Wentworth solutions chapter 3

Prévia do material em texto

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 
1.0A 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
I
d
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 xy
xy 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
I
for 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 sin1
sin 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
2cos
r
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 I
d 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 Wb
q
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 Wb
q 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 Wb
q 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 Wb
q x C x pN
s m
π− −⎛ ⎞= × = − ×− =⎜ ⎟
⎝ ⎠
F a a au b
0
7
12 2 2 1 2 240 10x y
L
Wb
I 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
N
x
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
Wb
A 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 I
B 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
Wb
mA
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 mWb
B
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 Wb
m
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 − ⎛ ⎞= = =⎜ ⎟
⎝ ⎠