0d× =B �u L 1 2 and 3 4 → →
2 3→
Fig. P4.18
4- !13
!
!
! , so for the ! section, the contributions to Vemf
cancel. This will also be the case for the section, and therefore Vemf = 0; no current
is induced.
P4.20: In Figure 4.16, replace the rectangular loop with a circular one of radius a and
rederive Vemf.
!
!
P4.21: A conductive rod, of length 6.0 cm, has one end fixed on a grounded origin and is
free to rotate in the x-y plane. It rotates at 60 revolutions per second in a magnetic field
B = 100. mT az. Find the voltage at the end of the bar.
!
!
( )
3
2
; , , o z
ddd B d d
dt dt
φ
ρ φ
ρ φ
ρ ρω= × = = = = =∫ ∫
a
B B a a a�u L L u l
( )
0 2
2
o
o
a
B ad B d ωω ρ ρ −× = =∫ ∫B �u L
( )
2
0 2
a
o
o
B ad B d ωω ρ ρ +× = =∫ ∫B �u L
2 3→
4 1→
( )2 2sin cos coso y o y x y od B d B a B aφ π φ φ π φ= = − + =∫ ∫B S a S = a a a� � �
2
2
sin ;
sin
flux
o
flux
emf o
d
B a
dt
d
V B a
dt
φ
ω π φ
φ
ω π φ
= −
= − =
2 60 120rev rad rad
s rev s
π
ω π⎛ ⎞⎛ ⎞= =⎜ ⎟⎜ ⎟
⎝ ⎠⎝ ⎠
( )emfV d= ×∫ B ��u L
Fig. P4.19bFig. P4.19a
4- !14
!
!
!
We can confirm the sign by observing that a
positive charge placed in the middle of the
bar would move to the ungrounded end by
the Lorentz force equation.
P4.22: Consider the rotating conductor shown in Figure 4.24. The center of the 2a
diameter bar is fixed at the origin, and can rotate in x-y plane with B = Boaz. The outer
ends of the bar make conductive contact with a ring to make one end of the electrical
contact to R; the other contact is made to the center of the bar. Given Bo = 100. mWb/m2,
a = 6.0 cm, and R = 50. Ω, determine I if the bar rotates at 1.0 revolution per second.
!
Figure P4.22 indicates one of the paths for the circulation integral.
!
!
!
d
dt φ φ
φ
ρ ωρ= =a au
( )
2
0 2
o
emf o z
BV B dφ ρ
ω
ωρ ρ= × =∫ a a a�
l l
( )22
1 120 0.1 0.06
2
68
emf
rad Wb VsV m
s m Wb
mV
π⎛ ⎞⎛ ⎞= ⎜ ⎟⎜ ⎟
⎝ ⎠⎝ ⎠
=
( ) ; , emf
dV d d d
dt φ φ ρ
φ
ρ ωρ ρ= × = = =∫ B a a a��u L u L
( )
2
0 0 2
a a
o
emf o z o
B aV B d B dφ ρ
ω
ωρ ρ ω ρ ρ⎡ ⎤= × = =⎣ ⎦∫ ∫a a a�
( )
2
23
2
1 11 2 100 10 0.06
2 2 50
emf oV B a rev rad Wb Vs AI x m
R R s rev m Wb V
ω
π −
Ω⎛ ⎞⎛ ⎞⎛ ⎞ ⎛ ⎞= = = ⎜ ⎟⎜ ⎟⎜ ⎟ ⎜ ⎟Ω⎝ ⎠⎝ ⎠⎝ ⎠ ⎝ ⎠
22.6I Aµ=
Fig. P4.21
Fig. P4.22
4- !15
P4.23: A Faraday Disk Generator is similar to the rotating conductor of P4.22, only now
the rotating element is a disk instead of a bar. Derive an expression of the Vemf produced
by a Faraday Disk Generator, and using the parameters given in problem 4.22, find I.
Worked exactly as P4.22.
P4.24: Consider a sliding rail problem where the conductive rails expand as they progress
in the y direction as shown in Figure 4.25. If w = 10. cm and the distance between the
rails increases at the rate of 1.0 cm in the x direction per 1.0 cm in the y direction, and uy
= 2.0 m/sec, find the Vemf across a 100. Ω resistor at the instant when y = 10. cm if the
field is Bo = 100. mT.
First we modify the figure so that the top rail is horizontal and all the spreading occurs
via the bottom rail. As before, our approach will be to find φ and then dφ /dt. We have:
!
Now, notice that x and y are not independent and are in fact related: x=y+w
So we have
!
!
Alternate Method:
!
o z zd B dxdyφ = =∫ ∫B S a a� �
( ) 2
0 0 0
1
2
y y w y
o o o
y x
B dxdy B y w dy B y wyφ
+
= =
⎛ ⎞= = + = +⎜ ⎟
⎝ ⎠∫ ∫ ∫
( ) ( ) ( )20.100 0.1 0.1 2
40
emf o o y
emf
d dy Wb mV B y w B y w u m m
dt dt m s
V mV
φ ⎛ ⎞ ⎛ ⎞= − = − + = − + = − +⎜ ⎟ ⎜ ⎟
⎝ ⎠ ⎝ ⎠
=
( ) ( )
( )
1
2
1
2
emf y y o z x
y
emf y o y o
w y
V d u B dx
V u B dx u B w y
−
+
= × = ×
= = − +
∫ ∫
∫
u B L a a a� �
Fig. P4.24
4- !16
5.
Displacement Current
P4.25: Suppose a vector field is given as
!
Verify that the divergence of the curl of this vector field is equal to zero.
!
!
!
!
P4.26: Suppose a vector field is given by
!
Verify that the divergence of the curl of this vector field is equal to zero.
!
!
P4.27: A pair of 60 cm2 area plates are separated by a 2.0 mm thick layer of ideal
dielectric characterized by εr = 9.0. If a voltage v(t) = 1.0 sin (2πx103t) V is placed
across the plates, determine the displacement current.
!
2 33 .xx yz=A a
verify that 0∇ ∇× =� A
( ) ( )2 3 2 3
2 3
3 3
3 0 0
x y z
y zx yz x yzx y z z y
x yz
∂ ∂ ∂ ∂ ∂∇× = = −∂ ∂ ∂ ∂ ∂
a a a
a aA
2 2 2 39 3y zx yz x z= −a a
( ) ( )2 2 2 3 2 2 2 29 3 9 9 0x yz x z x z x zy z
∂ ∂
∇ ∇× = − = − =
∂ ∂
� A
2 cos .zρ φ=A a
( ) ( )2 21 cos cos sin 2 cosρ φ ρ φρ φ ρ φ ρ φ ρ φρ φ ρ
∂ ∂
∇× = − = − −
∂ ∂
a a a aA
( )( ) ( )1 1sin 2 cos
1 2 sin 2sin 0
ρ ρ φ ρ φ
ρ ρ ρ φ
ρ φ φ
ρ
∂ ∂
∇ ∇× = − + −
∂ ∂
= − + =
� A
( )( )( )
( )
12 29 8.854 10 60 10 1; 239
2 100d
x F m cmdv S mm mi C C pF
dt d mm cm cm
ε
−
= = = =
4- !17
!
!
P4.28: Plot the loss tangent of seawater (σ = 4 S/m and εr = 81) versus log of frequency
from 1 Hz to 1 GHz. At what frequency is the magnitude of the displacement current
density equal to the magnitude of the conduction current density?
!
(Plot this in Figure P4.28)
Solving for f when tanδ = 1: f = 890MHz
% MLP0428
clear
clc
for i=1:9
for j=1:10
m=(i-1)*10+j;
f(m)=j*10^(i-1);
tand(m)=889e6/f(m);
end
end
loglog(f,tand)
xlabel('frequency (Hz)')
ylabel('loss tangent')
grid on
( )3 32 10 cos 2 10dv Vx x tdt sπ π=
( )31.5cos 2 10di x t Aπ µ∴ =
( )
( )( )( )( )
6
12
4 889 10tan
2 ( ) 81 8.854 10
S m x
ff Hz x F m
σ
δ
ωε π −
= = =
Fig. P4.28
4- !18
P4.29: A 1.0 m long coaxial cable of inner conductor diameter 2.0 mm and outer
conductor diameter 6.0 mm is filled with an ideal dielectric with εr = 10.2. A voltage v(t)
= 10.cos(6πx106 t) mV is placed on the inner conductor and the outer conductor is
grounded. Neglecting fringing fields at the ends of the coax, find the displacement
current between the inner and outer conductors.
!
!
!
so !
!
!
Now evaluate id with the given parameters:
!
7. Lossless TEM Waves
P4.30: Suppose in free space that E(z,t) = 5.0e-2zt ax V/m. Is the wave lossless? Find
H(z,t).
Since the wave has an attenuation term (e-2zt) it is clearly not lossless.
!
!
, so ( )QC Q Cv t
V
= =
( ) ( )
2 cos2for coax (from chapter 2): C= so
ln b a ln b a
oV tQ πε ωπε = ll
2d D d dz D Qρ ρ ρ ρρ φ πρ= = =∫ ∫D S a a� �� l
( ) ( ) ( )
2 cos cos cos and .
2 ln ln ln
o o oV t V t V tD
b a b a b aρ ρ
πε ω ε ω ε ω
πρ ρ ρ
= = =D all
( )
sin
ln
o
d
V t
t b a ρ
ωε ω
ρ
−∂
= =
∂
D a J
( ) ( )
2
0 0
sin 2 sin1
ln ln
o o
d
V t V ti d d dz
b a b a
πωε ω πωε ω
ρ φ
ρ
− −
= = =∫ ∫ ∫dJ S�
l l
( )( )( )( )( )
( )
( )
6 12
6
2 6 10 10.2 8.854 10 1 0.010 sin
ln 0.006 0.002
97sin 6 10
d
d
x rad s x F m m V t C Asi
FV C
i x t A
π π ω
π µ
−−
=
= −
( )2 2
2
5 10
5 0 0
x y z
zt zt
o y y
zt
e tex y z zt
e
µ − −
−
∂ ∂ ∂ ∂ ∂∇× = − = = = −∂ ∂ ∂ ∂∂
a a a
HE a a
2 210 10, =zt zty y
o o
td e dt te dt
µ µ
− −−=
− ∫H a H a
4- !19
This integral is solved by parts ! where we let !
We arrive at:
!
P4.31: An electric field propagating in a lossless non-magnetic media is characterized by
!
Find the wave amplitude, frequency, propagation velocity, wavelength, and the relative
permittivity of the media. (b) Find H(y,t).
!
!
!
!
Inserting the given values we find: !
P4.32: A magnetic field propagating in free space is given by
!
Find f, β, λ, and E(z,t).
!
!
!
( )udv uv vdu= −∫ ∫ 2 and .ztu t dv e dt−= =
2 2
2
10 10
2 4
zt zt
y
o o
t Ae e
z z mµ µ
− −⎛ ⎞−= −⎜ ⎟
⎝ ⎠
H a
( )6 .( , ) 100.cos 4 10 0.1257 Vy t t y mπ= × − zE a
62 1100 , 2 , , 50 , 100 10 ; 9o p r
V mE f MHz m u x
m s
π ω
β λ ε
λ β µε
= = = = = = = =
( )sino x oE t y t
β ω β µ
∂
∇× = − = −
∂
HE a
( ) ( )sin coso o x
o o
E Et y dt t yβ βω β ω β
µ ωµ
= − − = −∫H a
( )( , ) coso x
o
Ey t t yβ ω β
ωµ
= −H a
( )6( , ) 0.796cos 4 10 0.126 x Ay t x t y mπ= −H a
( )8( , ) 20.sin 10 .Az t t z
m
π β= × + xH a
8 812 10 , 10 50
2
f x f x MHzω π π= = = =
8
8
3 10 26 ,
0.5 10 3
c x radm
f x m
π π
λ β
λ
= = = = =
c (no )t
∂
∇× =
∂
DH J
4- !20
!
!
so !
inserting the given values we find: !
P4.33: Find the instantaneous expression for E for the magnetic field of problem P4.6.
! . We can write this wave
as ! , and can reinsert the values at the end.
First we have ! Assuming a nonmagnetic medium, we have !
or εr = 9. Now we evaluate the curl of H:
!
!
So !
Solving for E:
!
!
Now we can evaluate the variables:
( )
( )( )
( )
sin
sin 0 0
cos
x y z
o y
o
o y o
H t zx y z z
H t z
H t z
t
ω β
ω β
β ω β ε
∂ ∂ ∂ ∂∇× = = +∂ ∂ ∂ ∂
+
∂
= + =
∂
a a a
H a
Ea
( ) ( )cos sino oy y
o o
H Hd t z dt t zβ βω β ω β
ε ωε
= + = +∫ ∫E E = a a
( )( , ) sino y
o
H Vz t t z
m
β
ω β
ωε
= +E a
8( , ) 7.5sin 10
3 y
kVz t x t z
m
π
π⎛ ⎞= +⎜ ⎟
⎝ ⎠
E a
( )0.14 6From P4.6 we have 2.0 cos 20 10 0.2y xAe x t y mπ π
−= −H a
( )cosyo x
AH e t y
m
α ω β−= −H a
.
t
ε
∂
∇× =
∂
EH ,p
r
cu
ε
=
( )
( )cos .
cos 0 0
x y z
y
o z
y
o
H e t y
x y z y
H e t y
α
α
ω β
ω β
−
−
⎡ ⎤
⎢ ⎥
∂ ∂ ∂ ∂⎢ ⎥ ⎡ ⎤∇× = = − −⎣ ⎦⎢ ⎥∂ ∂ ∂ ∂
⎢ ⎥
−⎢ ⎥⎣ ⎦
a a a
H a
( ) ( ) ( )cos cos sin .y y yo z o zH e t y H e t y e t yy
α α αω β α ω β β ω β− − −
∂
⎡ ⎤ ⎡ ⎤− − = − − − + −⎣ ⎦ ⎣ ⎦∂
a a
( ) ( )cos sin .y yo z o zH e t y H e t y t
α αα ω β β ω β ε− −
∂
∇× = − − − =
∂
EH a a
( ) ( )cos siny yo oz z
H He t y dt e t y dtα αα βω β ω β
ε ε
− −= − − −∫ ∫E a a
( ) ( )sin cos .y yo oz z
H He t y e t yα αα βω β ω β
ωε ωε
− −= − + −E a a
4- !21
!
!
So
!
We can use a trigonometric identity to simplify this answer. Given that
!
we can let !
where b=cosy, c=siny, and tany=c/b. Now, since b/c = 251/56, then y=12.6° and A=257.
Thus
!
This result is easier to obtain using the phasor approach of the next section.
P4.34: Given, at some point distant from a point source at the origin in free space,
!
find frequency, phase constant and H(r,t).
!
!
!
!
Assume we can ignore the first term in the far-field for large r, then we have
( )( )
( )( )( )6 12
2 0.14
56,
20 10 9 8.854 10
oH
x x
α
ωε π −
= =
( )( )
( )( )( )6 12
2 0.2
251.
20 10 9 8.854 10
oH
x x
πβ
ωε π −
= =
( ) ( )0.14
6 0.14 656 sin 20 10 0.2 251 cos 20 10 0.2 .y yz z Ve x t y e x t y mπ π π π
− − − −= − + −E a a
( )cos cos cos sin sin ,x y x y x y± = �
( ) ( )( )cos sin ,y ze A b t y c t yα ω β ω β− ⎡ ⎤= − + −⎣ ⎦E a
( )0.14 6( , ) 257 cos 20 10 0.2 12.6 .y z Vy t e x t y mπ π
−= − −E a�
( )6( , ) 8.0cos 9 10 ,Vr t t r mθπ β= × −E a
22 , 4.5 , 66.7 , 0.094c radf f MHz m
f m
π
ω π λ β
λ
= = = = = =
ot t
µ
∂ ∂
∇× = − = −
∂ ∂
B HE
( ) ( )
( )( ) ( ) ( )
1 , cos ,
cos cos sin
o
o o o
rE E E t r
r r
E r t r E t r E r t r
r
θ φ θ ω β
ω β ω β β ω β
∂
∇× = = −
∂
∂
− = − + −
∂
E a
( ) ( )
( ) ( )
cos sin
cos
sin
o
o
o
E t r r t r
r
E t r
E t r
r
φ
φ φ
ω β β ω β
ω β
β ω β
∇× = − + −⎡ ⎤⎣ ⎦
−
= + −
E a
a a
4- !22
!
!
Inserting the given values we find
!
P4.35: In a lossless, non-magnetic media, the magnetic field at some point distant from a
source at the origin is given by
!
find the relative permittivity of the media, the frequency and phase constant of the wave,
and E(ρ,t).
!
!
!
!
!
In the far-field, we will ignore the first term (has 1/ρ) leaving us with
!
Now, !
!
Plugging in the given values we arrive at: ! .
P4.36: Suppose, in a non-magnetic medium of relative permittivity 3, that
( )sino oE t r tφ
β ω β µ
∂
∇× = − = −
∂
HE a
( ) ( )sin coso o
o o
E Ed t r dt t rφ φ
β β
ω β ω β
µ ωµ
= = − − = −∫ ∫H H a a
( )( , ) 21cos mAr t t r
mφ
ω β= −H a
( )8( , ) 6.0sin 3 10 10 ,At t mφρ ρ= × +H a
82 3 10 ; 48f x f MHzπ = =
80.3 10 ; 100p r
r
m cu x
s
ω
ε
β ε
= = = =
10r rad
c m
ω ε
β = =
t t
ε
∂ ∂
∇× = =
∂ ∂
D EH
( )( ) ( ) ( )( )1 1sin 10 sin coso z o o zH t H t H tρ ω ρ ω βρ β ρ ω βρρ ρ ρ
∂
∇× = + = + + +
∂
H a a
( )coso zH tβ ω βρ∇× = +H a
( )coso z
Hd t dtβ ω βρ
ε
= = +∫ ∫E E a
( )sin .o z
H tβ ω βρ
ωε
= +E a
8( , ) 226sin(3 10 10 ) z
Vt x t
m
ρ ρ= +E a
4- !23
!
Determine β and H(y,t).
!
!
!
Inserting the given values we arrive at:
!
8. Time-Harmonic Fields and Phasors
P4.37: Show that
!
We first show that :
!
Expanding the left side we have:
!
Now expanding the right side we have:
!
So since we have :
!
then
!
( ) ( ) ( )7 7, 4.0sin 10 9.0cos 10 .x z Vy t x t y x t y mπ β π β= − + −E a a
8 25 , 1.73 10 , 34.6 , 0.181pp
r
uc m radf MHz u x m
s f m
π
λ β
λε
= = = = = = =
( ) ( )9 sin 4 cosx z ot y t y t
β ω β β ω β µ
∂
∇× = − + − = −
∂
HE a a
( ) ( )9 4cos sinx z
o o
t y t yβ βω β ω β
ωµ ωµ
−
= − + −H a a
( ) ( )( , ) 41cos 18sin x z
mAy t t y t y
m
ω β ω β= − − −H a a
( ) ( )Re Re .j t j ts se j et
ω ωω
∂
⎡ ⎤ ⎡ ⎤− = −⎣ ⎦ ⎣ ⎦∂
B B
( )Re Rej t j ts se et t
ω ω∂ ∂⎡ ⎤⎡ ⎤− = −⎣ ⎦ ⎢ ⎥∂ ∂⎣ ⎦
B B
( ) [ ]( )Re Re cos( ) sin( )
cos( ) sin( )
j t
s s s
s s
e t j t
t t
t t
t
ω ω ω
ω ω ω
∂ ∂
⎡ ⎤− = − +⎣ ⎦∂ ∂
∂
= − =
∂
B B B
B B
( ) ( )( )
( ) ( ) ( )2
Re Re Re cos sin
Re cos sin sin
j t j t
s s s
s s
e j e j t j t
t
j t j t t
ω ωω ω ω ω
ω ω ω ω ω ω
∂⎡ ⎤ ⎡ ⎤⎡ ⎤− = − = − +⎣ ⎦⎢ ⎥ ⎣ ⎦∂⎣ ⎦
⎡ ⎤= − + =⎣ ⎦
B B B
B B
( )Re Re Re ,j t j t j ts s se e j et t
ω ω ωω
∂ ∂⎡ ⎤⎡ ⎤ ⎡ ⎤− = − = −⎣ ⎦ ⎣ ⎦⎢ ⎥∂ ∂⎣ ⎦
B B B
( )Re Re Rej t j t j ts s se e j et t
ω ω ωω
∂ ∂⎡ ⎤⎡ ⎤ ⎡ ⎤− = − = −⎣ ⎦ ⎣ ⎦⎢ ⎥∂ ∂⎣ ⎦
B B B
4- !24
P4.38: Derive the differential phasor form of (a) Gauss’s Law, and (b) Ampere’s Circuit
Law.
!
!
!
!
P4.39: Find H(y,t) in problem P4.31(b) using phasors.
!
!
Plugging in the values from P4.31, we find
!
P4.40: Find E(z,t) in problem P4.32 using phasors.
!
!
!
!
Plugging in the values from P4.32, we find
( ) ( ), , , , , , ,v vx y z t x y z tρ ρ∇ = ∇ =D D� �
Re Re ; Re Re ;j t j t j t j ts vs s vs
s vs
e e e eω ω ω ωρ ρ
ρ
⎡ ⎤ ⎡ ⎤ ⎡ ⎤ ⎡ ⎤∇ = ∇ =⎣ ⎦ ⎣ ⎦ ⎣ ⎦ ⎣ ⎦
∴∇ =
D D
D
� �
�
t t
ε
∂ ∂
∇× = =
∂ ∂
D EH
( ) ( ) ( ) ( )Re Re ; Re Rej t j t j t j ts s s s
s s
e e e j e
t
j
ω ω ω ωε ωε
ωε
∂
∇× = ∇× =
∂
∴∇× =
H E H E
H E
, j ys o z s sE e j
β ωµ−= ∇× = −E a E H
( ) and ( , ) cos .
j y
s o x s
j yo o
s x x
o o
j E e j
E Ee y t t y
β
β
β ωµ
β β
ω β
ωµ ωµ
−
−
∇× = − = −
∴ = = −
E a H
H a H a
( )6( , ) 0.796cos 4 10 0.126 .x Ay t x t y mπ= −H a
, j zs o x s o sH e j
β ωε= ∇× =H a H E
0 0
x y z
j z
s o y
j z
o
jH ex y z z
jH e
β
β
∂∂ ∂ ∂∇× = =∂ ∂ ∂ ∂
a a a
H a
2j z j z j z
s o y o y o y o sjH e j H e H e jz
β β ββ β ωε
∂
∇× = = = − =
∂
H a a a E
( ), ( , ) sin
j z
o y j zo o
s y y
o o o
H e H H Vj e z t t z
j m
β
ββ β β ω β
ωε ωε ωε
−
= = ∴ = +
a
E a E a
4- !25
!
P4.41: In free space,
!
Find H(z,t).
The phasor version of E is: !
!
!
!
!
Inserting the appropriate values we then find:
!
P4.42: Find H(y,t) in problem P4.36 using phasors.
In phasor form the field of P4.36 is: !
Now apply !
!
!
!
8( , ) 7.5sin 10 .
3 y
kVz t t z
m
π
π⎛ ⎞= +⎜ ⎟
⎝ ⎠
E a
( ) ( )6 6( , ) 10.cos 10 20.cos 10 .Vz t t z t z mπ β π β= × − + × −x yE a a
10 20j z j zs x ye e
β β− −= +E a a
20 10
10 20 0
x y z
j z j z
s x y
j z j z
e ex y z z z
e e
β β
β β
− −
− −
∂ ∂∂ ∂ ∂∇× = = − +∂ ∂ ∂ ∂ ∂
a a a
E a a
20 10j z j zs x y o sj e j e j
β ββ β ωµ− −∇× = − = −E a a H
20 10 20 10j z j z j z j z
s x y x y
o o o o
j je e e e
j j
β β β ββ β β β
ωµ ωµ ωµ ωµ
− − − −− −= + = +
− −
H a a a a
6
3
8
10; 10.47 10
3 10p p
x radu x
u x m
ω ω π
β
β
−= = = =
( ) ( )
53 27 ,
and ( , ) 53cos 27cos
j z j z
s x y
x y
mAe e
m
mAz t t z t z
m
β β
ω β ω β
− −= − +
= − − + −
H a a
H a a
4 9j y j ys x zj e e
β β− −= +E a a
s o sjωµ∇× = −E H
29 4
4 0 9
x y z
j y j y
s x z
j y j y
j e j ex y z
j e e
β β
β β
β β− −
− −
∂ ∂ ∂∇× = = − +∂ ∂ ∂
a a a
E a a
9 4j y j yx z o sj e e j
β ββ β ωµ− −− − = −a a H
9 4 9 4j y j y j y j y
s x z x z
o o o o
j e e e j e
j j
β β β ββ β β β
ωµ ωµ ωµ ωµ
− − − −− −= + = −
− −
H a a a a
4- !26
Inserting the appropriate values we then find:
!
!
P4.43: MATLAB: In MATLAB 4.4, a polar plot of the phasor corresponded to a location
on a sine wave for a particular time, at a fixed position in space. You can also make a
polar phasor plot for a snapshot in time, where you change position. Modify MATLAB
4.4 to provide an animation of the phasor versus sine wave as you change the position.
% M-File: MLP0443
%
% This program modifies ML0404, generating a
% position plot animation synchronized with
% a polar plot.
%
% Wentworth, 1/19/03
%
% Variables:
% Eo field amplitude (V/m)
% E electric field intensity (V/m)
%
E E-field for movie
% f frequency (Hz)
% theta angle
% theta1 angle for movie
% T period (1/f)
% t time (s)
% t1 time for movie
clc %clears the command window
clear %clears variables
clf
% Generate the reference frame
% Initialize Variables
c=3e8;
Eo=1;
f=1000;
lambda=c/f;
81.73 10 , 0.181p
pr
c m radu x
s u m
ω
β
ε
= = = =
( ) ( ) mA( , ) 41cos 18sin
mx z
y t t y t yω β ω β= − − −H a a
4- !27
beta=2*pi/lambda;
T=1/f;
% Perform Calculation
z=0:lambda/100:2*lambda;
theta=-beta*z;
E=Eo*cos(theta);
% Generate the Plot
subplot(211),plot(z,E,0,Eo,'ro');
subplot(212),polar(0,Eo,'ro');
pause
%Make the Movie
z1=0:lambda/50:2*lambda;
for n=1:100
theta1(n)=-beta*z1(n);
E1(n)=Eo*cos(theta1(n));
subplot(211),plot(z,E,z1(n),E1(n),'ro');
subplot(212),polar(theta1(n),Eo,'ro');
M(:,1)=getframe;
end
P4.44: MATLAB: Repeat problem P4.43, now accounting for attenuation. Run the
program assuming an attenuation of 2 x 10-6 Np/m.
Fig. P4.43 (at start of movie)
4- !28
% M-File: MLP0444
%
% Same as MLP0443 but add attenuation.
%
% This program modifies ML0404, generating a
% position plot animation synchronized with
% a polar plot and accounting for attenuation.`
%
% Wentworth, 1/19/03
%
% Variables:
% Eo field amplitude (V/m)
% E electric field intensity (V/m)
% E E-field for movie
% f frequency (Hz)
% theta angle
% theta1 angle for movie
% T period (1/f)
% t time (s)
% t1 time for movie
clc %clears the command window
clear %clears variables
clf
% Generate the reference frame
% Initialize Variables
atten=0.000002
;
c=3e8;
Eo=1;
f=1000;
lambda=c/f;
beta=2*pi/lambda;
T=1/f;
% Perform Calculation
z=0:lambda/100:2*lambda;
theta=-beta*z;
E=Eo.*exp(-atten.*z).*cos(theta);
% Generate the Plot
4- !29
subplot(211),plot(z,E,0,Eo,'ro');
subplot(212),polar(0,Eo,'ro');
hold on
pause
%Make the Movie
z1=0:lambda/50:2*lambda;
for n=1:100
theta1(n)=-beta*z1(n);
E2(n)=Eo.*exp(-atten.*z1(n));
E1(n)=E2(n).*cos(theta1(n));
E2(n)=Eo.*exp(-atten.*z1(n));
subplot(211),plot(z,E,z1(n),E1(n),'ro');
subplot(212),polar(theta1(n),E2(n),'ro');
M(:,1)=getframe;
end
Fig. P4.44 (at end of movie)
Wentworth solutions chapter 5.pdf
5- !1
Solutions for Chapter 5 Problems
1. General Wave Equations
P5.1: Starting with Maxwell’s equations for simple, charge-free media, derive the
Helmholtz equation for H.
!
Using a vector identity we also have: !
But ! , leading to
!
P5.2: Derive equation (5.10) by starting with the phasor point form of Maxwell’s
equations for simple, charge-free media.
For charge-free media the phasor form of Maxwell’s equations are:
!
Now we take the curl of both sides of Faraday’s Law,
!
Now since ! , and since ! , we have
!
P5.3: A wave with λ = 6.0 cm in air is incident on a nonmagnetic, lossless liquid media.
In the liquid, the wavelength is measured as 1.0 cm. What is the wave’s frequency (a) in
air? (b) in the liquid? (c) What is the liquid’s relative permittivity?
(a) !
(b) the frequency doesn’t change with the media (the wavelength does) so f = 5 GHz
(c)
( )
2
2
t t
t t
σ ε σ ε
µσ µε
∂ ∂⎛ ⎞∇× ∇× =∇× + = ∇× ∇×⎜ ⎟∂ ∂⎝ ⎠
∂ ∂
−
∂ ∂
EH E E+ E
H H= -
( ) 2∇× ∇× =∇ −∇H H H�
0∇ =H�
2
2
2t t
µσ µε
∂ ∂
∇ +
∂ ∂
H HH =
( )
0
0
s
s
s s
s s
j
j
ωµ
σ ωε
∇ =
∇ =
∇× = −
∇× = +
D
B
E H
H E
�
�
( ) ( ) ( )s s s sj j j jωµ ωµ ωµ σ ωε∇× ∇× =∇× − = − ∇× = − +E H H E
( ) 2∇× ∇× =∇ −∇s s sE E E� 0s∇ =E�
( )2 s sj jωµ σ ωε∇ = +E E
83 10 5
0.06
pu c x m sf GHz
mλ λ
= = = =
5- !2
!
P5.4: Suppose Hs(z) = Hys(z) ay. Start with (5.14) and derive (5.29).
Since Hs is only a function of z, (5.14) becomes
! (a)
If we let ! then !
So (a) becomes ! This has two solutions:
(1) for !
(2) for !
The general solution is the linear superposition of the two, or
!
P5.5: Given σ = 1.0x10-5 S/m , εr = 2.0, µr = 50., and f = 10. MHz, find γ, α, β, and η.
!
!
!
!
Inserting these into the expressions for γ and η,
!
These results are confirmed by ML0501.
P5.6: MATLAB: In some material, the constitutive parameters are constant over a large
frequency range and are given as σ = .10 S/m , εr = 4.0, and µr = 600. Write a MATLAB
( )9 7
28
8
15 10 0.01 5 10
3 10 36
0.5 10
p
r
r
m cu f x m x
s s
x
x
λ
ε
ε
⎛ ⎞= = = =⎜ ⎟
⎝ ⎠
⎛ ⎞
∴ = =⎜ ⎟
⎝ ⎠
2
2
2 0.
s
s
H H
z
γ
∂
− =
∂
,zsH Ae
λ=
2
2
2, and .
z zs sH HAe Ae
z z
λ λλ λ
∂ ∂
= =
∂ ∂
( )( )2 2 0, or 0.λ γ λ γ λ γ− = + − =
0, we have , , or .z zs s oH Ae H H e
γ γλ γ λ γ − + −+ = = − = =
0, we have , , or .z zs s oH Ae H H e
γ γλ γ λ γ −− = = = =
( ) .z zs o o yH e H eγ γ+ − −= +H a
( )r o r oj j jγ ωµ µ σ ωε ε α β= + = +
r o
r o
j
j
ωµ µ
η
σ ωε ε
=
+
( )( )( )6 72 10 10 50 4 10 3948r oj j x x jωµ µ π π −= =
( )( )( )5 6 12 5 31 10 2 10 10 2 8.854 10 1 10 1.11 10r oj x j x x x j xσ ωε ε π− − − −+ = + = +
3 3 2579.4 10 2.1 1 , 9.4 10 , 2.1 , 1880 jx j m x Np m rad m eγ α β η− −= + = = = Ω
�
5- !3
routine that will plot α, β, and η (magnitude and phase) versus the log of frequency from
1 Hz up to 100 GHz.
% M-File: MLP0506
%
% This program is a modification of ML0501.
% For a given material, it will plot the attenuation,
% phase constant and intrinsic impedance vs f.
%
% Wentworth, 1/23/03
%
clc %clears the command window
clear %clears variables
% Initialize Variables
uo=pi*4e-7;
eo=8.854e-12;
sig=0.10;
er=4;
ur=600;
% Perform Calculation
for i=1:10
for j=1:10
m=(i-1)*10+j;
f(m)=j*10^(i-1);
w(m)=2*pi*f(m);
A(m)=i*(w(m)*ur*uo);
B(m)=complex(sig,w(m)*er*eo);
gamma(m)=sqrt(A(m)*B(m));
alpha(m)=real(gamma(m));
beta(m)=imag(gamma(m));
eta(m)=sqrt(A(m)/B(m));
meta(m)=abs(eta(m));
aeta(m)=180*angle(eta(m))/pi;
end
end
subplot(3,1,1)
plot(f,alpha,'-o',f,beta,'-*')
ylabel('1/m')
xlabel('frequency (Hz)')a
legend('alpha','beta')
subplot(3,1,2)
semilogx(f,meta)
ylabel('magnitude of eta (ohms)')
5- !4
subplot(3,1,3)
semilogx(f,aeta)
ylabel('phase of eta (degrees)')
xlabel('frequency (Hz)')
P5.7: Suppose E(x,y,t) = 5.0 cos(πx106t – 3.0x + 2.0y) az V/m. Find the direction of
propagation, ap, and H(x,y,t).
!
We assume nonmagnetic material and therefore have
!
!
!
To find the direction of propagation,
3 25 j x j ys ze e
−=E a
3 2 3 210 15j x j y j x j ys s x yj j e e j e eωµ
− −∇× = − = +E H a a
3 2 3 2 3 2 3 210 15 2.53 3.8j x j y j x j y j x j y j x j ys x y x y
o
j je e e e e e e e
j jωµ ωµ
− − − −= + = − −
− −
H a a a a
( ) ( )6 6 A( , , ) 2.53cos 10 3 2 3.80cos 10 3 2 mx yx y t x t x y x t x yπ π= − − + − − +H a a
Fig. P5.6
5- !5
!
!
And with the exponential terms canceling in the top and bottom of the equation for ap, we
have:
!
P5.8: Suppose in free space, H(x,t) = 100.cos(2πx107t – βx + π/4) az mA/m. Find E(x,t).
!
!
Since free space is stated,
!
and then
!
2. Propagation in Lossless, Charge-Free Media
P5.9: Start with the Helmholtz equation (5.11), and using γ = jβ, derive (5.41), the
traveling wave equation.
!
Let !
Now we have
!
This can be factored: ! ,
suggesting two solutions. The first solution uses ! and
!
Likewise, the second solution uses ! and
!
s s
P
s s
×
=
×
E Ha
E H
6 4 6 419 12.65j x j y j x j ys s x ye e e e
− −× = −E H a a
0.83 0.55 .P x y= −a a a
( )0.100 , , 4
120 0.100 12
j x j
s z P x
j x j j x j
s P s x z y
e e
e e e e
β φ
β φ β φ
φ π
η π π
−
− −
= = =
= − × = − × =
H a a a
E a H a a a
( )12 cos yt xπ ω β φ= − +E a
2 2 2 30 rad m
c f
π π
β π
λ
= = =
7 212 cos 2 10
30 4 y
Vx t x
m
π π
π π⎛ ⎞= − +⎜ ⎟
⎝ ⎠
E a
2
2 2 2
20, let ( ) , and with =j we have 0.
xs
s s s xs x xs
EE z E
z
γ γ β β
∂
∇ − = = + =
∂
E E E a
2
2
2, so and
z x xxs xs
xs
E EE Ae A e A e
z z
λ λ λλ λ
∂ ∂
= = =
∂ ∂
2 2 2 20, or 0z zA e Aeλ λλ β λ β+ = + =
( )( )2 2 0j jλ β λ β λ β+ = + − =
jλ β= −
.j z j zxs oE Ae E e
β β− + −= =
jλ β= +
.j z j zxs oE Ae E e
β β−= =
5- !6
The complete solution is a linear superposition of these two solutions, or
!
P5.10: A 100 MHz wave in free space propagates in the y direction with an amplitude of
1 V/m. If the electric field vector for this wave has only an az component, find the
instantaneous expression for the electric and magnetic fields.
From the given information we have ! and !
or ! .
Now to find H.
!
So
!
or
!
P5.11: In a lossless, nonmagnetic material with εr = 16, H = 100 cos(ωt – 10y) az mA/m.
Determine the propagation velocity, the angular frequency, and the instantaneous
expression for the electric field intensity.
!
!
!
!
.j z j zs o o xE e E e
β β+ − − += +E a
62 200 10 radf x
s
ω π π= =
2 ,
3p
rad
u m
ω π
β = =
6 2( , ) 1cos 200 10
3 z
Vy t x t y
m
π
π⎛ ⎞= −⎜ ⎟
⎝ ⎠
E a
1 1 11 , 1
120 120
j y j y j y
s z s P s y z xe e e
β β β
η π π
− − −= = × = × =E a H a E a a a
( ) 61 2, cos 200 10
120 3 x
Ay t x t y
m
π
π
π
⎛ ⎞= −⎜ ⎟
⎝ ⎠
H a
( ) 6 2, 2.7cos 200 10 .
3 x
mAy t x t y
m
π
π⎛ ⎞= −⎜ ⎟
⎝ ⎠
H a
8
83 10 0.75 10
16p r
c x mu x
s
ω
β ε
= = = =
( )( )8 80.75 10 10 7.5 10p radu x x sω β= = =
( )8( , ) 100cos 7.5 10 10 z mAy t x t y m= −H a
0.100 ,
120 0.100 3
j y
s z
j y j y
s P s y z x
r
e
e e
β
β βπη π
ε
−
− −
=
−
= − × = × = −
H a
E a H a a a
5- !7
!
P5.12: Given E = 120π cos(6πx106t – 0.080πy) az V/m and H = 2.00cos(6πx106t –
0.080πy) ax A/m, find µr and εr.
!
!
so we know
!
Now, !
!
And now
!
!
3. Propagation in Dielectrics
P5.13: Work through the algebra to derive equation α and β equations (5.52) from
equations (5.50) and (5.51).
!
Comparing the imaginary parts, we see !
and comparing the real parts, ! .
Rearranging and inserting our value for β:
!
This is a quadratic expression (x2 + bx + c = 0), where here
( )8( , ) 9.4cos 7.5 10 10 x Vy t x t y m= − −E a
0.080120 , 2j y j ys z s xe e
β ππ − −= =E a H a
1 1 120 120
120
j y j y j yr
s P s y z x x
rr r
e e eβ β βεππ
η η µπ µ ε
− − −= × = × = =H a E a a a a
2r
r
ε
µ
=
6
61 6 10 75 10
0.080p r r
c xu xω π
β πµε µ ε
= = = = =
6 475 10r r
c
x
µ ε = =
(2)(4) 8r r r r
r
µ
µ ε ε
ε
= = =
4 2
2
r r
r
r r
µ ε
µ
µ ε
= = =
( )2 2 2 2 2 ;j jγ ω µε ωµσ α β αβ= − + = − +
2 , or ,
2
ωµσ
ωµσ αβ β
α
= =
2 2 2 0α β ω µε− + =
2 2 2
4 2 2 0
4
ω µ σ
α ω µεα+ − =
5- !8
!
Solving the quadratic:
!
Reinserting the a, b and c values:
!
!
Now for β:
!
Rearranging,
!
Solving this quadratic we find
!
P5.14: MATLAB: Write a routine to prompt the user for a material’s constitutive
parameters and an operating frequency, and calculate the α and β from (5.52). Verify the
program by running Drill 5.6.
% MLP0514
%
% Prompts user for material's constitutive
% parameters and an operating frequency, then
% calculates alpha(Np/m) and beta(rad/m).
%
% Wentworth, 1/24/03
%
clc
clear
ur=input('relative permeability: ');
2
2 2, ,
2
x b c ωµσα ω µε ⎛ ⎞= = = −⎜ ⎟
⎝ ⎠
2
2
2
4 1 1 1 44 1 1
2 2 2 2
b b c cx b c b b
b
⎛ ⎞− ± −
= = − − = − −⎜ ⎟⎜ ⎟
⎝ ⎠
22 2 2
2 2 2
4 2 2
1 41 1 1 1
2 4
ω µ σ σ
α ω µε ω µε
ω µ ε ωε
⎡ ⎤⎡ ⎤ ⎛ ⎞⎢ ⎥= + − = + −⎢ ⎥ ⎜ ⎟
⎢ ⎥⎝ ⎠⎢ ⎥⎣ ⎦ ⎣ ⎦
2
1 1σα ω µε
ωε
⎛ ⎞⎛ ⎞⎜ ⎟∴ = + −⎜ ⎟⎜ ⎟⎝ ⎠⎝ ⎠
2
2 2 2 2 20, = , so 0
2 2
ωµε ωµε
β α ω µε α β ω µε
β β
⎛ ⎞
− − = − − =⎜ ⎟
⎝ ⎠
2
4 2 2 0
2
ωµσ
β ω µεβ ⎛ ⎞− − =⎜ ⎟
⎝ ⎠
2
1 1σβ ω µε
ωε
⎛ ⎞⎛ ⎞⎜ ⎟= + +⎜ ⎟⎜ ⎟⎝ ⎠⎝ ⎠
5- !9
erp=input('real part of rel permittivity: ');
erdp=input('complex part of rel permittivity: ');
s=input('conductivity (S/m): ');
f=input('frequency (Hz): ');
w=2*pi*f;
uo=pi*4e-7;
eo=8.854e-12;
seff=s+w*erdp*eo;
A=sqrt(1+(seff/(w*erp*eo))^2);
B=ur*uo*erp*eo/2;
alpha=w*sqrt(B*(A-1))
beta=w*sqrt(B*(A+1))
Now run the program for Drill 5.6:
(a)
relative permeability: 1
real part of rel permittivity: 10
complex part of rel permittivity: .01
conductivity (S/m): 1e-12
frequency (Hz): 100
alpha =
3.3730e-009
beta =
6.6268e-006
(b)
relative permeability: 1
real part of rel permittivity: 10
complex part of rel permittivity: .01
conductivity (S/m): 1e-12
frequency (Hz): 1e6
alpha =
3.3134e-005
beta =
0.0663
These results agree with Drill 5.6.
5- !10
P5.15: Given a material with σ = 1.0x10-3 S/m, µr = 1.0, and εr’ = 3.0, εr’’ = 0.015,
compare a plot of α versus frequency from 1 Hz to 1 GHz using (5.52) to a similar plot
using (5.54). At what frequency does the % error exceed 2%?
% MLP0515
%
% Compares alpha calculated using (5.52) to
% that calculated using (5.54).
%
% Wentworth, 1/25/03
%
clc
clear
% Initialize variables
ur=1;
erp=3;
erdp=.015;
s=1e-3;
uo=pi*4e-7;
eo=8.854e-12;
B=ur*uo*erp*eo/2;
% Perform calculations
for i=1:10
for j=1:10
m=(i-1)*10+j;
f(m)=j*10^(i-1);
w(m)=2*pi*f(m);
seff(m)=s+w(m)*erdp*eo;
A(m)=sqrt(1+(seff(m)/(w(m)*erp*eo))^2);
alpha1(m)=w(m)*sqrt(B*(A(m)-1));
alpha2(m)=(seff(m)/2)*sqrt(ur*uo/(erp*eo));
diff(m)=abs(100*(alpha1(m)-alpha2(m))/alpha1(m));
C(m)=diff(m)<2;
if diff(m)<2
if diff(m-1)>2
fdiff=f(m);
Fstr=num2str(fdiff);
end
end
end
end
% generate plot
5- !11
loglog(f,alpha1,'-o',f,alpha2,'-*')
legend('(5.52)','(5.54)')
xlabel('frequency (Hz)')
ylabel('alpha(Np/m)')
S=strcat('Error drops below 2% when frequency > ',Fstr);
title(S)
grid on
P5.16: In a media with properties σ = 0.00964 S/m , εr = 1.0, µr = 100., and f = 100.
MHz, a 1.0 mA/m amplitude
magnetic field travels in the +x direction with its field
vector in the z direction. Find the instantaneous form of the related electric field
intensity.
!
!
!
!
Finally,
( )1 cos ; x x j xz s o z
mA e t x H e e
m
α α βω β− − −⎛ ⎞= − =⎜ ⎟
⎝ ⎠
H a H a
x j x x j x
s P s x o z o yH e e H e e
α β α βη η η− − − −= − × = − × =E a H a a a
( )( )( )
( )( )
6 7
30
6 12
2 100 10 100 4 10
2664
0.00964 2 100 10 8.854 10
j
j x xj e
j j x x
π πωµ
η
σ ωε π
−
−
= = = Ω
+ +
�
( ) 114.8 25.7 j j j
m
γ ωµ σ ωε= + = +
Fig. P5.15
5- !12
!
P5.17: MATLAB: Make a pair of plots similar to Figure 5.4 for the 3 materials of Table
5.1. Instead of loss tangent, one plot is to contain the magnitude of η and the other is to
have the phase of η.
%ML P5.17
clc;clear
%want to plot intrinsic impedance vs frequency for
%the data listed in table 5.1
%Here, we'll plot the magnitude and phase of the
%intrinsic impedance.
%enter data from Table 5.1
sigC=5.8e7; %conductivity of copper in S/m
sigS=4; % conductivity of seawater
sigG=1e-12; % conductivity of glass
er1C=1; %real part of rel perm for Copper
er1S=72; %real part of rel perm for seawater
er1G=10; %real part of rel perm for glass
er2C=0; %imag part of rel perm for Copper
er2S=12; %imag part of rel perm for seawater
er2G=0.010; %imag part of rel perm for glass
%enter constant values
eo=8.854e-12; %free space permittivity, F/m
uo=pi*4e-7; %free space permeability, H/m
%calculations
n=2:.2:14;
f=10.^n;w=2*pi*f;
seffC=sigC+w*er2C*eo;
seffS=sigS+w*er2S*eo;
seffG=sigG+w*er2G*eo;
etaC=sqrt(i*w*uo./(seffC+i*er1C*eo))
etaS=sqrt(i*w*uo./(seffS*er1S*eo))
etaG=sqrt(i*w*uo./(seffG+i*er1G*eo))
magC=abs(etaC);
angC=180*angle(etaC)/pi;
subplot(3,2,1)
( )15 6( , ) 2.66 cos 200 10 26 30x y Vx t e x t x mπ
−= − +E a�
5- !13
semilogx(f,magC)
ylabel('mag, ohms')
title('copper')
subplot(3,2,2)
semilogx(f,angC)
ylabel('phase, deg')
magS=abs(etaS);
angS=180*angle(etaS)/pi;
subplot(3,2,3)
semilogx(f,magS)
ylabel('mag, ohms')
title('seawater')
subplot(3,2,4)
semilogx(f,angS)
ylabel('phase, deg')
magG=abs(etaG);
angG=180*angle(etaG)/pi;
subplot(3,2,5)
semilogx(f,magG)
ylabel('mag, ohms')
xlabel('freq (Hz)')
title('glass')
subplot(3,2,6)
semilogx(f,angG)
xlabel('freq (Hz)')
ylabel('phase, deg')
4. Propagation in Conductors
P5.18: Starting with (5.13), show that α = β for a good conductor.
!
!
(Note: we get the same result starting with (5.52) and assuming !
( ) for a good conductorj j jγ ωµ σ ωε ωµσ= + ≈
( )1 , 1
2 2 22
2
j jj j jωµσ ωµσ ωµσγ
ωµσ
α β
+
= = + = +
= =
1.σ ωε >>
5- !14
P5.19: In seawater, a propagating electric field is given by E(z,t) = 20.e-αz cos(2πx106t –
βz + 0.5) ay V/m. Assuming ε’’=0, find (a) α and β, and (b) the instantaneous form of H.
For seawater we have εr = 72, σ = 5, and µr = 1.
So: !
!
!
!
!
!
7.896, 0.004o r oj j j jωµ ωε ε= =
44.981.257 jj e
j
ωµ
η
σ ωε
= = Ω
+
�
( ) 4.441 4.445 1 mj j jγ ωµ σ ωε= + = +
14.4
m
α β∴ = =
0.5 28.620 20z z j radians z z js y y
V Ve e e e e e
m m
α β α β− − − −= =E a a
�
28.6 28.61 1 2020 z z j z z js P s z y x
Ae e e e e e
m
α β α β
η η η
− − − −−= × = × =H a E a a a
� �
Fig. P5.17
5- !15
!
or with appropriate significant digits:
!
P5.20: Calculate the skin depth at 1.00 GHz for (a) copper, (b) silver, (c) gold, and (d)
nickel.
!
Table P5.19
P5.21: For Nickel (σ = 1.45 x 107, µr = 600), make a table of α, β, η, up, and δ for 1Hz,
1kHz, 1MHz, and 1 GHz.
For Ni we have σ = 1.45x107S/m, µr = 600
!
δ = 1/α
!
!
Table P5.21
( )4.4 6( , ) 15.9 cos 2 10 4.4 28.6 45z x Az t e x t z mπ
−= − − + −H a� �
( )4.4 6( , ) 16 cos 2 10 4.4 16z x Az t e x t z mπ
−= − − −H a�
6
9 7 7
1 ;as an example, for copper at 1 GHz:
1 2.1 10 2.1
1 11 10 4 10 5.8 10
f
x m m
H Vs Ax x x
s m m HA V
δ
π µσ
δ µ
π π
−
−
=
= = =
Ω⎛ ⎞⎛ ⎞⎛ ⎞
⎜ ⎟⎜ ⎟⎜ ⎟Ω⎝ ⎠⎝ ⎠⎝ ⎠
σ(S/m) µr δ(µm)
Cu 5.8x107 1 2.1
Ag 6.2x107 1 2.0
Au 4.1x107 1 2.5
Ni 1.5x107 600 0.17
( )( )( )( )7 7 3600 4 10 1.45 10 34.35 10 ( )f f Hz x x x f Hzα β π µσ π π −= = = =
45 6 452 18.08 10 ( )j je x f Hz eαη
σ
−= Ω = Ω
� �
612 10p
r r
c mu x
sµ ε
= =
f(Hz)= 1 103 106 109
5- !16
P5.22: A semi-infinite slab exists for z > 0 with σ = 300 S/m, εr = 10.2, and µr = 1.0. At
the surface (z = 0),
E(0,t) = 1.0 cos(π x
106t) ax V/m.
Find the instantaneous expressions for E and H anywhere in the slab.
The general expression for E is: !
!
!
Here, σ >> ωε (i.e. it is a good conductor), so
!
!
So now we have
!
To find B we’ll work in phasors.
!
!
!
P5.23: In a nonmagnetic material, E(z,t) = 10.e-200z cos(2π x 109t - 200z) ax mV/m.
α(Np/m) 185 5860 185x103 5.9x106
β(rad/m) 185 5860 185x103 5.9x106
η 18ej45ºµΩ 570ej45º µΩ 18ej45º mΩ 0.57ej45º Ω
δ 5.4mm 170µm 5.3µm 170nm
up(m/s) 12x106 12x106 12x106 12x106
( )6( , ) 1.0 cos 10z x Vz t e x t z m
α π β−= −E a
( )( )6 710 4 10 3.948j j x x jωµ π π −= =
( )( )( )6 12 610 10.2 8.854 10 284 10j j x x j xωε π − −= =
124.3f
m
α π µσ β= = =
45 452 0.115j je eαη
σ
= = Ω
� �
( )24 6( , ) 1.0 cos 10 24z x Vz t e x t z mπ
−= −E a
1 1 11 , 1z j z z j z z j zs x s P s z x ye e e e e e
α β α β α β
η η η
− − − − − −= = × = × =E a H a E a a a
( )24 61( , ) cos 10 24 450.115
z
y
Az t e x t z
m
π−= − −H a�
( )24 6( , ) 8.7 cos 10 24 45z y Az t e x t z mπ
−= − −H a�
5- !17
Find H(z,t).
Since α = β, the media is a good metal. With µr = 1 we have
!
!
!
!
P5.24: A 0.1 µm layer of copper is deposited atop a very thick slab of nickel. For a field
incident on the copper surface, (a) calculate Rs at 1.0 GHz. Compare this with Rs at 1.0
GHz for (b) a semi-infinite slab of copper and (c) for a 0.1 µm thickness of copper by
itself.
Refer to Figure P5.24..
In the copper portion the field is !
In the nickel portion, !
The current density in the copper is ! and in the nickel is
! The current is
( )
( )( )
22
9 7
200
, or 10.13
1 10 4 10o o
Sf
f mx x
α
α π µ σ σ
π µ π π −
= = = =
45 452 28j je eαη
σ
= = Ω
� �
1 1 1010 , 10z j z z j z z j zs x s P s z x ye e e e e e
α β α β α β
η η η
− − − − − −= = × = × =E a H a E a a a
( )200 9( , ) 360 cos 2 10 200 45z y mAz t e x t z mπ
−= − −H a�
Cu z
x xoE E e
α−=
( ) ( )NiCu z ttx xoE E e e αα − −−=
,Cu zxCu Cu xoJ E e
ασ −=
( ) ( ).NiCu z ttxNi Ni xoJ E e e αασ − −−=
Fig. P5.24
5- !18
! , or
! and upon evaluating
and with V=ExoL,
we have !
Now we’re ready to perform the calculations using the following data:
!
!
(a) 0.1µm Cu over Ni: Rs = 176 mΩ
(b) Semi-infinite Cu: Rs = 8.3 mΩ
(c) 0.1 µm Cu: Rs = 177 mΩ
P5.25: Calculate the DC resistance per meter length of a 4.0 mm diameter copper wire.
Now find the resistance at 1.0 GHz.
!
1 GHz: !
!
5. The Poynting Theorem and Power Transmission
P5.26: In air, H(z,t) = 12.cos(πx106t - βz + π/6) ax A/m. Determine the power density
passing through a 1.0
square meter surface that is normal to the direction of propagation.
!
( )Cu Cu Niz t z t
Cu xo Ni xoI E e dydz E e e dydz
α α ασ σ− − − −= +∫ ∫
( ) ,Cu Cu Ni
t
z t z t
Cu xo Ni xo
o t
I w E e dz w E e e dzα α ασ σ
∞
− − − −= +∫ ∫
( )1 ,Cu Cut tCu Nixo
Cu Ni
I wE e eα ασ σ
α α
− −⎡ ⎤= − +⎢ ⎥
⎣ ⎦
( )
1
, where = 1 .Cu Cut tCu Nis s
Cu Ni
LR R R e e
w
α ασ σ
α α
−
− −⎡ ⎤∴ = − +⎢ ⎥
⎣ ⎦
7 3
Cu5.8 10 , 1, 479 10Cu r
S Npx x
m m
σ µ α= = =
7 6
Cu1.5 10 , 600, 596 10Ni r
S Npx x
m m
σ µ α= = =
( ) ( )22 7
1 1 1 1DC: 1.37
5.8 10 0.002
R m
L a mxσ π π
Ω
= = =
61 ; 1 2.09 10
2 2
sRR f x m
L a a
δ π µσ
π σδ π
−= = = =
( )( ) ( )7 6
1 0.66
5.8 10 2.09 10 2 0.002
R
L mx x π−
Ω
= =
( )
2
2
2
1 1 120 12 27
2 2avg xo z z z
A kWH
m m
η π ⎛ ⎞= = Ω =⎜ ⎟
⎝ ⎠
P a a a
5- !19
P5.27: A 600 MHz uniform plane wave incident in the z direction on a thick slab of
Teflon (εr = 2.1, µr = 1.0) imparts a 1.0 V/m amplitude y-polarized electric field intensity
at the surface. Assuming σ = 0 for Teflon, find in the Teflon (a) E(z,t), (b) H(z,t) and (c)
Pav.
!
!
Teflon: σ = 0 so α = 0,
and !
(a) !
(b) !
!
(c) !
P5.28: Assume distilled water (σ = 10-4 S/m, εr = 81, µr = 1.0) fills the region z > 0. At
the surface, we have E(0,t) = 8.0cos(2πx108t) ax V/m. Determine, for z > 0, (a) E(z,t),
(b) H(z,t), and (c) Pav at z = 1.0 m. (d) Find the power passing through a 10 square meter
surface located at z = 1.0 m.
(a) The general expression for E is: !
and we can see from the given information that
! . Also
!
!
!
( )( )6(0, ) 1cos 2 600 10 y Vt x t z mπ β= −E a
( )( , ) 1 cosz y
Vz t e t z
m
α ω β−= −E a
( )6
8
2 600 10
2.1 18.2
3 10r
x rad
c x m
πω
β ω µε ε= = = =
( )9( , ) 1cos 1.2 10 18.2 y Vz t x t z mπ= −E a
1 2.1 1 ,
120
j z
s P s z y
Ve
m
β
η π
−= × = ×
Ω
H a E a a
( )9( , ) 3.8cos 1.2 10 18.2 x mAz t x t z mπ= − −H a
( )2
2
1 2.11 1.9
2 120avg z z
mW
mπ
= =P a a
( )( , ) cos ,zo x
Vz t E e t z
m
α ω β φ−= − +E a
8 88 , 2 10 , 10 , 0o
V radE x f Hz
m s
ω π φ= = = =
( )( )( )8 12 42 10 81 8.854 10 0.45, 10 , so 1 (low loss dielectric).x x σωε π σ
ωε
− −= = = <<
( )
410 1 120 0.0021
2 2 81
Np
m
σ µ
α π
ε
−
= = =
18.8r
rad
c m
ω
β ω µε ε= = =
5- !20
!
so !
(b)
!
so !
(c) !
(d) !
P5.29: The density of solar radiation is approximately 150 W/m2 at some locations on the
earth’s surface. How much solar power is incident on a typical “100 Watt” solar panel (.6
m x 1.6 m area) if the panel is normal to the radiation propagation direction? How much
power is incident if the panel is tilted 45° to the radiation propagation direction?
!
P5.30: A 200 MHz uniform plane wave incident on a thick copper slab imparts a 1.0 mV/
m amplitude at the surface. How much power passes through a square meter at the
surface? How much power passes through a square meter area 10. µm beneath the
surface?
!
Cu: !
!
Now at 10 µm beneath the surface, we have
1120 41.9
81
µ
η π
ε
= = = Ω
( )0.0021 8( , ) 8 cos 2 10 18.8z x Vz t e x t z mπ
−= −E a
0.0021 18.8
0.0021 18.8 0.0021 18.8
8 ,
1 8 191
41.9
z j z
s x
z j z z j z
s P s y y
Ve e
m
mAe e e e
mη
− −
− − − −
=
= × = =
E a
H a E a a
( )0.0021 8( , ) 191 cos 2 10 18.8z y mAz t e x t z mπ
−= −H a
2
2 2(0.0021)(1)
2
1 0.764 0.761
2
zxo
avg z z
E We e
m
α
η
− −= = =P a a
2(10 ) 7.6avgP P m W= =
144 , cos 45 102avg avgP P S W P P S W= = = =
�
21200 , 1 ,
2
o
o avg
EmVf MHz E P
m η
= = =
45 3 452 , 214 10 , so 5.22j jNpe f x e m
m
α
η α π µσ η
σ
= = = = Ω
� �
( )23
3 2
101 96 ; 96
2 5.22 10avg avg
WP P P S W
x m
µ
µ
−
−
= = = =
5- !21
!
!
6. Wave Polarization
P5.31: Suppose E(z,t) = 10.cos(ωt-βz)ax + 5.0cos(ωt-βz)ay V/m. What is the wave
polarization and tilt angle?
The figure indicates linear polarization.
The tilt angle is:
!
P5.32: Given E(z,t) = 10.cos(ωt-βz)ax - 20.cos(ωt-βz-45°)ay V/m, find the polarization
and handedness.
The field can be rewritten as E(z,t) = 10.cos(ωt-βz)ax + 20.cos(ωt-βz-45°-180°)ay
or E(z,t) = 10.cos(ωt-βz)ax + 20.cos(ωt-βz+135°)ay
Running ML0503:
Polarization Plot
3(10 ) 3 (214 10 )(10 ) 6( 10 ) 10 118 10m x mo
VE z m E e e x
m
α µ µµ − − − −= = = =
( )26
3 2
118 101 1.3 ; 1.3
2 5.22 10avg
x WP P W
x m
µ
µ
−
−
= = =
1 5tan 27
10
τ − ⎛ ⎞= =⎜ ⎟
⎝ ⎠
�
Fig. P5.31 Fig. P5.32
5- !22
enter x-amplitude: 10
enter x-phase angle (degrees): 0
enter y-amplitude: 20
enter y-phase angle (degrees): 135
To determine direction of polarization,
move from the o to + along the plot.
>>
From the figure, we have left-hand elliptical polarization.
P5.33: Given H(z,t) = 2.0cos(ωt-βz)ax + 6.0cos(ωt-βz-120°)ay A/m, find the polarization
and handedness.
Convert to E(z,t):
!
!
With this we can run ML0503:
Polarization Plot
enter x-amplitude: 6
enter x-phase angle (degrees): -120
enter y-amplitude: 2
enter y-phase angle (degrees): -180
To determine direction of polarization,
move from the o to + along the plot.
>> From the figure, we have right-hand elliptical polarization.
P5.34: Given
!
we say that Ey leads Ex for 0° < φ < 180°, and that Ey lags Ex when –180° < φ < 0°.
Determine the handedness for each of these two cases.
For 0 < φ < 180°, we have LHP
( )120 1202 6 2 6j z j z j j z j z js P s o z x y o y o xe e e e e eβ β β βη η η η− − − − − −= − × = − × + = − +E a H a a a a a� �
( ) ( )( )( , ) 6cos 120 2cos 180o x yE z t t z t zη ω β ω β= − − + − −a a� �
( ) ( )( , ) cos cos ,xo yoz t E t z E t zω β ω β φ= − + − +x yE a a
5- !23
For 180° < φ < 360°, we have RHP
P5.35: MATLAB: For a general elliptical polarization represented by
! the axial ratio and tilt angle can be
found from the following formulas (from K. R. Demarest, Engineering Electromagnetics,
Prentice-Hall, 1998, pp. 451-453):
a=|Exo|, b=|Eyo|
MAJ = length of majority-axis
MIN = length of minority-axis
!
axial ratio=MAJ/MIN
! .
Compose a program that not only draws a polarization plot like MATLAB 5.3, but that
also calculates the axial ratio and tilt angle. Run the program on Drill 5.11.
% M-File: MLP0535
%
% This program modifies ML0503. As before, it will
% trace polarization ellipses, given the amplitude
% and phase of a pair of linearly polarized waves.
% Now it will also calculate axial ratio and tilt %angle.
%
( ) ( )( , ) cos cos ,xo yoz t E t z E t zω β ω β φ= − + − +x yE a a
2 2 4 4 2 2
2 2 4 4 2 2
12 2 cos2
2
12 2 cos2
2
MAJ a b a b a b
MIN a b a b a b
φ
φ
⎡ ⎤= + + + +
⎣ ⎦
⎡ ⎤= + − + +
⎣ ⎦
1
2 2
1 2tan cos
2
ab
a b
τ φ− ⎡ ⎤= ⎢ ⎥−⎣ ⎦
Fig. P5.33
5- !24
% Wentworth 1/28/03
% Variables:
% Exo,Eyo amplitudes for the pair of waves
% fxd,fyd phase angle for each wave
% fx,fy phase (radians) for each wave
% wtd ang freq * time, in degrees
% wtr ang freq
* time, in radians
% x,y superposed position
% x0,y0 position at wtd=0 degrees
% x45,y45 position at wtd=45 degrees
% a,b shorthand for Exo,Eyo
% MAJ,MIN majority,minority axis length
% AR,tiltangle axial ration, tilt angle
%
clc %clears the command window
clear %clears variables
% Prompt for input values
disp('Polarization Plot')
disp(' ')
Exo=input('enter x-amplitude: ');
fxd=input('enter x-phase angle (degrees): ');
fx=fxd*pi/180;
Eyo=input('enter y-amplitude: ');
fyd=input('enter y-phase angle (degrees): ');
fy=fyd*pi/180;
disp(' ')
disp('To determine direction of polarization,')
disp('move from the o to + along the plot.')
disp(' ')
%Perform calculations
wtd=0:360; %wt in degrees
wtr=wtd*pi/180;
x=Exo*cos(wtr+fx);
y=Eyo*cos(wtr+fy);
x0=Exo*cos(fx);
y0=Eyo*cos(fy);
x45=Exo*cos(fx+pi/4);
y45=Eyo*cos(fy+pi/4);
fdiff=fy-fx;
a=abs(Exo);b=abs(Eyo);
temp=sqrt(a^4+b^4+2*a^2*b^2*cos(2*fdiff));
MAJ=2*sqrt(0.5*(a^2+b^2+temp));
MIN=2*sqrt(0.5*(a^2+b^2-temp));
5- !25
AR=MAJ/MIN
temp2=(2*a*b/(a^2-b^2))*cos(fdiff);
tiltangle=(0.5*atan(temp2)*180/pi)
%Make the plot
plot(x,y,x0,y0,'ok',x45,y45,'+k')
xlabel('x')
ylabel('y')
title('Polarization Plot')
axis('equal')
Now we run the program for Drill 5.11.
Polarization Plot
enter x-amplitude: 3
enter x-phase angle (degrees): -30
enter y-amplitude: 8
enter y-phase angle (degrees): 90
To determine direction of polarization,
move from the o to + along the plot.
AR = 3.1997
tiltangle = 11.7874
7. Reflection and Transmission at Normal Incidence
P5.36: Starting with (5.107) and (5.109), derive (5.110) and (5.111).
(1) !
(2) !
Add (1) and (2): !
Now subtract (2) from (1):
!
i r t
o o oE E E+ =
1
2
i r t
o o oE E E
η
η
− =
1 1 2
2 2 1 2
22 1 , so i t t t t io o o o o oE E E E E E
η η η
η η η η
⎛ ⎞
= + = + =⎜ ⎟ +⎝ ⎠
1 1 2 2 1
2 2 1 2 2 1
22 1 1 , r t i r io o o o oE E E E E
η η η η η
η η η η η η
⎛ ⎞ ⎛ ⎞ −
= − = − =⎜ ⎟ ⎜ ⎟ + +⎝ ⎠ ⎝ ⎠
Fig. P5.35
5- !26
P5.37: A UPW is normally incident from media 1 (z < 0, σ = 0, µr = 1.0, εr = 4.0) to
media 2 (z > 0, σ = 0, µr = 8.0, εr = 2.0). Calculate the reflection and transmission
coefficients seen by this wave.
!
!
!
P5.38: Suppose media 1 (z < 0) is air and media 2 (z > 0) has εr = 16. The transmitted
magnetic field intensity is known to be Ht = 12 cos (ωt-β2z)ay mA/m. (a) Determine the
instantaneous value of the incident electric field. (b) Find the reflected average power
density.
!
!
!
!
!
!
!
!
P5.39: Suppose a UPW in air carrying an average power density of 100 mW/m2 is
normally incident on a nonmagnetic material with εr = 11. What is the time-averaged
power density of the reflected and transmitted waves?
2 1
1 2
2 1
120 8; 60 , 120 240
24
η η µ π µ
η π η π π
η η ε ε
−
Γ = = = = Ω = = = Ω
+
240 60 3 0.60
240 60 5
π π
π π
−
Γ = = =
+
1 1.60τ = +Γ =
2 2
2
12
t
j z j zt o
s y y
EmA mAe e
m m
β β
η
− −= =H a a
2t t
2 o s
2
30 , so 12 , E 0.36 , and 1.13
t
j zo
x
E mA V Ve
m m m
βη π π
η
−= Ω = = ∴ =E a
( ) 2 1
2 1
3 21 ; , 1
5 5
t i i
o o oE E E
η η
τ τ
η η
−
= = +Γ Γ = = − = +Γ =
+
12.83, so 2.83
t
j zi io
o s x
EE e β
τ
−= = ∴ =E a
( )1( , ) 2.83cos .x
Vz t t z
m
ω β∴ = −E a
11.70, so 1.70 j zr i ro o s xE E e
β+= Γ = − = −E a
( ) ( )1 11 1 1.70 4.5120
j z j zr r
s P s z x y
mAe e
m
β β
η π
+ += × = − × − =H a E a a a
( )( )( ) ( )3 2
1 1.70 4.5 10 3.8
2
r
avg z z
mWx
m
−= − =P a -a
5- !27
!
!
!
!
P5.40: A UPW in a lossless nonmagnetic εr = 16 media (for z < 0) is given by
E(z,t) = 10.cos(ωt-β1z)ax + 20.cos(ωt-β1z+π/3)ay V/m.
This is incident on a lossless media characterized by µr = 12, εr = 6.0 (for z > 0). Find the
instantaneous expressions for the reflected and transmitted electric field intensities.
!
!
!
!
!
!
! or ! , so
! .
P5.41: The wave Ei = 100 cos(π x 106t - β1z + π/4) ax V/m is incident from air onto a
perfect conductor. Find Er and Et.
For the perfect conductor, η2 = 0. So Γ = -1 and
Er = -100 cos(π x 106t + β1z + π/4) ax V/m
Et = 0
1 2
1 1120 11120 ; ; 0.537
11 1 1
11
o
π
η η π η
⎛ ⎞ −⎜ ⎟Ω ⎝ ⎠= = Ω = Γ = = −
⎛ ⎞+⎜ ⎟
⎝ ⎠
1 0.463τ = +Γ =
2
2P P 28.8
r i
avg avg
mW
m
= Γ =
2
2
2
2
1P P 11 71.2
2
xot i
avg avg
E mW
m
τ
η
= = =
1 1 310 20j z j zi js x ye e e
β β π− −= +E a a
1 1 310 20j z j zr js x ye e e
β β π= Γ + ΓE a a
1 2
120 1230 ; 120 120 2
616
π
η π η π π= = Ω = = Ω
2 1
2 1
0.700; 1 1.70η η τ
η η
−
Γ = = = +Γ =
+
1 1 37 14j z j zr js x ye e e
β β π= +E a a
( )( , ) 1 17 cos 14cos 3
r
z t x y
Vt z t z
m
π
ω β ω β⎛ ⎞= + + + +⎜ ⎟
⎝ ⎠
E a a
2 2 310 20 ,j z j zt js x ye e e
β β πτ τ− −= +E a a 2 2 317 34j z j zt js x ye e e
β β π− −= +E a a
( )( , ) 2 217cos 34cos 3
t
z t x y
Vt z t z
m
π
ω β ω β⎛ ⎞= − + − +⎜ ⎟
⎝ ⎠
E a a
5- !28
P5.42: A UPW given by E(z,t) = 10.cos(ωt-β1z)ax + 20.cos(ωt-β1z+π/3)ay V/m is incident
from air (for z < 0) onto a perfect conductor (for z > 0). Find the instantaneous
expression for the reflected electric field intensity and the SWR.
As in the previous problem, Γ = -1. We then have
E(z,t) = -10.cos(ωt+β1z)ax - 20.cos(ωt+β1z+π/3)ay V/m
!
P5.43: The wave Ei = 10.cos(2π x 108t - β1z) ax V/m is incident from air onto a copper
conductor. Find Er, Et and the time-averaged power density transmitted at the surface.
For copper we have
!
so !
We find !
So Er = -10.cos(2π x 108t + β1z) ax V/m
! and !
!
P5.44: Given a UPW incident from medium 1 (σ = 0, µr = 1.0, εr = 25.) to medium 2 (σ =
0.0080, µr = 1.0, εr = 81.), calculate Γ, SWR and τ at 1 kHz, 1 MHz, and 1 GHz.
!
Table P5.44
1
1
SWR
+ Γ
= = ∞
− Γ
( )( )( )
452
2
2
8 7 7 3
2 2 2 2
2
where 10 4 10 5.8 10 151 10
je
Npf x x x
m
α
η
σ
α π µ σ π π β−
= Ω
= = = =
�
45
2 3.7
je mη = Ω
�
6 452 2
1 2 1
2 21, and = 19.6 10 jx eη ητ
η η η
−Γ ≈ − ≈ =
+
�
( ) 2 2 45196 ,z j zt js x
Ve e e
m
α β µ− −=E a
� ( )2 2196 cos 45zt x Ve t z m
α ω β−= − +E a�
( )
( ) ( )
26
23
196 101 cos 45 3.7 .
2 3.7 10
t
avg z z
x V m W
mx
µ
−
−
= =
Ω
P a a�
6
1 2 9
120 7.896 10 ( )24 ;
0.008 4.506 10 ( )25
j j x f Hz
j j x f Hz
π ωµ
η π η
σ ωε
−
−
= = Ω = =
+ +
5- !29
P5.45: MATLAB: Write a program that prompts the user for the constitutive parameters
in medium 1 and medium 2 separated by a planar surface. You are to assume a wave is
normally incident from media 1 to media 2. The program is to plot Γ and τ versus a
frequency range supplied by the user. Use this program to plot Γ and τ from 100 Hz to
10 GHz for the pair of media specified in the previous problem.
%ML P0545
clear
clc
%prompt user for constit parameters of media 1 & 2
%then plot ref & trans coeff over a freq range.
%We'll plot mag and angle of each.
%enter constant values
eo=8.854e-12; %free space permittivity, F/m
uo=pi*4e-7; %free space permeability, H/m
%enter media 1 values
er1=input('enter er1: ');
ur1=input('enter ur1: ');
s1=input('enter s1: ');
%enter media 2 values
er2=input('enter er2: ');
ur2=input('enter ur2: ');
s2=input('enter s2: ');
%calculations
n=2:.5:10;
f=10.^n;
w=2*pi.*f;
eta1=sqrt(i*w*ur1*uo./(s1+i*w*er1*eo));
eta2=sqrt(i*w*ur2*uo./(s2+i*w*er2*eo));
Gamma=(eta2-eta1)./(eta2+eta1);
f η2(Ω)
1kHz 0.994ej44.98° 0.9815ej178.9° 107.3
1MHz 29.3ej30.3° 0.513ej155° 3.11
1GHz 41.9ej0.05° 0.286ej179.9° 1.80
!
1
1
SWR
+ Γ
=
− Γ!
2 1
2 1
η η
η η
−
Γ =
+
5- !30
Gmag=abs(Gamma);
Gang=180*angle(Gamma)/pi;
Tau=1+Gamma;
Tmag=abs(Tau);
Tang=180*angle(Tau)/pi;
subplot(2,1,1)
semilogx(f,Gmag,'-o',f,Tmag,'-*')
xlabel('frequency (Hz)')
ylabel('magnitude')
legend('reflection','transmission')
subplot(2,1,2)
semilogx(f,Gang,'-o',f,Tang,'-*')
xlabel('frequency (Hz)')
ylabel('phase angle (degrees)')
legend('reflection','transmission')
Run the program:
enter er1: 25
enter ur1: 1
enter s1: 0
enter er2: 81
enter ur2: 1
enter s2: .008
>>
P5.46: A wave specified by Ei = 100.cos(πx107t-β1z)ax V/m is incident from air (at z < 0)
to a nonmagnetic media (z > 0, σ = 0.050 S/m, εr = 9.0). Find Er, Et and SWR. Also find
the average power densities for the incident, reflected and transmitted waves.
!
In this problem we find in medium 2 (z > 0) that ωε = 0.0025 and σ = 0.05. These values
are too close to allow for simplifying assumptions. Using (5.13) and (5.31), we calculate:
7 6
1 1
1
2120 , 10 so 5 10 , 0.105rad radx f x Hz
s c m
π ω
η π ω π β
λ
= Ω = = = = =
Fig. P5.45
5- !31
! .
Then,
!
!
! ,
so !
! ,
so !
!
!
!
!
(check: 13.3 W/m2 = 10.7 W/m2 + 2.6 W/m2)
P5.47: A wave specified by Ei = 12π cos(2πx107t-β1z+π/4)ax V/m is incident from a
nonmagnetic, lossless, εr = 9.0 media (at z < 0) to a media (z > 0) with σ = 0.020 S/m, µr
= 2.0, and εr = 16.). Find Hi, Er, Hr, Et, Ht, and the average power densities for the
incident, reflected and transmitted waves.
We use ML0501 in each media to find:
!
!
We also will need reflection and transmission coefficients:
43.6
2 2 20.969 , 1.019 , 28.1
jNp rad e
m m
α β η= = = Ω
�
174 40.82 1
2 1
1
0.898 , 18.6, 1 0.141
1
j je SWR eη η τ
η η
+ Γ−
Γ = = = = = +Γ =
+ − Γ
� �
1100 j zis x
Ve
m
β−=E a
1 1 174100 89.8j z j zr js x x
V Ve e e
m m
β β+ += Γ =E a a
�
( )7( , ) 89.8cos 10 0.105 174 .r x Vz t x t z mπ= + +E a
�
2 2 40.8100 14.1j z j zt js x x
V Ve e e
m m
β βτ − −= =E a a
�
( )7( , ) 14.1cos 10 1.02 40.8 .t x Vz t x t z mπ= − +E a
�
2 243.6 40.8 2.814.1 0.502
28.1
j z j zt j j j
s y y
A Ae e e e e
m m
β β− −− −= =H a a
� � �
( )( ) ( ) 2
1 14.1 0.502 cos 40.8 2.8 2.6
2
t
z z
W
m
= + =P a a� �
( )
( )
2
2
100
13.3
2 120
i
z
W
mπ
= =P a
( )
( )
( )
2
2
89.8
10.7
2 120
r
z
W
mπ
−
= =P -a
1 1 10; 0.628 ; 40
rad
m
α β η π= = = Ω
33
2 2 21.01 ; 1.56 ; 84.9
jNp rad e
m m
α β η= = = Ω
�
5- !32
!
Incident:
!
! ,
!
!
Reflected:
!
!
!
!
!
Transmitted:
! ,
!
! ,
!
!
(Check: 5.655W/m2 = 0.704W/m2 + 4.954W/m2)
126 19.82 1
2 1
0.353 ; 1 0.84j je eη η τ
η η
−
Γ = = = +Γ =
+
� �
1 412 j zi js x
Ve e
m
β ππ −=E a
1 14 4
1
12 0.300j z j zi j js y y
A Ae e e e
m m
β βπ ππ
η
− −= =H a a
7( , ) 0.300cos 2 10 0.628 .
4
i
y
Az t x t z
m
π
π⎛ ⎞= − +⎜ ⎟
⎝ ⎠
H a
( )2
2
1
121 5.655
2
i
avg z z
W
m
π
η
= =P a a
( ) 1 1 14 45 126 17112 13.3 13.3j z j z j zr j j j js x x x
V V Ve e e e e e e
m m m
β β βππ + + += Γ = =E a a a
� � �
( )7( , ) 13.3cos 2 10 0.628 171 .r x Vz t x t z mπ= + +E a
�
1 1171 17113.3 0.106
40
j z j zr j j
s y y
A Ae e e e
m m
β β
π
+ +−= =H a a
� �
( )7( , ) 0.106cos 2 10 0.628 171 .r y Az t x t z mπ= − + +H a
�
( ) ( )
2
2
13.31 0.704
2 40
r
avg z z
W
mπ
= − =P a -a
( ) 1 24 64.812 31.67j z j zt j js x x
V Ve e e e
m m
β βπτ π − −= =E a a
�
( )7( , ) 31.7cos 2 10 1.56 64.8 .t x Vz t x t z mπ= − +E a
�
2 233 64.8 31.831.67 0.373
84.9
j z j zt j j j
s y y
A Ae e e e e
m m
β β− −−= =H a a
� � �
( )7( , ) 0.373cos 2 10 1.56 31.8 .t y Az t x t z mπ= − +H a
�
( )( ) ( ) 2
31.67 0.373
cos 64.8 31.8 4.954
2
t
avg z z
W
m
= − =P a a� �
5- !33
8. Reflection and Transmission at Oblique Incidence
P5.48: A 100 MHz TE polarized wave with amplitude 1.0 V/m is obliquely incident from
air (z < 0) onto a slab of lossless, nonmagnetic material with εr = 25 (z > 0). The angle of
incidence is 40°. Calculate (a) the angle of transmission, (b) the reflection and
transmission coefficients, and (c) the incident, reflected and transmitted fields.
(a) !
!
(b) Now we need to calculate the reflection and transmission coefficients.
!
!
(c) The fields,
Incident:
!
!
!
!
!
Reflected:
!
!
!
( )6
1 28
2 100 10
2.09 , 10.45 .
3 10
r
x rad rad
c x m c m
π ω εω
β β= = = = =
1
2 2
1 1 1; sin sin 40 ; 7.4
5 5t tr
β
θ θ
β ε
= = = =� �
1 2
120120 ; 24
25
π
η π η π= Ω = Ω = Ω
2 1
2 1
cos cos 0.732; 1 0.268
cos cos
i t
TE TE TE
i t
η θ η θ
τ
η θ η θ
−
Γ = = − = +Γ =
+
( )2.09 sin 40 cos40 1.34 1.601 1j x zi j x j zs y y
Ve e e
m
− + − −= =E a a
� �
( )( , ) 1cos 1.34 1.60i y
Vz t t x z
m
ω= − −E a
( )1.34 1.601 cos 40 sin 40120
i j x j z
s x ze eπ
− −= − +H a a� �
( ) 1.34 1.602.03 1.71i j x j zs x z
mAe e
m
− −= − +H a a
( ) ( )( , ) 2.03 1.71 cos 1.34 1.60i x z
mAz t t x z
m
ω= − + − −H a a
0.732r io TE oE E= Γ = −
( ) 1.34 1.600.732r j x j zs y
Ve e
m
− += −E a
( )( , ) 0.732cos 1.34 1.60r y
Vz t t x z
m
ω= − − +E a
5- !34
!
!
transmitted:
!
!
!
!
!
P5.49: A 100 MHz TM polarized wave with amplitude 1.0 V/m is obliquely incident from
air (z < 0) onto a slab of lossless, nonmagnetic material with εr = 25 (z > 0). The angle of
incidence is 40°. Calculate (a) the angle of transmission, (b) the reflection and
transmission coefficients, and (c) the incident, reflected and transmitted fields.
(a) The material parameters in this problem are the same as for P5.48. So, once again we
have θt = 7.4°. Also, β1 = 2.09 rad/m and β2 = 10.45 rad/m.
(b)
!
!
(c)
Incident:
!
!
!
( ) ( )
( )
1.34 1.60
1.34 1.60
0.732
cos 40 sin 40
120
1.49 1.25
r j x j z
s x z
r j x j z
s x z
Ae e
m
mAe e
m
π
− +
− +
−
= +
= − −
H a a
H a a
� �
( ) ( )( , ) 1.49 1.25 cos 1.34 1.60r x z
mAz t t x z
m
ω= − − − +H a a
0.268t io TE oE Eτ= =
( )2 sin cos 1.35 10.40.268 0.268t tj x zt j x j zs y y
Ve e e
m
β θ θ− + − −= =E a a
( )1.35 10.40.268 cos7.4 sin 7.424
t j x j z
s x z
Ae e
mπ
− −= − +H a a� �
( ) 1.35 10.43.5 0.46t j x j zs x z
mAe e
m
− −= − +H a a
( ) ( )( , ) 3.5 0.46 cos 1.35 10.4t x z
mAz t t x z
m
ω= − + − −H a a
2 1
2 1
cos cos 0.589
cos cos
t i
TM
t i
η θ η θ
η θ η θ
−
Γ = = −
+
2
2 1
2 cos 0.318
cos cos
i
TM
t i
η θ
τ
η θ η θ
= =
+
( )1.34 1.601 cos 40 sin 40i j x j zs x ze e− −= −E a a� �
( ) ( )( , ) 0.766 0.643 cos 1.34 1.60i x z
Vz t t x z
m
ω= − − −E a a
1.34 1.601
120
i j x j z
s y
Ae e
mπ
− −=H a
5- !35
!
Reflected:
!
!
!
!
transmitted:
!
!
!
P5.50: A randomly polarized UPW at 200 MHz is incident at the Brewster’s angle from
air (z < 0) onto a thick slab of lossless, nonmagnetic material with εr = 16 (z > 0). The
wave can be decomposed into equal TE and TM parts, each with an incident electric field
amplitude of 10. V/m. Find expressions for the instantaneous value of the incident,
reflected and transmitted electric fields.
First we calculate the Brewster’s angle: !
Also, we calculate β1 = 4.19 rad/m, β2 = 16.8 rad/m, η1 = 120π Ω, and η2 = 30 π Ω.
TE
!
At the Brewster’s angle of incidence, we have from Snell’s Law:
!
!
( )( , ) 2.65cos 1.34 1.60i y
mAz t t x z
m
ω= − −H a
( )1.34 1.600.589 cos 40 sin 40r j x j zs x ze e− += − +E a a� �
( ) ( )( , ) 0.452 0.379 cos 1.34 1.60r x z
Vz t t x z
m
ω= − − − +E a a
1.34 1.600.589
120
r j x j z
s y
Ae e
mπ
− +−=H a
( )( , ) 1.56cos 1.34 1.60r y
mAz t t x z
m
ω= − − +H a
( )1.35 10.40.318 cos7.4 sin 7.4t j x j zs x ze e− −= −E a a� �
( ) ( )( , ) 0.315 0.041 cos 1.35 10.4t x z
Vz t t x z
m
ω= − − −E a a
( )( , ) 4.22cos 1.35 10.4t y
mAz t t x z
m
ω= − −H a
1sin ; 76
1 1 16BA BA
θ θ= =
+
�
( )1 sin cos 4.06 1.0110 10i ij x zi j x j zs y y
Ve e e
m
β θ θ− + − −= =E a a
1 1
2
sin sin 14t i
β
θ θ
β
− ⎛ ⎞= =⎜ ⎟
⎝ ⎠
�
2 1
2 1
cos cos 0.883; 1 0.117
cos cos
i t
TE TE TE
i t
η θ η θ
τ
η θ η θ
−
Γ = = − = +Γ =
+
5- !36
!
!
!
TM:
At the Brewster’s angle, ΓTM = 0 and
! ,
!
! ,
!
Combining the results we arrive at:
!
!
!
V V=-8.83 ; =1.17
m m
r i t i
o TE o o TE oE E E Eτ= Γ =
4.06 1.018.83r j x j zs y
Ve e
m
− += −E a
( ) ( )2 sin cos 4.06 16.310 1.17t tj x zt j x j zs TE y y
Ve e e
m
β θ θτ − + − −= =E a a
0.rs =E
( )4.06 1.0110 cos sini j x j zs i x i ze e θ θ− −= −E a a
( ) 4.06 1.012.42 9.70 .i j x j zs x z
Ve e
m
− −= −E a a
( )4.06 16.310 cos sint j x j zs t x t ze e θ θ− −= −E a a
( ) 4.06 16.39.70 2.40 .t j x j zs x z
Ve e
m
− −= −E a a
( ) ( )( , ) 2.4 10 9.7 cos 4.06 1.01i x y z
Vz t t x z
m
ω= + − − −E a a a
( )( , ) 8.83cos 4.06 1.01r y
Vz t t x z
m
ω= − − +E a
( ) ( )( , ) 9.7 1.2 2.4 cos 4.06 16.3t x y z
Vz t t x z
m
ω= + − − −E a a a
Wentworth solutions chapter 6.pdf
6- !1
Solutions for Chapter 6 Problems
1. Distributed Parameters Model
P6.1: RG-223/U coax has an inner conductor radius a = 0.47 mm and inner radius of the
outer conductor b = 1.435 mm. The conductor is copper, and polyethylene is the
dielectric. Calculate the distributed parameters at 800 MHz.
!
!
!
!
!
P6.2: MATLAB: Modify MATLAB 6.1 to account for a magnetic conductive material.
Apply this program to problem P6.1 if the copper conductor is replaced with nickel.
!
Note that this program has also been modified for P6.04 as well.
%Coax distributed parameters
%
% Modified: P0602
% add rel permeability
% also modified for P0604
%
clear
clc
7
16
for copper: 5.8 10
for polyethylene: 2.26, 10
Cu
r
Sx
m
S
m
σ
ε σ −
=
= =
( )( )
( )
6 7
3 3 7
1 1 1'
2
800 10 4 101 1 1 3.32
2 0.47 10 1.435 10 5.8 10
c
fR
a b
x x
x x mx
π µ
π σ
π π
π
−
− −
⎛ ⎞= +⎜ ⎟
⎝ ⎠
Ω⎛ ⎞= + =⎜ ⎟
⎝ ⎠
74 10 1.435' ln ln 223
2 2 0.47
b x nHL
a m
µ π
π π
−⎛ ⎞ ⎛ ⎞= = =⎜ ⎟ ⎜ ⎟
⎝ ⎠ ⎝ ⎠
( )
( )
( )
16
18
2 102' 560 10
ln ln 1.435 0.47 0
SG x
b a m
ππσ
−
−= = =
≈
( )
( )( )
( )
122 2.26 8.854 102' 112
ln ln 1.435 0.47
x pFC
b a m
ππε
−
= = =
7for Nickel we have 1.5 10 and 600.Ni r
Sx
m
σ µ= =
6- !2
disp('Calc Coax Distributed Parameters')
%Some constant values
muo=pi*4e-7;
eo=1e-9/(36*pi);
%Prompt for input values
a=input('inner radius, in mm, = ');
b=input('outer radius, in mm, = ');
er=input('relative permittivity, er= ');
sigd=input('dielectric conductivity, in S/m, = ');
sigc=input('conductor conductivity, in S/m, = ');
ur=input('conductor rel. permeability, = ');
f=input('input frequency, in Hz, = ');
%Perform calulations
G=2*pi*sigd/log(b/a);
C=2*pi*er*eo/log(b/a);
L=muo*log(b/a)/(2*pi);
Rs=sqrt(pi*f*ur*muo/sigc);
R=(1000*((1/a)+(1/b))*Rs)/(2*pi);
omega=2*pi*f;
RL=R+i*omega*L;
GC=G+i*omega*C;
Gamma=sqrt(RL*GC);
Zo=sqrt(RL/GC);
alpha=real(Gamma);
beta=imag(Gamma);
loss=exp(-2*alpha*1);
lossdb=-10*log10(loss);
%Display results
disp(['G/h = ' num2str(G) ' S/m'])
disp(['C/h = ' num2str(C) ' F/m'])
disp(['L/h = ' num2str(L) ' H/m'])
disp(['R/h = ' num2str(R) ' ohm/m'])
disp(['Gamma= ' num2str(Gamma) ' /m'])
disp(['alpha= ' num2str(alpha) 'Np/m'])
disp(['beta= ' num2str(beta) 'rad/m'])
disp(['Zo = ' num2str(Zo) ' ohms'])
disp(['loss=' num2str(loss) ' /m'])
disp(['lossdb=' num2str(lossdb) ' dB/m'])
6- !3
Now run the program for Nickel:
Calc Coax Distributed Parameters
inner radius, in mm, = 0.47
outer radius, in mm, = 1.435
relative permittivity, er= 2.26
dielectric conductivity, in S/m, = 1e-16
conductor conductivity, in S/m, = 1.5e7
conductor rel. permeability, = 600
input frequency, in Hz, = 800e6
G/h = 5.6291e-016 S/m
C/h = 1.1249e-010 F/m
L/h = 2.2324e-007 H/m
R/h = 159.7792 ohm/m
Gamma= 1.78881+25.252i /m
alpha= 1.7888Np/m
beta= 25.252rad/m
Zo = 44.6608-3.1637i ohms
loss=0.027942 /m
lossdb=15.5374 dB/m
>>
Summarizing the distributed parameter data from this routine we have:
!
P6.3: Modify (6.3) to include internal inductance of the conductors. To simplify the
calculation, assume current is evenly distributed across the conductors. Find the new
value of L’ for the coax of Drill 6.1.
From Ampere’s Circuit Law we can find H versus ρ:
!
!
!
!
18' 160 , ' 223 , ' 560 10 , ' 112 pFnH SR L G x Cm m m m
−Ω= = = =
2 for 2
IH a
aφ
ρ
ρ
π
= ≤
for a
2
IH bφ ρπρ
= ≤ ≤
2 2
2 2
c for b
2
IH c
c bφ
ρ
ρ
πρ
−
= ≤ ≤
−
0 for H cφ ρ= ≥
6- !4
Using the energy approach, ! , we find
!
Inserting the given values we find
!
With two significant digits we therefore have L’ = 330 nH/m.
2. Time Harmonic Waves on Transmission Line
P6.4: MATLAB: Modify MATLAB 6.1 to also calculate γ, α, β and Zo. Confirm the
program using Drill 6.2.
See the solution for P6.2.
Calc Coax Distributed Parameters
inner radius, in mm, = 0.45
outer radius, in mm, = 1.47
relative permittivity, er= 2.26
dielectric conductivity, in S/m, = 1e-16
conductor conductivity, in S/m, = 5.8e7
conductor rel. permeability, = 1
input frequency, in Hz, = 1e9
G/h = 5.3078e-016 S/m
C/h = 1.0606e-010 F/m
L/h = 2.3675e-007 H/m
R/h = 3.8112 ohm/m
Gamma= 0.0403332+31.4857i /m
alpha= 0.040333Np/m
beta= 31.4857rad/m
Zo = 47.246-0.0605221i ohms
loss=0.9225 /m
lossdb=0.35033 dB/m
>>
This agrees with the results of Drill 6.2.
2 21
2 2
o
mW LI H dv
µ
= = ∫
22 2 2 2
2 2 2 2 2 2
1' ln ln
2 8 2 4
o o ob c c c c bL
a c b b c b c b
µ µ µ
π π π
⎡ ⎤⎛ ⎞ ⎛ ⎞ ⎛ ⎞+⎛ ⎞= + + − +⎢ ⎥⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎜ ⎟− − −⎝ ⎠⎢ ⎥⎝ ⎠ ⎝ ⎠ ⎝ ⎠⎣ ⎦
( ) nH nH' 237 50 41.2 328
m m
L = + + =
6- !5
P6.5: The impedance and propagation constant at 100 MHz for a T-Line are determined
to be Zo = 18.6 – j0.253 Ω and γ = 0.0638 + j4.68 /m. Calculate the distributed
parameters.
!
!
!
P6.6: The specifications for RG-214 coaxial cable are as follows:
• 2.21 mm diameter copper inner conductor
• 7.24 mm inner diameter of outer conductor
• 9.14 mm outer diameter of outer conductor
• Teflon dielectric (εr = 2.10)
Calculate the characteristic impedance and the propagation velocity for this cable.
!
!
P6.7: For the RG-214 coax of problem P6.6 operating at 1 GHz, how long is this T-line in
terms of wavelengths if its physical length is 50 cm?
!
P6.8: If 1 watt of power is inserted into a coaxial cable, and 1 microwatt of power is
measured 100 m down the line, what is the line’s attenuation in dB/m?
( )( )' ' , ' ' ' '
' 'o
R j LZ R j L G j C
G j C
ω
γ ω ω
ω
+
= = + +
+
' ' 2.37 87.0
' 2.37 , ' 87.0 ' 139
oZ R j L j
nHR L so L
m m
γ ω
ω
= + = +
Ω
∴ = = =
6' ' 7.63 10 0.252,
' 7.63 , ' 0.252 ' 401
o
G j C x j
Z
S pFG and C so C
m m
γ
ω
µ
ω
−= + = +
∴ = = =
60 60 3.62ln ln 49.1
1.1052.1o r
bZ
aε
⎛ ⎞ ⎛ ⎞= = = Ω⎜ ⎟ ⎜ ⎟
⎝ ⎠ ⎝ ⎠
82.07 10p
r
c mu x
sε
= =
( )
8
9
2.07 10, 0.207
1 10
1( ) 50 2.4
0.207 100
p
p
u xu f m
f x
mcm
m cm
λ λ
λ
λ λ
= = = =
⎛ ⎞⎛ ⎞= =⎜ ⎟⎜ ⎟
⎝ ⎠⎝ ⎠
l
6- !6
!
P6.9: Starting with a 1 mm diameter solid copper wire, you are to design a 75 Ω coaxial
T-Line using mica as the dielectric. Determine (a) the inner diameter of the outer copper
conductor, (b) the propagation velocity on the line and (c) the approximate attenuation, in
dB/m, at 1 MHz.
!
So the inner diameter of the outer conductor is 18 mm.
!
To calculate α, will need γ. Therefore we calculate R’, L’, G’ and C’.
!
!
!
!
Now, with ω = 2πf,
!
Finally, !
This is confirmed using MLP0602.
P6.10: MATLAB: A coaxial cable has a solid copper inner conductor of radius a = 1mm
and a copper outer conductor of inner radius b. The outer conductor is much thicker than
110log 60
1
60' 0.6
100
WA dB
W
dB dBA
m m
µ⎛ ⎞= − = +⎜ ⎟
⎝ ⎠
= =
( )( ) ( ) ( )( )o60 ln , b=a exp Z 60 0.5 exp 75 5.4 60 9.1o r
r
bZ mm mm
a
ε
ε
⎛ ⎞= = =⎜ ⎟
⎝ ⎠
8
8 82.998 10 1.29 10 , so 1.3 10
5.4p pr
c x m mu x u x
s sε
= = = =
( )( )6 7
3 3 7
1 10 4 101 1 1' 87.6
2 0.5 10 9.1 10 5.8 10
x x mR
x x x m
π π
π
−
− −
Ω⎛ ⎞= + =⎜ ⎟
⎝ ⎠
74 10 9.1' ln 580
2 0.5
x nHL
m
π
π
− ⎛ ⎞= =⎜ ⎟
⎝ ⎠
( )
( )
15
15
2 10
' 2.17 10
ln 9.1 0.5
SG x
m
π −
−= =
( )( )
( )
122 5.4 8.854 10
' 103.5
9.1ln 0.5
x pFC
m
π −
= =
( )( ) 6 1' ' ' ' 585 10 0.049R j L G j C x j
m
γ ω ω −= + + = +
6 38.686585 10 5.1 10Np dB dBx x
m Np m
α − −⎛ ⎞= =⎜ ⎟
⎝ ⎠
6- !7
a skin depth. The dielectric has εr = 2.26 and σeff = 0.0002 at 1 GHz. Letting the ratio b/
a vary from 1.5 to 10, generate a plot of the attenuation (in dB/m) versus the line
impedance. Use the lossless assumption to calculate impedance.
% MLP0610
%
% Plot of alpha vs Zo for a particular coax
clear
clc
%Some constant values
muo=pi*4e-7;
eo=8.854e-12;
a=1;
er=2.26;
sigd=0.0002;
sigc=5.8e7;
f=1e9;
%Perform calulations
b=1.5:.1:10;
G=2*pi*sigd./log(b./a);
C=2*pi*er*eo./log(b./a);
L=muo*log(b./a)/(2*pi);
Rs=sqrt(pi*f*muo/sigc);
R=(1000*((1./a)+(1./b))*Rs)/(2*pi);
w=2*pi*f;
RL=R+i*w*L;
GC=G+i*w*C;
Gamma=sqrt(RL.*GC);
Zo=abs(sqrt(RL./GC));
alpha=real(Gamma);
loss=exp(-2*alpha*1);
lossdb=-10*log10(loss);
plot(Zo,lossdb)
xlabel('Characteristic Impedance (ohms)')
ylabel('attenuation (dB/m)')
grid on
3. Terminated T-Lines
P6.11: Start with equation (6.54) and derive (6.55).
6- !8
!
With ! we then have
!
We also know that
!
So now we have
!
and with rearranging,
!
We can convert the exponential terms into hyperbolic functions, given
!
This leads to
o o
in o
o o
V e V eZ Z
V e V e
γ γ
γ γ
+ + − −
+ + − −
+
=
−
l l
l l
,o L oV V
− += Γ
( )
( )
L
in o
L
e e
Z Z
e e
γ γ
γ γ
+ −
+ −
+Γ
=
−Γ
l l
l l
,L oL
L o
Z Z
Z Z
−
Γ =
+
( ) ( )
( ) ( )
L o
L o L o L o
in o o
L o L oL o
L o
Z Ze e
Z Z Z Z e Z Z e
Z Z Z
Z Z e Z Z eZ Ze e
Z Z
γ γ
γ γ
γ γ
γ γ
+ −
+ −
+ −
+ −
⎛ ⎞−
+ ⎜ ⎟+ + + −⎝ ⎠= =
+ − −⎛ ⎞−
− ⎜ ⎟+⎝ ⎠
l l
l l
l l
l l
( ) ( )
( ) ( )
.L oin o
L o
Z e e Z e e
Z Z
Z e e Z e e
γ γ γ γ
γ γ γ γ
+ − + −
+ − + −
+ + −
=
− + +
l l l l
l l l l
( ) ( )1 1 sinh(x)sinh( ) , cosh( ) , and tanh(x)= .2 2 cosh(x)
x x x xx e e x e e− −= − = +
Fig. P6.10
6- !9
!
or finally
!
P6.12: Derive (6.56) from (6.55) for a lossless line.
! and ! since α = 0 for
lossless line. Using the hyperbolic definitions, we have
!
Now using Euler’s formula,
!
Plugging this in, we find,
!
P6.13: A 2.4 GHz signal is launched on a 1.5 m length of T-Line terminated in a matched
load. It takes 6.25 ns to reach the load and suffers 1.2 dB of loss. Find the propagation
constant.
!
!
!
!
So
!
( ) ( )
( ) ( )
2 cosh 2 sinh
,
2 sinh 2 cosh
L o
in o
L o
Z Z
Z Z
Z Z
γ γ
γ γ
+
=
+
l l
l l
( )
( )
tanh
.
tanh
L o
in o
o L
Z Z
Z Z
Z Z
γ
γ
+
=
+
l
l
( )
( )
tanh
,
tanh
L o
in o
o L
Z Z
Z Z
Z Z
γ
γ
+
=
+
l
l ( ) ( ) ( )tanh tanh tanhj jγ α β β= + =l l l l
( ) ( )
( )
( )
( )
sinh
tanh .
cosh
j j
j j
e ej
j
j e e
β β
β β
β
β
β
+ −
+ −
−
= =
+
l l
l l
ll l
( ) ( ) ( )
( ) ( )
( )
( )
cos sin( ) - cos sin( ) 2sin
tanh tan( )
cos sin( ) cos sin( ) 2cos
j j j
j j
j j
β β β β β
β β
β β β β β
+ − − −
= = =
+ + − +
l l l l ll ll l l l l
( )
( )
tan
.
tan
L o
in o
o L
Z jZ
Z Z
Z jZ
β
β
+
=
+
l
l
jγ α β= +
1.2 1 0.092
1.5 8.686
dB Np Np
m dB m
α = =
81.5: 2.4 10
6.25p
m mu x
t ns s
ω
β
β
= = = =
l
( )9
8
2 2.4 10
62.8
2.4 10p
x rad
u x m
πω
β = = =
10.092 62.8 j
m
γ = +
6- !10
P6.14: A source with 50 Ω source impedance drives a 50 Ω T-Line that is 1/8 of a
wavelength long, terminated in a load ZL = 50 – j25 Ω. Calculate ΓL, VSWR, and the
input impedance seen by the source.
!
!
!
!
P6.15: A 1 m long T-Line has the following distributed parameters: R’ = 0.10 Ω/m, L’ =
1.0 µH/m, G’ = 10.0 µS/m, and C’ = 1.0 nF/m. If the line is terminated in a 25 Ω resistor
in series with a 1 nH inductor, calculate, at 200 MHz, ΓL and Zin.
!
Now, MLP0615 is used to solve the problem.
% MLP0615
%
% calculate gamma and char impedance
% given the distributed parameters
% Then, calculate gammaL and Zin
%
% define variables
clc
clear
R=0.1;
L=1.0e-6;
G=10e-6;
C=1.0e-9;
f=200e6;
7650 25 50 0.242
50 25 50
jL o
L
L o
Z Z j e
Z Z j
−− − −Γ = = =
+ − +
�
1
1.64
1
L
L
VSWR
+ Γ
= =
− Γ
2 , tan 1
8 4 4
π λ π π
β
λ
⎛ ⎞= = =⎜ ⎟
⎝ ⎠
l
( )
( )
tan
tan
50 25 5050
50 50 25
30.8 3.8
L o
in o
o L
Z jZ
Z Z
Z jZ
j j
j
j
β
β
+
=
+
− +
=
+ +
= − Ω
l
l
( )( )6 925 2 200 10 10 25 1.257 LZ j x jπ −= + = + Ω
Fig. P6.14
6- !11
w=2*pi*f;
length=1;
ZL=25+j*1.257;
% Perform calcuations
A=R+i*w*L;
B=G+i*w*C;
gamma=sqrt(A*B) %Propagation Constant
Zo=sqrt(A/B)
gammaL=(ZL-Zo)/(ZL+Zo) %Reflection coefficient
TGL=tanh(gamma*length);
Zin=Zo*((ZL+Zo*TGL)/(Zo+ZL*TGL))
Running the program,
Gamma = 0.0017 +39.7384i
Zo = 31.6228 - 0.0011i
gammaL = -0.1164 + 0.0248i
Zin = 34.0192 - 7.4618i
>>
So the answers are, with the appropriate significant digits,
!
P6.16: The reflection coefficient at the load for a 50 Ω line is measured as ΓL = 0.516ej8.2°
at f = 1 GHz. Find the equivalent circuit for ZL.
Rearranging ! we find ! .
This is a resistor in series with an inductor. The inductor is found by considering
! ,
So the load is a 150 Ω resistor in series with a 4.8 nH inductor.
P6.17: The input impedance for a 30 cm length of lossless 100 Ω impedance T-line
operating at 2 GHz is Zin = 92.3 – j67.5 Ω. The propagation velocity is 0.7c. Determine
the load impedance.
Rearranging ! we find !
1680.12 and 34 7.5 jL ine Z jΓ = = − Ω
�
,L oL
L o
Z Z
Z Z
−
Γ =
+
1 150 30
1
L
L o
L
Z Z j+Γ= = + Ω
−Γ
( )9
3030, or 4.8
2 1 10
j L j L nH
x
ω
π
= = =
( )
( )
tan
,
tan
L o
in o
o L
Z jZ
Z Z
Z jZ
β
β
+
=
+
l
l
( )
( )
tan
tan
in o
L o
o in
Z jZ
Z Z
Z jZ
β
β
−
=
−
l
l
6- !12
!
Evaluating, we have
! or L = 1.3 pH.
This is a very small inductance, so we have !
P6.18: For the lossless T-Line circuit shown in Figure 6.51, determine the input
impedance Zin and the instantaneous voltage at the load end vL.
!
!
!
!
!
!
! , so !
P6.19: Referring to Figure 6.10, a lossless 75 Ω T-Line has up = 0.8c and is 30 cm long.
The supply voltage is vs = 6.0 cos(ωt) V with Zs = 75 Ω. If ZL = 100 + j125 Ω at 600
MHz, find (a) Zin, (b) the voltage at the load end of the T-Line, and (c) the voltage at the
sending end of the T-Line.
!
!
Referring to Fig P6.19,
( )
( )
( ) ( )
9
8
2 2 10
59.84 ; tan tan 59.84 0.3 1.254
0.7 0.7 3 10
x rad rad m
c m mx
πω
β β
⎛ ⎞⎛ ⎞= = = = = −⎜ ⎟⎜ ⎟
⎝ ⎠⎝ ⎠
l
( )950 0.016 50 2 2 10 ,LZ j j x Lπ= + Ω = +
50 .LZ ≈ Ω
25 50 1 2, , tan 0
25 50 3 2L
π λ
β π π
λ
−
Γ = = − = = =
+
l
0 25
0
L
in o L
L
ZZ Z Z
Z
+
= = = Ω
+
25 8 2
25 75
j z j z
in o oV V V V e V e
β β+ − − += = = +
+
( )2 j jo LV e eβ β+ −= +Γl l
cos sin 1, 1,j je j eπ ππ π −= + = − = −
( )1 21 1 2; 3
3 3o o o
V V V V+ + +−⎛ ⎞− − − = = = −⎜ ⎟
⎝ ⎠
( ) 11 3 1 2
3L o L
V V V+ ⎛ ⎞= +Γ = − − = −⎜ ⎟
⎝ ⎠ ( )2cos 180Lv t Vω= + �
, 15.7 , 4.71, tan 418.6p
p
radu
u m
ω ω
β β β
β
= = = = =l l
( )
( )( )
100 125 75 418.6
75
75 100 125 418.6
22 28
in
j j
Z
j j
j
+ +
=
+ +
= − Ω
Fig. P6.19
6- !13
!
!
!
!
!
!
P6.20: Suppose the T-Line for Figure 6.10 is characterized by the following distributed
parameters at 100 MHz: R’ = 5.0 Ω/m, L’ = 0.010 µH/m, G’ = 0.010 S/m, and C’ = 0.020
nF/m. If ZL = 50 – j25 Ω,vs = 10cos(ωt)V, Zs = 50Ω, and the line length is 1.0 m, find the
voltage at each end of the T-line.
The following MATLAB routine was used to find the required parameters.
% MLP0620
%
% calculate gamma and char impedance
% given the distributed parameters
% Then, calculate gammaL and Zin
%
% define variables
clc
clear
R=5;
L=.010e-6;
G=.01;
C=.020e-9;
f=100e6;
w=2*pi*f;
length=1;
ZL=50-j*25;
% Perform calcuations
A=R+i*w*L;
366 2.1
75
jin
in
in
ZV e V
Z
−= =
+
�
( )2.1cos 36inv t Vω∴ = − �
430.593 jL oL
L o
Z Z e
Z Z
−
Γ = =
+
�
( ) 126 360.70 2.1j j j jin o L oV V e e e V e Vβ β+ + − − + −= +Γ = =
� �l l
36
90
126
2.1 3
0.70
j
j
o j
eV e V
e
−
+
−
= =
�
�
�
( )
( )
105.81 4.47
4.5cos 106
j
L o L
L
V V e V
v t Vω
+= +Γ =
= +
�
�
6- !14
B=G+i*w*C;
gamma=sqrt(A*B)
Zo=sqrt(A/B)
gammaL=(ZL-Zo)/(ZL+Zo)
TGL=tanh(gamma*length);
Zin=Zo*((ZL+Zo*TGL)/(Zo+ZL*TGL))
Running the program,
gamma = 0.2236 + 0.2810i
Zo = 22.3607
gammaL = 0.4479 - 0.1908i
Zin = 27.2079 -15.4134i
>>
!
!
so !
!
4. The Smith Chart
P6.21: Locate on a Smith Chart the following load impedances terminating a 50 Ω T-
Line. (a) ZL = 200 Ω , (b) ZL = j25 Ω, (c) ZL = 50 + j50 Ω, and (d) ZL = 25 – j200 Ω.
P6.22: Repeat problem P6.14 using the Smith Chart.
First we locate the normalized load, zL = 1 – j0.5 (point a). By inspection of the Smith
Chart, we see that this point corresponds to ! Also, after drawing the
constant Γ circle we can see VSWR = 1.66. Finally, we move from point a, at 0.356λ on
( )18.23.97 , 4.0cos 18.2jinin SS in
in S
ZV V e V v t V
Z Z
ω−= = ∴ = −
+
� �
( ) ( ) ( )3.841.504 0.101 1.507 jin o L o oV V e e V j V eγ γ+ − + += +Γ = + = �l l
18.2
22
3.84
3.97 2.63
1.507
j
j
o j
eV e
e
−
+ −= =
�
�
�
( ) ( )29.61 3.85 , 3.9cos 30jL o L LV V e v t Vω+ −= +Γ = ∴ = −
� �
760.245 .jL e
−Γ =
�
6- !15
the WTG scale, clockwise (towards the generator) a distance 0.125 λ to point b, at 0.481
λ. At this point we see zin = 0.62 – j0.07. Denormalizing we find:
Zin = 31 – j3.5 Ω.
P6.23: A 0.690λ long lossless Zo = 75 Ω T-Line is terminated in a load ZL = 15 + j67 Ω.
Use the Smith Chart to find (a) ΓL, (b) VSWR, (c) Zin and (d) the distance between the
input end of the line and the first voltage maximum from the input end.
Fig. P6.22a
Fig. P6.21
6- !16
After normalizing ZL and locating it on the chart (point a), we see After
drawing the constant Γ circle, we see that VSWR = 9 (point c). We locate the input
impedance by moving from the load (point a at WTG = 0.118λ) clockwise towards the
generator to the input point (point b at WTG = 0.118 λ + 0.690 λ – 0.500 λ = 0.308 λ).
At this point, zin = 0.8 – j2.4, so Zin = 60 – j180 Ω. Finally, the distance from the input
end of the line (point b) to the first voltage maximum (point c) is simply 0.308 λ – 0.250
λ = 0.058 λ. Or, using the WTL scale, it is 0.250 λ – 0.192 λ = 0.058 λ.
P6.24: A 0.269λ
long lossless Zo = 100 Ω T-Line is terminated in a load ZL = 60 + j40 Ω.
Use the Smith Chart to find (a) ΓL, (b) VSWR, (c) Zin and (d) the distance from the load to
the first voltage maximum.
(a) zL = 0.6 + j0.4 located at WTG=0.082λ.
We read off the Smith Chart that this point corresponds to: ! After drawing
the constant Γ circle we notice the VSWR = 2.05 (point c).
Moving from this point a distance 0.269 λ (clockwise, towards generator), we find the
input point (point b at WTG = 0.351 λ). At this point we have zin = 0.96-j0.72, or Zin =
96-j72 Ω. Finally, we move from point a towards the generator at point c to reach the
voltage maximum, a distance 0.168λ.
950.80 .jL eΓ =
�
1210.34 .jL eΓ =
�
Fig. P6.22b Fig. P6.23
6- !17
P6.25: The input impedance for a 100 Ω
lossless T-Line of length 1.162λ is
measured as 12 + j42 Ω. Determine the
load impedance.
We first locate the normalized input
impedance, zin = 0.12 + j0.42, at point a
(WTL=0.436λ). Then we move a distance
1.162 λ towards the load to point b, at WTL
= 0.436 λ + 1.162 λ =1.598 λ; 1.598 λ –
1.500 λ = 0.098 λ. At this point, we read zL
= 0.15-j0.7, or ZL = 15 – j70 Ω.
P6.26: On a 50 Ω lossless T-Line, the VSWR is measured as 3.4. A voltage maximum is
located 0.079λ away from the load. Determine the load.
We can use the given VSWR to draw a constant Γ circle as shown in the figure. Then we
move from Vmax at WTG = 0.250λ to point a at WTG = 0.250 λ - 0.079 λ = 0.171 λ. At
this point we have zL = 1 +j1.3, or ZL = 50 + j65 Ω.
Fig. P6.25
Fig. P6.26
6- !18
VSWR when the load is attached and (c) the
load impedance.
From the locations of minima on the
shorted line we find λ:
!
!
(b) From the voltage maxima and voltage
minimum on the loaded line, we have
!
Using VSWR=2 we draw the constant |Γ|
circle on the Smith Chart. Point a on the
circle represents the 1.9 cm minimum. We
move from this point towards the load at
the 1.25 cm reference location, a move of
!
At this point (point b on the circle) we have
zL = 0.55 – j0.25, and upon denormalizing
we have (c) ZL = 28 – j12 Ω.
P6.28: Figure 6.53 is generated for a 50 Ω
slotted coaxial air line terminated in a short
circuit and then in an unknown load.
Determine (a) the measurement frequency,
(b) the VSWR when the load is attached and
(c) the load impedance.
From the location of the maxima on the
shorted line, we find λ:
!
!
(b) From the load line,
!
( )2 7.55 1.25 12.6cm cm cmλ = − =
( ) 2.4ca f GHz
λ
= =
4 2
2
VSWR = =
1.9 1.25 0.0516
12.6
cm cm
cm
λ
λ
⎛ ⎞−
=⎜ ⎟
⎝ ⎠
( )2 9.3 1.7 15cm cm cmλ = − =
( ) 2.0ca f GHz
λ
= =
10 2.5
4
VSWR = =
Fig. P6.27
Fig. P6.28
6- !19
Using VSWR=2.5 we draw the constant |Γ| circle on the Smith Chart. Point a on the
circle represents the minimum at 7.9 cm. We move from this point towards the load at
the 5.5 cm reference location, a move of
!
At this point (point b on the circle) we have zL = 1 – j0.95, and upon denormalizing we
have (c) ZL = 50 – j48 Ω.
P6.29: Referring to Figure 6.20, suppose we measure Zinsc = +j25 Ω and ZinL = 35 + j85
Ω. What is the actual load impedance? Assume Zo = 50 Ω.
We normalize the short circuit impedance to zinsc = 0+j0.5 and locate this on the Smith
Chart to determine the length of the T-Line is 0.074λ. Then we normalize ZinL to
zinL=0.70+j1.70, locate this on the chart at 0.326λ (WTL scale) and draw a constant |Γ|
circle. We then move towards the load, or to 0.336λ + 0.074 λ = 0.400 λ, and find this
point on the Smith Chart (zL = 0.25+j0.7). Denormalizing, we find ZL = 12+j35 Ω.
P6.30: MATLAB: Modify MATLAB 6.3 to draw the normalized load point and the
constant circle, given Zo and ZL. Demonstrate your program with the values from
Drill 6.11.
Add this to the end of the Matlab 6.3 program:
%now add constant gamma circles
ZL=50;
fudge=0.001+i*0.001;
newZL=ZL+fudge;
Zo=50;
zL=newZL/Zo;
gamma=(zL-1)/(zL+1);
plot(gamma,'-o');
constgamma(zL);
You must change the value of ZL for each load point. Notice the addition of a ‘fudge
factor’. This ensures that gamma has both a nonzero and finite real and imaginary part to
work with in the plot.
You’ll also need to add an additional function:
function [h]=constgamma(zL)
%constgamma(zL) draws the constant gamma circle;
7.9 5.5 0.16
15
cm cm
cm
λ
λ
⎛ ⎞−
=⎜ ⎟
⎝ ⎠
LΓ
6- !20
phi=1:1:360;
theta=phi*pi/180;
a=abs((zL-1)/(zL+1));
Re=a*cos(theta);
Im=a*sin(theta);
z=Re+i*Im;
h=plot(z,'--k');
axis('equal')
axis('off')
The program is run for each point of
Drill 6.11 by changing the ZL value.
Since the MATLAB routine has the
‘hold on’, each new point is added to
the plot.
5. Impedance Matching
P6.31: A matching network, using a reactive element in series with a length d of T-Line,
is to be used to match a 35 – j50 Ω load to a 100 Ω T-Line. Find the through line length
d and the value of the reactive element if (a) a series capacitor is used, and (b) a series
inductor is used.
First we normalize the load and locate it on the Smith Chart (point a, at zL = 0.35-j0.5,
WTG = 0.419λ).
(a) need to move to point b, at z = 1+j1.4, so that a capacitive element of value jx = -j1.4
can be added to provide an impedance match. Moving to this point b gives d = 0.500λ
+0.173 λ -0.419 λ = 0.254 λ. The capacitance is
Fig. P6.30
Fig. P6.31a Fig. P6.31b
6- !21
!
(b) Now we need to move to point c, at z = 1-j1.4, so that an inductive element of value jx
= +j1.4 can be added. Moving to this point c gives d = 0.500 λ + 0.327 λ – 0.419 λ =
0.408 λ. The inductance is
!
P6.32: A matching network consists of a length of T-Line in series with a capacitor.
Determine the length (in wavelengths) required of the T-Line section and the capacitor
value needed (at 1.0 GHz) to match a 10 – j35 Ω load impedance to the 50 Ω line.
We find the normalized load, zL = 0.2 – j0.7, located at point a (WTG = 0.400λ). Now
we move from point a clockwise (towards the generator) until we reach point b, where we
have z = 1 + j2.4. Moving from a to b corresponds to d = 0.500λ+0.194λ-0.400λ =
0.294λ. For the series capacitance we have
!
P6.33: You would like to match a 170 Ω load to a 50 Ω T-Line. (a) Determine the
characteristic impedance required for a
quarter-wave transformer. (b) What
through-line length and stub length are
required for a shorted shunt stub matching
network?
( )( )( )9
1.4,
1 1.14
2 1 10 100 1.4
o
j j
CZ
C pF
x
ω
π
−
= −
= =
( )( )
( )9
1.4 100
1.4, 22.3
2 1 10o
j L j L nH
Z x
ω
π
= = =
( )( )( )9
2.4 ,
1or 1.33
2 1 10 50 2.4
o
jj
CZ
C pF
x
ω
π
−
− =
= =
Fig. P6.32a
Fig. P6.32b
6- !22
(a) !
(b)
(1)Normalize the load (point a, zL = 3.4 + j0).
(2) locate the normalized load admittance: yL (point b)
(3) move from point b to point c, at the y=1+jb circle (d = 0.170λ)
(4) move from the shorted end of the stub (normalized admittance point c) to the point y
= 0 – jb. (l = 0.354 λ – 0.250 λ = 0.104 λ.)
Note in step 3 we could have gone to the
point y = 1-jb. This would have resulted in
d = 0.329 λ and l = 0.396 λ.
P6.34: A load impedance ZL = 200 + j160 Ω is to be matched to a 100 Ω line using a
shorted shunt stub tuner. Find the solution that minimizes the length of the shorted stub.
Refer to Figure P6.33a for the shunt stub
circuit.
(1)Normalize the
load (point a, zL = 2.0 +
j1.6).
(2) locate the normalized load admittance:
yL (point b)
(3) move from point b to point c, at the
y=1+jb circle(0.500λ + 0.170λ -0.458λ =
0.212λ)
(4) move from the shorted end of the stub
(normalized admittance point) to the point
y = 0 – jb. (l = 0.354 λ – 0.250 λ = 0.104
λ.)
P6.35: Repeat P6.34 for an open-ended shunt stub tuner.
92s o LZ Z R= = Ω
Fig. P6.33a
Fig. P6.33b
Fig. P6.34
6- !23
(1)Normalize the load (point a, zL = 2.0 + j1.6).
(2) locate the normalized load admittance: yL (point b)
(3) move from point b to point c, at the y=1-jb circle(0.500λ + 0.330λ -0.458λ = 0.372λ).
We choose this point for c so as to minimize the length of the shunt stub.
(4) move from the open end of the stub (normalized admittance point) to the point y = 0 +
jb. (l = 0.146 λ)
P6.36: A load impedance ZL = 25 + j90 Ω is to be matched to a 50 Ω line using a shorted
shunt stub tuner. Find the solution that minimizes the length of the shorted stub.
Refer to Figure P6.33a for the shunt stub
circuit.
(1)Normalize the load (point a, zL = 0.5 +
j1.8).
(2) locate the normalized load admittance: yL
(point b)
(3) move from point b to point c, at the
y=1+jb circle(0.500λ + 0.198λ -0.423λ =
0.275λ)
(4) move from the shorted end of the stub
(normalized admittance point) to the point y
= 0 – jb. (l = 0.308 λ – 0.250 λ = 0.058 λ.)
Fig. P6.35a
Fig. P6.35b
Fig. P6.36
6- !24
P6.37: Repeat P6.36 for an open-ended shunt stub tuner.
Refer to Figure P6.35a.
(1)Normalize the load (point a, zL = 0.5 +
j1.8).
(2) locate the normalized load admittance: yL
(point b)
(3) move from point b to point c, at the
y=1+jb circle(0.500λ + 0.392λ -0.423λ =
0.379λ)
(4) move from the open end of the stub
(normalized admittance point) to the point y
= 0 + jb. (l = 0.191 λ)
P6.38: (a) Design an open-ended shunt stub matching network to match a load ZL = 70 +
j110 Ω to a 50 Ω impedance T-Line. Choose the solution that minimizes the length of the
through line. (b) Now suppose the load turns out to be ZL = 40 + j100 Ω. Determine the
reflection coefficient seen looking into the matching network.
(a) Refer to Figure P6.35a.
(1)Normalize the load (point a, zL = 1.4 + 2.2).
Fig. P6.37
Fig. P6.38a Fig. P6.38b
6- !25
(2) locate the normalized load admittance: yL (point b)
(3) move from point b to point c, at the y=1+jb circle(0.500λ + 0.185λ -0.448λ = 0.237λ)
(4) move from the open end of the stub (normalized admittance point) to the point y = 0 -
jb. (l = 0.328 λ)
(b)
(1) Normalize the load (point a: zL = 0.8 + j2.0)
(2) locate yL (point b)
(3) Move a distance 0.237λ to point c (0.434 λ + 0.237 λ = 0.671 λ; or WTG = 0.171 λ)
(4) Move from yopen to 0.328 λ (point d)
(5) add admittances of point c and d to get ytot = 0.6 – j0.2.
(6) locate the corresponding ztot (point f) and read the reflection coefficient as:
!
6. Microstrip
P6.39: A 6.00 cm long microstrip transmission line is terminated in a 100. Ω resistive
load. The signal line is 0.692 mm wide atop a 0.500 mm thick polyethylene substrate.
What is the input impedance of this line at 1.0 GHz? What is the maximum frequency at
which this microstrip can operate?
This can be solved using either the Smith Chart or ML0604 in conjunction with the
Zinput function from Matlab 6.2. Using the latter approach we have:
Microstrip Analysis
enter width & thickness in the same units
enter the line width: 0.692e-3
enter the substrate thickness: 0.500e-3
enter substrate rel permittivity: 2.26
eeff = 1.8326
up = 221461941.7986m/s
Zo = 80.2454ohms
To run the Zinput routine, we also need the propagation constant. Assuming lossless line,
we have !
>> Zinput(80.2,100,j*28.4,0.06)
ans = 64.7278 + 3.7906i
>>
340.28 jeΓ =
�
( )9
8
2 1 10 128.4 .
2.215 10p
x
j j j j
u x m
πω
γ β= = = =
6- !26
So we have Zin = 65 + j3.8 Ω
To find fmax, we have
!
P6.40: A 75 Ω impedance microstrip line is to be designed on a 2.0 mm thick Teflon
substrate using copper metallization. What is the maximum operating frequency for this
microstrip? Now determine w, and the physical length of a quarter wave section of line
at 800. MHz.
!
Using ML0605:
Microstrip Design
width & thickness will be in the same units
enter the desired impedance: 75
enter the substrate thickness: 2
enter substrate rel permittivity: 2.1
w = 3.2929
eeff = 1.741
up = 227209857.0703m/s
>>
so w = 3.29mm and εeff = 1.741.
The guide wavelength is:
!
The quarter wave section length is then: !
P6.41: Analysis of a 2.56 cm long microstrip line reveals that it has a 50 Ω characteristic
impedance and an effective relative permittivity of 5.49. It is terminated in a 60 Ω
resistor in series with a 1.42 pF capacitor. Determine the input impedance looking into
this terminated line at 1.60 GHz.
( )
8
max 3
3 10 100
4 4 0.5 10 2.26r
c x m sf GHz
h x mε −
= = =
( )
8
max 3
3 10 26
4 4 2 10 2.1r
c x m sf GHz
h x mε −
= = =
8 63 10 800 10 0.284
1.741
o
G
eff eff
c f x x mλλ
ε ε
= = = =
0.071 7.1
4
G m cmλ= = =l
6- !27
This problem may be solved analytically or with the Smith Chart. For the analytical
solution we have:
! !
!
The load capacitance has an impedance:
!
so the total load impedance is ZL = 60-j70 Ω.
Then, the input impedance is
!
With the Smith Chart, the answer is !
P6.42: A 100 Ω impedance microstrip line is to be designed using copper metallization on
a 0.127 cm thick dielectric of relative permittivity 3.8. Determine (a) w, (b) fmax, and at 2
GHz find (c) up and λG.
From ML0605:
Microstrip Design
width & thickness will be in the same units
enter the desired impedance: 100
enter the substrate thickness: 0.127
enter substrate rel permittivity: 3.8
w = 0.066625
eeff = 2.6865
up = 182909468.597m/s
>>
(a) So we have w = 0.0666 cm = 0.67 mm.
(b)
!
(c) We know up, from the program (up = 1.83x108 m/s), so at 2 GHz
( )8
9
3 10 5.49
0.080 ,
1.6 10
effp xcu m
f f x
ε
λ = = = =
.0256 0.320 .
0.080
m
m
λ
λ
= =l
( )2 0.320 , tan 2.125πβ λ β
λ
= = −l l
70c
jZ j
Cω
−
= = − Ω
( )
( )( )
60 70 50 2.125
50 31.8 48.2
50 60 70 2.125in
j j
Z j
j j
− + −
= = + Ω
+ − −
32 48 inZ j≈ + Ω
( )
8
max 3
3 10 30
4 4 1.27 10 3.8r
c x m sf GHz
h x mε −
= = =
6- !28
!
P6.43: MATLAB: Modify MATLAB 6.4 to calculate attenuation. Try out your program
using the parameters of Drill 6.21 and Drill 6.22.
% M-File: MLP0640
%
% Microstrip Analysis
%
% Given the physical dimensions and er, this
% will calculate eeff, Zo and up for microstrip.
%
% Wentworth, 8/3/02
% modified ML0604 on 9/5 to calculate attenuation
%
% Variables:
% w line width (m)
% h substrate thickness (m)
% t conductor thickness (m)
% sigc conductor conductivity (S/m)
% ur conductor rel permeability
% er substrate relative permittivity
% eeff effective relative permittivity
% up propagation velocity (m/s)
% Zo characteristic impedance (ohms)
% ad dielectric attenuation(dB/m)
% ac conductor attenuation dB/m)
% atot total attenuation (dB/m)
% ds skin depth (m)
% Rs skin effect resistance (ohms/square)
% tand dielectric loss tangent
clc %clears the
command window
clear %clears variables
disp('Microstrip Analysis')
disp(' ')
% Prompt for input values
0.0915 9.15pG
u
m cm
f
λ = = =
6- !29
w=input('enter the line width (m): ');
h=input('enter the substrate thickness(m): ');
er=input('enter substrate rel permittivity: ');
t=input('enter conductor thickness (m): ');
sigc=input('enter conductor conductivity (S/m): ');
ur=input('enter conductor relative permeability: ');
tand=input('enter dielectric loss tangent: ');
f=input('enter frequency (Hz): ');
uo=pi*4e-7;
eo=8.854e-12;
c=2.998e8;
u=ur*uo;
e=er*eo;
% Perform Calculations
eeff=((er+1)/2)+(er-1)/(2*sqrt(1+12*h/w));
up=2.998e8/sqrt(eeff);
if w/h<=1
Zo=(60/sqrt(eeff))*log((8*h/w)+(w/(4*h)));
else if w/h>1
Zo=120*pi/(sqrt(eeff)*((w/h)+1.393+0.667*log((w/h)
+1.444)));
end
end
ds=1/sqrt(pi*f*u*sigc);
Rs=1/(sigc*ds*(1-exp(-t/ds)));
ac=8.686*Rs/(Zo*w);
ad=8.686*2*pi*f*er*(eeff-1)*tand/(c*2*sqrt(eeff)*(er-1));
atot=ac+ad;
% Display results
disp(['eeff = ' num2str(eeff) ])
disp(['up = ' num2str(up) 'm/s'])
disp(['Zo = ' num2str(Zo) 'ohms'])
disp(['ac = ' num2str(ac) 'dB/m'])
disp(['ad = ' num2str(ad) 'dB/m'])
Now we run the program using the information from Drill 6.21 and Drill 6.22. Note that
we’ve changed the dimensions to metric units.
6- !30
Microstrip Analysis
enter the line width (m): 9.8e-4
enter the substrate thickness(m): 1.016e-3
enter substrate rel permittivity: 9.9
enter conductor thickness (m): 6e-6
enter conductor conductivity (S/m): 5.8e7
enter conductor relative permeability: 1
enter dielectric loss tangent: 0.0001
enter frequency (Hz): 1e9
eeff = 6.6638
up = 116137011.5308m/s
Zo = 49.8369ohms
ac = 1.5554dB/m
ad = 0.022214dB/m
>>
P6.44: A 50 Ω impedance microstrip line is desired for operation at 2.4 GHz. It is to be
built on a 20 mil thick mica substrate using a 10 µm thick copper conductor. Calculate
(a) w, (b) αc, (c) αd, and (d) αtot at this frequency.
Mica has εr = 5.4 and tanδ = 0.0003. Using ML0605:
Microstrip Design
width & thickness will be in the same units
enter the desired impedance: 50
enter the substrate thickness: 20
enter substrate rel permittivity: 5.4
w = 32.6859
eeff = 3.9617
up = 150623255.8531m/s
>>
To use this information in ML0640, we need to convert h and w to metric units. We find
h = 5.08x10-4m and w = 8.3x10-4m.
Microstrip Analysis
enter the line width (m): 8.3e-4
enter the substrate thickness(m): 5.08e-4
6- !31
enter substrate rel permittivity: 5.4
enter conductor thickness (m): 10e-6
enter conductor conductivity (S/m): 5.8e7
enter conductor relative permeability: 1
enter dielectric loss tangent: 0.0003
enter frequency (Hz): 2.4e9
eeff = 3.9616
up = 150624957.4337m/s
Zo = 50.1512ohms
ac = 2.6687dB/m
ad = 0.11967dB/m
>>
With more appropriate significant digits we find:
w = 33 mils
αc = 2.67 dB/m
αd = 0.12 dB/m
αtot = 2.79 dB/m
P6.45: One type of board routinely used to build microwave circuits is 50 mils thick
Rogers Corporation RT/Duroid©, with εr = 10.8 and tanδ = 0.0028. It is coated on both
sides by “1/4 oz copper”. This translates to a 0.35 mil thickness of copper. Find w and up
for a 50 Ω line. Then determine the αc, αd and αtot at three frequencies: 1, 10 and 20
GHz. What is the maximum frequency of operation for this microstrip?
Using ML0605:
Microstrip Design
width & thickness will be in the same units
enter the desired impedance: 50
enter the substrate thickness: 50
enter substrate rel permittivity: 10.8
w = 44.3241
eeff = 7.1852
up = 111844030.4942m/s
>>
Now convert w and h and t to metric (1.125x10-3m, 1.27x10-3m, and 8.89x10-6m,
respectively), and run MLP0640 at each frequency. For instance, at 1 GHz we have:
Microstrip Analysis
6- !32
enter the line width (m): 1.125e-3
enter the substrate thickness(m): 1.27e-3
enter substrate rel permittivity: 10.8
enter conductor thickness (m): 8.89e-6
enter conductor conductivity (S/m): 5.8e7
enter conductor relative permeability: 1
enter dielectric loss tangent: 0.0028
enter frequency (Hz): 1e9
eeff = 7.1847
up = 111847474.32m/s
Zo = 49.8031ohms
ac = 1.2975dB/m
ad = 0.64805dB/m
>>
Tabulating the results for each frequency:
The maximum frequency is
!
P6.46: A 1.5 inch length of microstrip line of width 48.86 mils sits atop a 50 mil thick
substrate with dielectric constant 4. Determine the impedance looking into this circuit at
2 GHz if it is terminated in a 300 Ω resistor. Assume ideal conductors and lossless
dielectric.
From ML0604 we find:
Microstrip Analysis
enter width & thickness in the same units
enter the line width: 48.86
enter the substrate thickness: 50
enter substrate rel permittivity: 4
α in dB/m 1 GHz 10 GHz 20 GHz
αc= 1.3 4.0 5.7
αd= 0.65 6.5 13.0
αtot= 2.0 10.5 18.7
( )
8
max 6
3 10 1 18
25.2 104 4 50 10.8r
c x m s milf GHz
x mh milsε −
= = =
6- !33
eeff = 2.9116
up = 175697087.6994m/s
Zo = 74.9641ohms
>>
So we have Zo = 75 Ω. Also, we find
!
We can also calculate β: !
Now either a Smith Chart or the Zinput equation may be used to evaluate Zin. From the
Zinput function we defined earlier in MATLAB,
>> Zo=75;
>> ZL=300;
>> G=j*71.5;
>> L=0.0381;
>> Zinput(Zo,ZL,G,L)
ans = 8.6567e+001 +1.2031e+002i
>>
So Zin = 87 + j120 Ω.
A Smith Chart approach is more approximate, yielding Zin 90 + j120 Ω.
P6.47: The top-down view of a microstrip circuit is shown in Figure 6.54. If the
microstrip is supported by a 40 mil thick alumina substrate, (a) determine the line width
required to achieve a 50 Ω impedance line. (b) What is the guide wavelength on this
microstrip line at 2 GHz? (c) Suppose at this frequency the load impedance is ZL = 150 -
j100 Ω. Determine the length of the stubs (dthru and lstub) required to impedance match the
load to the line.
We first apply ML0605:
Microstrip Design
width & thickness will be in the same units
enter the desired impedance: 50
enter the substrate thickness: 40
( ) 0.02541.5 0.0381 , 0.08785 ,
0.0381so 0.434
0.08785
p
G
G
G
umin m m
in f
m
m
λ
λ
λ
⎛ ⎞= = = =⎜ ⎟
⎝ ⎠
= =
l
l
171.5 .
pu m
ω
β = =
≈
6- !34
enter substrate rel permittivity: 9.9
w = 38.6273
eeff = 6.6644
up = 116131354.2486m/s
>>
From this we have (a) w = 38.6 mils.
Also,
(b) !
(c) Now we use a Smith Chart to determine
the open-ended shunt stub matching network.
(1) Normalize the load (point a: zL = 3.0 -
j2.0)
(2) locate yL (point b)
(3) Move to point c (0.180λ - 0.025λ = 0.155λ; or dthru = 0.155λ = 9 mm (354 mils))
(4) Move from yopen to 0.336λ, so lstub = 0.336λ = 19.5 mm (768 mils)
P6.48: Suppose the microstrip circuit shown in Figure 6.54 is realized atop the RT/
Duroid© board of problem P6.45. Assuming the board material is lossless, (a) determine
the line width required to achieve a 75 Ω impedance line. (b) Now suppose at 1.0 GHz
the load impedance is ZL = 150 + j150 Ω. Find the length of the stubs (dthru and lstub)
required to impedance match the load to the line.
Applying ML0605:
Microstrip Design
width & thickness will be in the same units
enter the desired impedance: 75
enter the substrate thickness: 50
enter substrate rel permittivity: 10.8
w = 15.7817
eeff = 6.6844
up = 115957584.2884m/s
>>
So we have (a) w = 15.8 mils.
We also have: !
(b) Now we employ a Smith Chart to
determine the open-ended shunt stub
8
9
1.161
10 0.058 2.29
2 10
p
G
u x m in
f x
λ = = = =
0.116pG
u
m
f
λ = =
Fig. P6.47
Fig. P6.48
6- !35
matching network.
(1) Normalize the load (point a: zL = 2.0 + j2.0)
(2) locate yL (point b)
(3) Move to point c (0.178λ + 0.500λ - 0.459λ = 0.219λ; or dthru = 0.219λ = 2.54cm (1
in))
(4) Move from yopen to 0.339λ, so lstub = 0.339λ = 3.93cm (1.55 in)
7. Transients
P6.49: Consider Figure 6.33 with the following values: Vs = 10 V, Zs = 30 Ω, Zo = 50 Ω,
up = 0.666c, ZL = 150 Ω, l = 10 cm. Plot, out to 2 ns, (a) the voltage at the source end, (b)
the voltage at the middle, and (c) the voltage at the load end of the T-Line.
!
!
!
P6.50: Repeat problem P6.49 for a 10 V pulse of duration 0.4 ns.
P6.51: MATLAB: Consider a 12 cm long 50 Ω transmission line terminated in a 25 Ω
load and having a matched source impedance (Zs = 50 Ω). Propagation velocity on the T-
8
0.1 0.5
0.666 3 10
mt ns
mx
s
= =
⎛ ⎞
⎜ ⎟
⎝ ⎠
l
150 50 1 30 50 1;
150 50 2 30 50 4L S
− −
Γ = = Γ = = −
+ +
5010 6.25
50 30o
V V= =
+
Fig. P6.49b
6- !36
Line is 0.67c. The source is a 0.4 ns square pulse of amplitude 6 V. Modify MATLAB
Fig. P6.50a
Fig. P6.50b
6- !37
6.6 to plot v(z,t) at two points: z = 2 cm and z = 10 cm.
% M-File: ML0651
% modifies ML0606
% replaces triangular pulse with rectangular pulse
%
% ML0606:
% Analysis of a triangular pulse
% (matched source impedance) traveling
% down a T-Line and reflecting off a
% resistive load. We want to be able to
% trace the voltage at an arbitrary
% point along the line.
%
% Wentworth, 4/25/03
% Variables
% Vo pulse height (V)
% t1 pulse start (ns)
% t2 pulse end (ns)
% L line length (cm)
% T transit time (ns)
% z location to find pulse (cm)
% tau time "location" to find pulse (ns)
% up propagation velocity (m/s)
% Zo,ZL line,load impedance (ohms)
% N number of points
% GL load reflection coefficient
clc
clear
%enter variables
Vo=6;
t1=0;
t2=.4;
L=12;
z=2;
up=2e8;
Zo=50;
ZL=25;
T=1e9*(L/up)/100;
tau=1e9*(z/up)/100;
N=500;
GL=(ZL-Zo)/(ZL+Zo);
%initialize array
for i=1:N+1
v(i)=0;
end
dt=2*T/N;
6- !38
%enter rectangular pulse function
t=0:dt:2*T;
vo=0.5*Vo*(step(t,t1)-step(t,t2));
%Note that “step” is a function defined
%in ML0606.
%Generate + wave data
for i=1:N+1
ta=i*dt;
if ta>tau
j=ceil((ta-tau)/dt);
vplus(i)=vo(j);
end
end
%Generate - wave data
for i=1:N+1
ta=i*dt;
tb=2*T-tau;
if ta>tb
j=ceil((ta-tb)/dt);
vmin(i)=GL*vo(j);
end
end
%Sum the data
for i=1:N+1
v(i)=vplus(i)+vmin(i);
end
plot(t,v)
xlabel('time (ns)')
ylabel('voltage')
AXIS([0 2*T -Vo Vo])
grid on
P6.52: MATLAB: Modify MATLAB 6.6 to plot v(z,t) at z = 4.5 cm if the source pulse is
as indicated in Figure 6.55.
% M-File: MLP0652
% Analysis of a triangular pulse
% (matched source impedance) traveling
% down a T-Line and reflecting off a
% resistive load. We want to be able to
% trace the voltage at an arbitrary
% point along the line.
%
% Wentworth, 4/25/03
% Variables
Fig. P6.51 (plot at z = 2 cm)
6- !39
% Vo pulse height (V)
% t1 pulse start (ns)
% t2 pulse middle (ns)
% t3 pulse end (ns)
% L line length (cm)
% T transit time (ns)
% z location to find pulse (cm)
% tau time "location" to find pulse (ns)
% up propagation velocity (m/s)
% Zo,ZL line,load impedance (ohms)
% N number of points
% GL load reflection coefficient
clc
clear
%enter variables
Vo=10;
t1=0;
t2=1;
t3=2;
L=6;
z=4.5;
up=3e7;
Zo=50;
ZL=0;
T=1e9*(L/up)/100;
tau=1e9*(z/up)/100;
N=500;
GL=(ZL-Zo)/(ZL+Zo);
%initialize array
for i=1:N+1
v(i)=0;
end
dt=2*T/N;
%enter triangular pulse function
m1=0.5*Vo/(t2-t1);
b1=0.5*Vo-m1*t2;
m2=0.5*Vo/(t2-t3);
b2=0.5*Vo-m2*t2;
for i=1:N+1
t(i)=i*dt;
if t(i)<t1
vo(i)=0;
end
if and(t(i)>t1,t(i)<=t2)
vo(i)=m1*t(i)+b1;
end
if and(t(i)>t2,t(i)<=t3)
vo(i)=m2*t(i)+b2;
end
if t(i)>t3
6- !40
vo(i)=0;
end
end
%Generate + wave data
for i=1:N+1
ta=i*dt;
if ta>tau
j=ceil((ta-tau)/dt);
vplus(i)=vo(j);
end
end
%Generate - wave data
for i=1:N+1
ta=i*dt;
tb=2*T-tau;
if ta>tb
j=ceil((ta-tb)/dt);
vmin(i)=GL*vo(j);
end
end
%Sum the data
for i=1:N+1
v(i)=vplus(i)+vmin(i);
end
plot(t,v)
xlabel('time (ns)')
ylabel('voltage')
AXIS([0 2*T -Vo Vo])
grid on
P6.53: The expression for iL(t) and vL(t) of equations (6.106) and (6.107) were derived for
a T-Line terminated in an inductor. Find similar expressions for a T-Line terminated in a
capacitor.
!
Suppressing U(τ), we have
!
!
This is in the form:
( ) ( )o
( )( ) , ( ) ( ), Z ( ) ( )i r i rLL L o o L o o
dv ti t C v t V V U i t V V U
dt
τ τ= = + = −
( )2 ( ) ( ) ( )i Lo L o L L o
dv tV v t Z i t v t Z C
dt
= + = +
2 2( ) ( )1 1( ) , or ( ) .
i i
o oL L
L L
o o o o
V Vdv t dv tv t v t
dt Z C Z C dt Z C Z C
−
+ = = +
Fig P6.52
6- !41
!
solving the integral
!
we have
!
Now since at t = 0, v = 0, we have !
!
Solving for v: !
!
Now for iL(t),
!
!
P6.54: For Figure 6.42, Zo = 100 Ω and up = 0.1c. Estimate L if the VL vs t is given in
Figure 6.56.
Using (6.107), ! here we have τ = t – 2 ns. Choosing the voltage
at 2 ns we have ! Then, at approximately 2.2 ns we have 0.2
V, so
!
Solving for L:
!
21, where , and
i
o
o o
Vdv Av B A B
dt Z C Z C
−
= + = =
dv dt
Av B
=
+∫ ∫
( )1 ln .Av B t C
A
+ = +
1 ln( ), and C B
A
=
1 ln .Av Bt
A B
+⎛ ⎞= ⎜ ⎟
⎝ ⎠
( )1 2 1o
t
Z CAt i
o
Bv e V e
A
−⎛ ⎞= − = − −⎜ ⎟
⎝ ⎠
( ) 2 1 ( ).o
t
Z Ci
L ov t V e U τ
−⎛ ⎞∴ = −⎜ ⎟
⎝ ⎠
2( ) 1( ) 2 o o
it t
Z C Z Ci oL
L o
o o
Vdv ti t C C V e e
dt Z C Z
− −⎛ ⎞
= = =⎜ ⎟
⎝ ⎠
2( ) ( ).o
i t
Z Co
L
o
Vi t e U
Z
τ
−
∴ =
( ) 2 ( ),
oZ
Li
L ov t V e U
τ
τ
−⎛ ⎞
⎜ ⎟
⎝ ⎠=
0.9 2 , or 0.45 .i io oV V V V= =
( )100 0.2
200.20.2 2 , or .
0.45
ns
Li ns L
oV V e e
−⎛ ⎞
⎜ ⎟ −⎝ ⎠= =
( )
20 25 .
ln 0.2 0.45
nsL nH−= =
6- !42
P6.55: A 50 Ω T-Line with up = 0.5c is terminated in some load such that the TDR is
given by Figure 6.57. Determine the location and the value of the load.
From the figure we see the two-way travel time is 7 ns, so therefore one-way is 3.5 ns and
we have
!
Also from the figure we can calculate Γ:
!
and then
!
So we have a 12.5 Ω terminating resistor located 0.525 m along the line.
P6.56: The TDR plot for a 75 Ω T-Line with up = 0.2c is given in Figure 6.58. What type
components terminate the line? Estimate the component values.
By inspection we see it is a series combination of resistance and inductance. The location
is:
!
The resistance is found by determining the reflection coefficient:
!
and then the resistance is: !
Only a very rough estimate can be found for the inductance. At τ = 10 ns – 10 ns = 0,. we
have Then, at (very roughly) τ = 11 ns – 10 ns = 1ns,, vL(t) has dropped to
1.75 V,
so
!
So our rough estimate is an inductance between 500 and 600 nH.
( )( ) 83.5 0.5 3 10 0.525 .ml ns x m
s
⎛ ⎞= =⎜ ⎟
⎝ ⎠
0.4 1 0.6
1
totV V
V
+
+
− −
Γ = = = −
1 0.650 12.5 .
1 0.6
R −= = Ω
+
( )( )( )9 81 10 10 sec 0.2 3 10 sec 0.302l x x m m
−= =
1.75 1 0.75,
1
−
Γ = =
1.7550 350 .
0.25
R = Ω = Ω
2 2 .ioV V=
( )
( )
75 1 751.75 2 , or 560 .
ln 1.75 2
ns L nse L nH− −= = =
6- !43
8. Dispersion
P6.57: MATLAB: Use Fourier Series to construct a 5 V pulse of duration 5 ns that
repeats every 10 ns.
% M-File: MLP0657
%
% This program assembles a pulse using
% Fourier series. It modifies ML0607.
%
% Wentworth, 8/3/02
%
% Variables:
% N number of Fourier coefficients
% aO avg value of the waveform (volts)
% T period (s)
% fo fundamental frequency (Hz)
% wo fund angular freq (rad/s)
% t time (sec)
% ftot fourier sum at a particular time(volts)
clc %clears the command window
clear %clears variables%MstripDesign
% Initialize variables
clear
N=1000;
a0=2.5;
T=10e-9;
fo=1/T;
wo=2*pi*fo;
% Evaluate Fourier Series Coefficients
for n=1:N
a(n)=(10/(pi*n))*sin(n*pi/2);
end
% Generate data and plot
for i=1:180
t(i)=i*T/90;
for n=1:N
f(n)=a(n)*cos(n*wo*t(i));
end
ftot(i)=a0+sum(f);
end
plot(t,ftot)
6- !44
xlabel('time(s)')
ylabel('volts')
grid on
P6.58: MATLAB: Actual pulses have some slope to the leading and trailing edge.
Suppose a symmetrical pulse is 5 V from –2 ns to +2ns, and has a linear slope to 0 V on
each edge of duration 0.2 ns. The pulse repeats every 20 ns. Construct this pulse using
Fourier Series for N = 10, 100 and 1000. Comment on how this pulse compares to the
one of Figure 6.48.
We must first find the Fourier coefficients:
ao = 1.05
!
% M-File: MLP0658
%
% This program assembles a pulse using
% Fourier series. It modifies ML0607.
%
% Wentworth, 8/3/02
%
( ) ( )( )
( )
( ) ( )( ) ( ) ( )( )2
10 50sin 2.2sin 0.22 2.0sin 0.20
5
500 110cos 0.22 cos 0.20 sin 0.22 sin 0.20
n
na n n
n n
n n n n
nn
π
π π
π π
π π π π
ππ
−⎛ ⎞= + −⎜ ⎟
⎝ ⎠
− − + −
Fig. P6.57
6- !45
% Variables:
% N number of Fourier coefficients
% aO avg value of the waveform (volts)
% T period (ns)
% fo fundamental frequency (Hz)
% wo fund angular freq (rad/s)
% t time (ns)
% ftot fourier sum at a particular time(volts)
clc %clears the command window
clear %clears variables%MstripDesign
% Initialize variables
clear
N=1000;
a0=1.05;
T=20;
fo=1/T;
wo=2*pi*fo;
% Evaluate Fourier Series Coefficients
for n=1:N
a1(n)=(10/(pi*n))*sin(n*pi/5);
a2a(n)=(-50/(pi*n))*(2.2*sin(.22*pi*n)-2.0*sin(.
20*pi*n));
a2b(n)=(-500/((pi*n)^2))*(cos(.22*pi*n)-cos(.20*pi*n));
a3(n)=(110/(pi*n))*(sin(.22*pi*n)-sin(.2*pi*n));
a(n)=a1(n)+a2a(n)+a2b(n)+a3(n);
end
% Generate data and plot
for i=1:180
t(i)=i*T/90;
for n=1:N
f(n)=a(n)*cos(n*wo*t(i));
end
ftot(i)=a0+sum(f);
end
plot(t,ftot)
xlabel('time(ns)')
ylabel('volts')
grid on
save 'Fourier1000' t ftot -ascii
6- !46
To see the difference between the N = 100 and N = 1000 cases requires looking at the
plot over a reduced portion of time, as shown in Figure P6.59b.
P6.59: MATLAB: A material has a constant εr = 4 from DC up to 20 GHz. Then
Fig. P6.58a (the N=100 and N=100 cases are indistinguishable)
Fig. P6.58b
6- !47
!
for 20 GHz < f < 50 GHz. Show the pulse from problem P6.58 after it has traveled along
a coaxial T-Line with this dielectric.
% M-File: MLP0659
%
% This program modifies ML0607.
% Wentworth, 2/2/03
clc %clears the command window
clear %clears variables%MstripDesign
% Initialize variables
clear
N=1000;
a0=1.2;
T=20e-9;
fo=1/T;
wo=2*pi*fo;
z=10;
%evaluate Fourier Series Coefficients
for n=1:N
a(n)=(12/(pi*n))*sin(n*pi/5);
end
%Generate data
for i=1:180
t(i)=i*T/90;
for n=1:N
f(n)=n*50e6;
er(n)=4;
if f(n)>20e9
er(n)=4*cos((f(n)-20e9)/60e9);
end
beta(n)=2*pi*f(n)*sqrt(er(n))/3e8;
V(n)=a(n)*cos(n*wo*t(i)-beta(n)*z);
end
Vtot(i)=a0+sum(V);
end
plot(t,Vtot)
xlabel('time(ns)')
9
9
20 104cos ,
60 10r
f x
x
ε
⎛ ⎞−
= ⎜ ⎟
⎝ ⎠
6- !48
ylabel('volts')
grid on
save 'dispoff' t Vtot -ascii
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