Soluções de Problemas de Magnetostática em Meios Materiais

Prévia do material em texto

Solucionário Notaros -
capitulo 05
Engenharia Elétrica
Universidade Federal do Rio Grande do Norte (UFRN)
17 pag.
Document shared on www.docsity.com
https://www.docsity.com/?utm_source=docsity&utm_medium=document&utm_campaign=watermark
P5 SOLUTIONS TO PROBLEMS
MAGNETOSTATIC FIELD IN MATERIAL
MEDIA
Section 5.3 Magnetization Volume and Surface
Current Densities
PROBLEM 5.1 Nonuniformly magnetized parallelepiped. (a) Combining
Eqs.(5.28) and (4.81), the volume magnetization current density vector inside the
parallelepiped is obtained to be
Jm = ∇× M = −
∂My
∂z
x̂ − ∂Mz
∂x
ŷ − ∂Mx
∂y
ẑ
= −πM0
(
1
c
cos
πz
c
x̂ +
1
a
cos
πx
a
ŷ +
1
b
cos
πy
b
ẑ
)
. (P5.1)
(b) Eq.(5.32) tells us that the surface magnetization current density vector over
the back (x = 0) and front (x = a) sides of the parallelepiped (assuming that its
position in space is as in Fig.5.7) are
Jms1 = M × n̂ = M(0+, y, z) ×(− x̂) = M0 sin
πz
c
ŷ ×(− x̂) = M0 sin
πz
c
ẑ ,
and Jms2 = M(a
−, y, z) × x̂ = M0 sin
πz
c
ŷ × x̂ = −M0 sin
πz
c
ẑ , (P5.2)
respectively. On the left- (y = 0) and right-hand (y = b) sides of the parallelepiped,
Jms3 = M(x, 0
+, z) ×(− ŷ) = M0 sin
πx
a
ẑ×(− ŷ) = M0 sin
πx
a
x̂ ,
and Jms4 = M(x, b
−, z) × ŷ = M0 sin
πx
a
ẑ × ŷ = −M0 sin
πx
a
x̂ , (P5.3)
and on the bottom (z = 0) and top (z = c) sides,
Jms5 = M(x, y, 0
+) ×(− ẑ) = M0 sin
πy
b
x̂×(− ẑ) = M0 sin
πy
b
ŷ ,
and Jms6 = M(x, y, c
−) × ẑ = M0 sin
πy
b
x̂ × ẑ = −M0 sin
πy
b
ŷ . (P5.4)
PROBLEM 5.2 Hollow cylindrical bar magnet. (a) According to Eq.(5.29),
there is no magnetization volume current inside the magnet, Jm = 0. In the cylin-
drical coordinate system adopted such that its z-axis is along the magnet axis,
Eq.(5.32) gives the following magnetization surface current densities on the inner
(r = a) and outer (r = b) lateral (cylindrical) sides of the magnet, similarly to
Eq.(5.36) and Fig.5.8:
Jmsa = M ẑ × (− r̂) = −Mφ̂ and Jmsb = M ẑ × r̂ = Mφ̂ , (P5.5)
140
© 2011 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be 
obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, 
mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, 
Pearson Education, Inc., Upper Saddle River, NJ 07458.
Document shared on www.docsity.com
https://www.docsity.com/?utm_source=docsity&utm_medium=document&utm_campaign=watermark
P5. Solutions to Problems: Magnetostatic Field in Material Media 141
whereas Jms = 0 on both the upper and the lower bases of the bar.
(b) Similarly to the analysis in Example 5.3, the magnetic flux density vector along
the axis of the magnet can be found as B due to two l long cylindrical current sheets
of radii a and b in a vacuum with the surface current densities in Eqs.(P5.5), or due
to two corresponding solenoids in Fig.4.10. So, using Eq.(4.36), twice, with NI/l
substituted, based on Eq.(4.30), by Jmsa = −M and Jmsb = M , respectively, we
obtain
B = Ba + Bb =
µ0Jmsa
2
(sin θ2 − sin θ1) ẑ +
µ0Jmsb
2
(sin θ′2 − sin θ′1) ẑ
=
µ0M
2
(− sin θ2 + sin θ1 + sin θ′2 − sin θ′1) ẑ , (P5.6)
where the angles θ1 and θ2 are defined exactly as those in Fig.4.10(b), while θ
′
1 and
θ′2 are defined in the same way but for the solenoid radius b.
PROBLEM 5.3 Uniformly magnetized square ferromagnetic plate. This
is similar to the analysis in Example 5.2 (Fig.5.8), the only difference being the
shape of the equivalent current loop; namely, the ferromagnetic plate in Fig.5.36 is
replaced by a square loop, with edge length a and the current intensity [Eq.(5.37)]
Im = Jmsd = M0d , (P5.7)
assumed to be in a vacuum. The magnetic flux density vector at the center of a
square current loop is computed in Example 4.3, whereas B at an arbitrary point
along the axis (z-axis) of a rectangular loop, and thus along the z-axis in Fig.5.36,
is found in Problem 4.1, and the result for a = b (square loop) and I = Im = M0d
is
B =
2
√
2µ0M0a
2d
π(4z2 + a2)
√
2z2 + a2
ẑ . (P5.8)
PROBLEM 5.4 Magnetization parallel to plate faces. Now we have two
magnetization current sheets of square shapes and densities
Jms1 = M× n̂ = M0 ŷ× ẑ = M0 x̂ and Jms2 = M0 ŷ× (− ẑ) = −M0 x̂ , (P5.9)
on the upper and lower faces, respectively, of the ferromagnetic plate in Fig.5.36.
We neglect end effects, and thus disregard magnetization surface currents on the
two lateral sides (thin strips) of the plate perpendicular to the x-axis and, moreover,
find the magnetic flux density vector (B) at the plate center (point O) as that due
to two infinite planar sheets with surface currents of the same density (Js) running
in opposite directions. To this end, we can employ Eq.(4.47) for a single infinite
planar current sheet and the superposition principle or use the result of Problem
4.14, part (b), to obtain
B = B1 + B2 = 2B1 = µ0Js ŷ = µ0Jms1 ŷ = µ0M0 ŷ = µ0M . (P5.10)
© 2011 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be 
obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, 
mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, 
Pearson Education, Inc., Upper Saddle River, NJ 07458.
Document shared on www.docsity.com
https://www.docsity.com/?utm_source=docsity&utm_medium=document&utm_campaign=watermark
142 Branislav M. Notaroš: Electromagnetics (Pearson Prentice Hall)
PROBLEM 5.5 Nonuniformly magnetized ferromagnetic disk. (a) Using
Eq.(5.28) in conjunction with the formula for the curl in cylindrical coordinates,
Eq.(4.84), the volume magnetization current density vector inside the ferromagnetic
disk in Fig.5.37 is given by
Jm = ∇× M = −
∂Mz(r)
∂r
φ̂ = −2M0r
a2
φ̂ , (P5.11)
and there is also a magnetization surface current flowing circumferentially along the
side disk surface (Fig.P5.1), with density [Eq.(5.32)]
Jms = M(a
−) ẑ × r̂ = M0 φ̂ . (P5.12)
dr
a
rO
v
Jms
Jm
d
dB
z
Rz
P
Figure P5.1 Evaluation of the magnetic field due to the nonuniformly magnetized
thin ferromagnetic disk in Fig.5.37.
(b) To evaluate its magnetic field, we subdivide the disk in Fig.5.37 into elementary
hollow disks of radii r (0 ≤ r ≤ a), width dr, and thickness (height) d, as shown
in Fig.P5.1. As d ≪ a, each such disk can be replaced by an equivalent circular
current loop (wire) of radius r and current intensity
dIm(r) = Jm(r) d dr (P5.13)
(cross section of the hollow disk through which the current of density Jm flows is a
small rectangle of side lengths d and dr, and surface area d dr). From Eq.(4.19), the
magnetic flux density vector of this loop at an arbitrary point at the z-axis (defined
by a coordinate z) in Fig.5.1 is
dB =
µ0 dIm(r) r
2
2R3
ẑ = −µ0M0d r
3 dr
a2R3
ẑ , R =
√
r2 + z2 , (P5.14)
and the resultant field B is found by the superposition of dB to sum the con-
tributions of all equivalent loops over the volume of the thin disk (in Fig.P5.1),
with the integral in r being solved by modifying the integration in Eqs.(1.61)-(1.63)
(substitution r dr = R dR is used),
B = −µ0M0d
a2
∫ a
r=0
r2
dR
R2
ẑ = −µ0M0d
a2
∫ a
r=0
(R2−z2) dR
R2
ẑ = −µ0M0d
a2
(∫ a
r=0
dR
© 2011 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be 
obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, 
mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, 
PearsonEducation, Inc., Upper Saddle River, NJ 07458.
Document shared on www.docsity.com
https://www.docsity.com/?utm_source=docsity&utm_medium=document&utm_campaign=watermark
P5. Solutions to Problems: Magnetostatic Field in Material Media 143
−z2
∫ a
r=0
dR
R2
)
ẑ = −µ0M0d
a2
(
√
a2 + z2 − |z|+ z
2
√
a2 + z2
− z
2
|z|
)
ẑ . (P5.15)
Finally, we add the B field due to the equivalent circular loop with radius a and
current Im(a) = Jmsd = M0d, representing the surface current in Eq.(P5.12), and
this field is given in Eq.(5.38), where we replace M by M0, so that the total B
amounts to
Btot = −
µ0M0d
a2
[
√
a2 + z2 − |z| + z
2
√
a2 + z2
− z
2
|z| −
a4
2 (z2 + a2)
3/2
]
ẑ . (P5.16)
PROBLEM 5.6 Infinite cylinder with circular magnetization. (a) By
means of Eqs.(5.28) and (4.84), the volume magnetization current density vector in
the ferromagnetic cylinder comes out to be
Jm = ∇× M =
1
r
∂
∂r
[rMφ(r)] ẑ =
2M0
a
ẑ . (P5.17)
Note that this distribution of the magnetization current, axial and uniform through-
out the cylinder (Jm = Jm ẑ and Jm = const), has the same form as the distribution
of the conduction current in the cylindrical copper conductor in Fig.4.15.
(b) The surface magnetization current density vector, Eq.(5.32), on the cylinder
surface is
Jms = Mφ(a
−) φ̂ × r̂ = −M0 ẑ . (P5.18)
(c) Because of symmetry, the B field in the cylinder (due to its magnetization
currents assumed to reside in a vacuum) is circular (magnetic-field lines are circles
centered at the cylinder axis), and of the form in Eq.(4.53). To find this field, we
apply Ampère’s law, Eq.(4.48), as if the magnetization currents found in (a) and
(b) were conduction currents in a nonmagnetic medium, to the circular contour C
of radius r, as in Fig.4.15(a),
B 2πr = µ0Jmπr
2 −→ B = µ0Jmr
2
φ̂ =
µ0M0r
a
φ̂ = µ0M (0 ≤ r ≤ a) .
(P5.19)
(d) For the observation point outside the cylinder, the right-hand side of Ampère’s
law includes the surface magnetization current of density Jms, Eq.(P5.18), as well,
and this current amounts, using Eq.(3.13), to Jms times the circumference of the
cylinder. Hence, B outside the ferromagnetic cylinder is computed as follows:
B 2πr = µ0
(
Jmπa
2 + Jms 2πa
)
−→ B = µ0a
r
(
Jma
2
+ Jms
)
=
µ0a
r
(M0 − M0) = 0 (a < r < ∞) . (P5.20)
© 2011 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be 
obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, 
mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, 
Pearson Education, Inc., Upper Saddle River, NJ 07458.
Document shared on www.docsity.com
https://www.docsity.com/?utm_source=docsity&amp;utm_medium=document&amp;utm_campaign=watermark
144 Branislav M. Notaroš: Electromagnetics (Pearson Prentice Hall)
Section 5.4 Generalized Ampère’s Law
PROBLEM 5.7 Magnetic field intensity vector. (a) The magnetic flux
density vector, B, along the axis (z-axis) of the magnetized disk in Fig.5.8 is given
in Eq.(5.38), from which the magnetic field intensity vector, H, along the axis is
found using Eq.(5.50) for points in the disk and Eq.(5.61) for the part of the z-axis
in air, as follows:
H =
B
µ0
− M = B
µ0
− M ẑ = M
[
d a2
2 (z2 + a2)
3/2
− 1
]
ẑ (|z| ≤ d/2)
and H =
B
µ0
=
Mda2
2 (z2 + a2)3/2
ẑ (|z| > d/2) . (P5.21)
(b) By means of Eqs.(5.50) and (5.39), the magnetic field intensity vector at the
center of the cylindrical bar magnet (from Example 5.3) amounts to
H =
B
µ0
− M =
(
l√
l2 + 4a2
− 1
)
M . (P5.22)
(c) The magnetic field intensity is zero both inside and outside the nonuniformly
magnetized cylinder in Fig.5.9, which is obtained from Eq.(5.41) for points inside
the cylinder and the fact that B = 0 outside it, respectively,
H =
B
µ0
− M = M − M = 0 (r ≤ a) and H = B
µ0
= 0 (r > a) . (P5.23)
PROBLEM 5.8 Total (conduction plus magnetization) current density.
(a) A combination of Eqs.(5.52), (5.28), and (5.50) yields that the total (conduction
plus magnetization) volume current density in a ferromagnetic material can, indeed,
be expressed in terms of the magnetic flux density vector, B, in the material only,
as follows:
Jtot = J+Jm = ∇×H+∇×M = ∇× (H + M) = ∇×
B
µ0
=
1
µ0
∇×B . (P5.24)
(b) For the given function B(x, y, z), we use the formula for the curl in Cartesian
coordinates, Eq.(4.81), to obtain
Jtot =
1
µ0
∇×B = 1
µ0
[(
∂Bz
∂y
− ∂By
∂z
)
x̂ +
(
∂Bx
∂z
− ∂Bz
∂x
)
ŷ +
(
∂By
∂x
− ∂Bx
∂y
)
ẑ
]
=
1
µ0
[
2 x̂ +
(
1
y2
− 1
)
ŷ +
2(2x + z)
y3
ẑ
]
(A/m2) (x, y, z in m) . (P5.25)
© 2011 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be 
obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, 
mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, 
Pearson Education, Inc., Upper Saddle River, NJ 07458.
Document shared on www.docsity.com
https://www.docsity.com/?utm_source=docsity&amp;utm_medium=document&amp;utm_campaign=watermark
P5. Solutions to Problems: Magnetostatic Field in Material Media 145
PROBLEM 5.9 Constant flux density vector in a magnetic region. From
the relationship J + Jm = ∇×B/µ0 (proved in the previous problem) and the fact
that any spatial derivative of B in this magnetic region is zero (B does not vary
with spatial coordinates), the magnetization volume current density vector (Jm) in
the region equals the negative of the conduction volume current density vector (J)
at the same point, namely,
Jm =
1
µ0
∇× B− J = −J (B = const) . (P5.26)
PROBLEM 5.10 Closed path in a uniform field. Applying Ampère’s law
in Eq.(5.47) to an arbitrary contour C in this region, along which the magnetic
flux density vector is constant, and can thus be taken out of the line integral, and
through which there is no current passing (current-free region), we obtain
∮
C
B · dl = µ0 (IC + ImC) −→ B ·
∮
C
dl = 0 (B = const , Jtot = 0) .
(P5.27)
The resulting relationship is satisfied (the dot product is zero) for an arbitrary
direction of the vector B and a fixed vector
∮
C dl, and this is possible only if
∮
C
dl = 0 , (P5.28)
which concludes our proof of this vector identity. Note an alternative proof, using
the head-to-tail rule for vector addition, in Fig.4.37.
PROBLEM 5.11 Circulation of the magnetic flux density vector. Com-
bining Eqs.(5.50) and (5.51), the circulation of the magnetic flux density vector
along a closed path C situated entirely inside the magnetic body can be expressed
as ∮
C
B · dl =
∮
C
µ0 (H + M) · dl = µ0
(∮
C
H · dl +
∮
C
M · dl
)
= µ0
(∫
S
J · dS +
∮
C
M · dl
)
, (P5.29)
where S is a surface bounded by C, adopted to also be entirely inside the body.
Section 5.5 Permeability of Magnetic Materials
PROBLEM 5.12 Total magnetization and conduction current. Based
on Eqs.(5.24), (5.68), and (5.49), the total magnetization current enclosed by the
contour C can be obtained from the total conduction current enclosed by C as
follows:
ImC =
∮
C
M · dl =
∮
C
(µr − 1)H · dl = (µr − 1)
∮
C
H · dl = (µr − 1) IC , (P5.30)
© 2011 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be 
obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, 
mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, 
Pearson Education, Inc., Upper Saddle River, NJ 07458.
Document shared on www.docsity.comhttps://www.docsity.com/?utm_source=docsity&amp;utm_medium=document&amp;utm_campaign=watermark
146 Branislav M. Notaroš: Electromagnetics (Pearson Prentice Hall)
where µr−1 can be taken out of the integral sign because µr = const (homogeneous
magnetic material).
PROBLEM 5.13 Thin toroidal coil with a linear ferromagnetic core. (a)
From the generalized Ampère’s law, Eq.(5.49), the circulation of the magnetic field
intensity vector through the core, along its mean length (contour C), amounts to
(see Fig.5.11)
∮
C
H · dl = NI . (P5.31)
(b) By means of Eqs.(5.60) and (P5.31), the circulation along C of the magnetic
flux density vector is
∮
C
B · dl =
∮
C
µH · dl = µ
∮
C
H · dl = µNI . (P5.32)
(c) Finally, with the use Eqs.(5.68) and (P5.31), the circulation of the magnetization
vector through the core turns out to be
∮
C
M · dl =
∮
C
(µr − 1)H · dl = (µr − 1)
∮
C
H · dl = µ − µ0
µ0
NI . (P5.33)
PROBLEM 5.14 Solenoidal coil with a linear ferromagnetic core. (a) The
magnetic field intensity vector inside the solenoid, H, is axial, while it is zero outside
it. Adopting a cylindrical coordinate system with the z-axis along the solenoid axis,
as in Fig.4.10(b), and applying the generalized Ampère’s law, Eq.(5.49), to the
same rectangular contour C as in Fig.4.19, we obtain, for an arbitrary point in the
ferromagnetic core,
Hl = N ′Il −→ H = N ′I ẑ . (P5.34)
(b)-(c) Using Eqs.(5.60) and (5.68), the magnetic flux density and magnetization
vectors in the core are
B = µH = µN ′I ẑ and M = (µr − 1)H =
µ − µ0
µ0
N ′I ẑ , (P5.35)
respectively.
(d)-(e) According to Eq.(5.29), since M = const, the volume magnetization current
density vector in the core is Jm = 0. By means of Eq.(5.32), the surface magneti-
zation current density vector over the surface of the core equals
Jms = M × n̂ = M ẑ× r̂ = Mφ̂ =
µ − µ0
µ0
N ′I φ̂ . (P5.36)
This magnetization current flows in the same way as the conduction current through
the solenoid turns.
© 2011 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be 
obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, 
mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, 
Pearson Education, Inc., Upper Saddle River, NJ 07458.
Document shared on www.docsity.com
https://www.docsity.com/?utm_source=docsity&amp;utm_medium=document&amp;utm_campaign=watermark
P5. Solutions to Problems: Magnetostatic Field in Material Media 147
PROBLEM 5.15 Ferromagnetic cylinder with a conduction current. (a)
In the cylindrical coordinate system adopted as in Fig.4.15(a), the magnetic field
intensity vector (H) in the ferromagnetic cylinder and outside it is circular and of
the form as in Eq.(4.53). From the generalized Ampère’s law, Eq.(5.51), applied as
in Fig.4.15(a), we obtain
H(r) 2πr = Jπr2 −→ H(r) = Jr
2
φ̂ , for r ≤ a ;
H(r) 2πr = Jπa2 −→ H(r) = Ja
2
2r
φ̂ , for r > a . (P5.37)
(b) With the use of Eq.(5.60) for points in the cylinder and Eq.(5.61) for those
outside it, the magnetic flux density vector in the two regions turns out to be
B(r) = µH(r) =
µrµ0Jr
2
φ̂ (r ≤ a) ; B(r) = µ0H(r) =
µ0Ja
2
2r
φ̂ (r > a) .
(P5.38)
(c) The magnetization vector inside the cylinder is found invoking Eq.(5.68), while
it is zero in air, and hence
M(r) = (µr − 1)H(r) =
(µr − 1)Jr
2
φ̂ (r ≤ a) ; M(r) = 0 (r > a) .
(P5.39)
(d) There is no volume magnetization current inside the cylinder [Eq.(5.29)]. By
means of Eq.(5.32), the surface magnetization current density vector on the cylinder
surface (r = a) amounts to
Jms = M(a
−) × n̂ = (µr − 1)Ja
2
φ̂ × r̂ = − (µr − 1)Ja
2
ẑ . (P5.40)
Section 5.6 Maxwell’s Equations and Boundary
Conditions for the Magnetostatic Field
PROBLEM 5.16 Magnetic-magnetic boundary conditions. (a) From the
boundary condition in Eq.(5.75), as there is no surface conduction current on the
boundary (Js = 0), the tangential components of H1 and H2, namely, their x- and
y-components, must be the same on the two sides of the boundary (plane z = 0),
so
H2x = H1x = 5 A/m and H2y = H1y = −3 A/m . (P5.41)
Eq.(5.76) gives the following for the normal component (z-component) of H2 for
z = 0−:
µ1H1n = µ2H2n −→ H2z = H2n =
µ1
µ2
H1n =
µr1
µr2
H1z = 4.8 A/m , (P5.42)
with which the magnetic field intensity vector in medium 2 near the boundary equals
H2 = (5 x̂− 3 ŷ + 4.8 ẑ) A/m (Js = 0) . (P5.43)
© 2011 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be 
obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, 
mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, 
Pearson Education, Inc., Upper Saddle River, NJ 07458.
Document shared on www.docsity.com
https://www.docsity.com/?utm_source=docsity&amp;utm_medium=document&amp;utm_campaign=watermark
148 Branislav M. Notaroš: Electromagnetics (Pearson Prentice Hall)
(b) Now Js in the plane z = 0 is nonzero, and we use the boundary condition in
Eq.(5.74) in place of that in Eq.(5.75),
ẑ× H1 − n̂× H2 = Js = Js ŷ . (P5.44)
Since
ẑ × H1 = ẑ × (H1x x̂ + H1y ŷ + H1z ẑ) = H1x ŷ − H1y x̂ , (P5.45)
and analogously for ẑ × H2, Eq.(P5.44) becomes
H1x ŷ − H2x ŷ = Js ŷ −→ H2x = H1x − Js = 2 A/m . (P5.46)
The other components of H2 are the same as in Eq.(P5.43), and hence
H2 = (2 x̂ − 3 ŷ + 4.8 ẑ) A/m (Js 6= 0) . (P5.47)
Section 5.7 Image Theory for the Magnetic Field
PROBLEM 5.17 Force on a conductor above an infinite PMC corner.
We first eliminate the horizontal PMC plane in Fig.5.38 utilizing the image theory
for the magnetic field, in Fig.5.14, which leaves us with two wires (line currents),
both of intensities I, on the right of the vertical PMC plane, which is then removed
as well by another application of the image theory. The result is a structure with
four wires, with currents of equal intensities, I, as shown in Fig.P5.2.
I
I I
I
1
2
3
4
Fm1'
Fm21'
Fm31'
Fm41'
h
h
h h
Figure P5.2 Cross section of a system of four wire conductors with equal currents
in free space equivalent, by virtue of the image theory for the magnetic field –
applied twice, to the structure with one wire parallel to a PMC corner in Fig.5.38.
Using the principle of superposition, the per-unit-length magnetic force on wire
1 in Fig.P5.2 (the original wire in Fig.5.38) is computed as the vector sum of partial
p.u.l. forces due to currents in wires 2, 3, and 4, respectively,
F′m1 = F
′
m21 + F
′
m31 + F
′
m41 , (P5.48)
© 2011 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be 
obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, 
mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, 
Pearson Education, Inc., Upper Saddle River, NJ 07458.
Document shared on www.docsity.com
https://www.docsity.com/?utm_source=docsity&amp;utm_medium=document&amp;utm_campaign=watermark
P5. Solutions to Problems: Magnetostatic Field in Material Media 149
where the partial forces are computed based on Eq.(4.166). As can be seen in
Fig.P5.2, the resultant force is along the line connecting wires 1 and 4, that is,
along the symmetry line in the cross section of the original system in Fig.5.38, and
its p.u.l. magnitude is
F ′m1 = F
′
m21 cos 45
◦+F ′m31 cos 45
◦+F ′m41 = 2
µ0I
2
2π(2h)
√
2
2
+
µ0I
2
2π(2h
√
2)
=
3
√
2µ0I
2
8πh
.
(P5.49)
PROBLEM 5.18 Uniformly magnetizedhollow disk on a PMC plane. In
the cylindrical coordinate system adopted such that its z-axis is the one in Fig.5.39,
Eq.(5.32) tells us that magnetization surface currents flow circumferentially along
the the inner and outer lateral surfaces of the hollow ferromagnetic disk, and their
densities are, respectively,
Jmsa = M ẑ × (− r̂) = −Mφ̂ (r = a) and Jmsb = M ẑ× r̂ = Mφ̂ (r = b) .
(P5.50)
As d ≪ a, b, these circumferential sheets of magnetization current can be regarded as
equivalent circular current loops with radii a and b and current intensities [Eq.(5.37)]
∓Im = ∓Jmsb d = ∓Md (P5.51)
(currents in the two loops flow in opposite directions). By the image theory for
the magnetic field (Fig.5.15), the current loops right on the plane (at z = 0+) and
their positive images right below the plane (at z = 0−) add to each other, so the
currents intensities ∓Im in fact double, and are considered to reside in a vacuum.
Using Eq.(4.19) or (5.38), with current intensities −2Im and 2Im, and radii a and
b, respectively, the magnetic flux density vector along the z-axis above the PMC
plane in Fig.5.39 comes out to be
B = Ba + Bb =
µ0(−2Im) a2
2 (z2 + a2)
3/2
ẑ +
µ0(2Im) b
2
2 (z2 + b2)
3/2
ẑ
= −µ0Md
[
a2
(z2 + a2)
3/2
− b
2
2 (z2 + b2)
3/2
]
ẑ (0 < z < ∞) . (P5.52)
PROBLEM 5.19 Magnetized cylinder between two PMC planes. We
can replace the ferromagnetic cylinder in Fig.5.40 by a cylindrical magnetization-
current sheet of radius a, hight (length) h, and surface current density in Eq.(5.36),
where M = M0, in a vacuum. By multiple applications of the image theory for the
magnetic field (Fig.5.15), much like those in Fig.5.17, the cylindrical current sheet
of length h becomes infinite, and thus equivalent to an infinitely long solenoid,
Fig.4.19. This means that the magnetic field outside it (for r > a), so in air in
Fig.5.40, is zero, while that inside it is axial, given by Eq.(4.37), which we modify
according to Eq.(4.30). Hence the magnetic flux density vector in the space between
the PMC planes in Fig.5.40, inside and outside the cylinder,
B = µ0N
′I ẑ = µ0Jms ẑ = µ0M0 ẑ = µ0M (r < a , 0 < z < h)
and B = 0 (r > a , 0 < z < h) . (P5.53)
© 2011 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be 
obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, 
mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, 
Pearson Education, Inc., Upper Saddle River, NJ 07458.
Document shared on www.docsity.com
https://www.docsity.com/?utm_source=docsity&amp;utm_medium=document&amp;utm_campaign=watermark
150 Branislav M. Notaroš: Electromagnetics (Pearson Prentice Hall)
Section 5.9 Magnetic Circuits – Basic Assumptions
for the Analysis
PROBLEM 5.20 Magnetic flux through a thick toroid. (a) The magnetic
field value corresponding to the “knee” point between the two parts of the piece-
wise linear magnetization curve in Fig.5.30(b) is Hk = 1000 A/m, the same as in
Fig.5.27(b). Therefore, the radial distance c in Fig.5.27(a) representing the bound-
ary between the two parts of the core, one in which the second (farther) part of
the curve in Fig.5.30(b) applies and the other with B(H) given by the first part
(containing the point B = 0) of the curve, is the same as in Eq.(5.91), so c = 3.2 cm.
In specific, from Eq.(5.103), the magnetic flux density in the first part of the core is
B1(r) = 0.4+2.5×10−5[H(r)−1000] (T) =
(
0.375 +
7.96 × 10−4
r
)
T (a ≤ r ≤ c)
(P5.54)
[H(r) is given in Eq.(5.83)], whereas we observe in Fig.5.30(b) that B in the second
part can be expressed as
B2(r) = 4 × 10−4H(r) (T) =
1.27 × 10−2
r
T (c < r ≤ b) . (P5.55)
Hence, in place of Eq.(5.92), the magnetic flux through the core amounts to
Φ =
∫ b
r=a
B(r) h dr
︸︷︷︸
dS
= h
∫ c
a
B1(r) dr + h
∫ b
c
B2(r) dr = 77.1 µWb . (P5.56)
(b) The assumption of a uniform field distribution in the core results in the following
approximate value of the magnetic flux through the core [see Eq.(5.93)]:
Φapprox = B1(rmean) (b − a)h
︸ ︷︷ ︸
S
= 80.3 µWb
[
rmean =
a + b
2
= 3 cm
]
, (P5.57)
and the relative error is now δΦ = 4.2%.
Section 5.10 Kirchhoff’s Laws for Magnetic Circuits
PROBLEM 5.21 Simple nonlinear magnetic circuit. The equation of the
load line for the magnetic circuit in Fig.5.41, Eq.(5.101), for the given numerical
data becomes
H + 497.4B = 2000 (H in A/m ; B in T) . (P5.58)
© 2011 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be 
obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, 
mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, 
Pearson Education, Inc., Upper Saddle River, NJ 07458.
Document shared on www.docsity.com
https://www.docsity.com/?utm_source=docsity&amp;utm_medium=document&amp;utm_campaign=watermark
P5. Solutions to Problems: Magnetostatic Field in Material Media 151
It is obvious that the intersection of this line with the magnetization curve of the
core material, in Fig.5.27(b), belongs to the saturation part of the curve, so B = 1 T,
which, substituted in Eq.(P5.58), then gives the magnetic field intensity in the core,
H = 1.503 kA/m. Combining Eqs.(5.98) and (5.100), the magnetic field intensity
in the air gap turns out to be
H0 =
B
µ0
= 795.8 kA/m . (P5.59)
Note that the alternative assumption in solving this magnetic circuit, that the
operating point for the circuit belongs to the linear (rising) part of the curve in
Fig.5.27(b), so that B = µaH , where µa = 0.001 H/m, results, upon substitu-
tion in Eq.(P5.58), in H = 1.336 kA/m > Hk = 1 kA/m, which, contradicts the
assumption, and is thus impossible.
PROBLEM 5.22 Complex nonlinear magnetic circuit. Let us orient the
branches of the magnetic circuit in Fig.5.42(a) as in Fig.P5.3. Kirchhoff’s “current”
and “voltage” laws for magnetic circuits, Eqs.(5.94) and (5.95), applied to the node
N1 and closed paths C1 and C2 (in Fig.P5.3) are:
B1S1 − B2S2 + B3S3 = 0 −→ B2 = B1 + B3 , (P5.60)
H1l1 + H2l2 = −N1I1 + N2I2 , (P5.61)
H2l2 + H3l3 = N2I2 . (P5.62)
Depending on whether the operating points for the individual branches of the circuit
belong to the linear part or to the saturation part of the magnetization curve in
Fig.5.42(b), we may have
Bi =
{
µaHi for Hi ≤ 1000 A/m
1.5 T for Hi > 1000 A/m
, i = 1, 2, 3 , (P5.63)
where the initial permeability of the material is µa = 1.5/1000 H/m = 0.0015 H/m.
l S3 3,
N2
l S1 1,
l S2 2,
N1
I1
I2
B1
B3
B2
N1
C1 C2
Figure P5.3 Analysis of the nonlinear magnetic circuit with three branches and
two mmf’s in Fig.5.42 using Kirchhoff’s “current” and “voltage” laws for magnetic
circuits.
Suppose, first, that none of the magnetizations in the branches in Fig.P5.3 is in
saturation (B = µaH in all three branches). Eq.(P5.60) then becomes
H2 = H1 + H3 , (P5.64)
© 2011 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be 
obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, 
mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, 
Pearson Education, Inc., Upper Saddle River, NJ 07458.
Document shared on www.docsity.com
https://www.docsity.com/?utm_source=docsity&amp;utm_medium=document&amp;utm_campaign=watermark
152 Branislav M. Notaroš: Electromagnetics (Pearson Prentice Hall)
and the solution of the system with Eqs.(P5.64), (P5.61),and (P5.62) is H1 =
375 A/m, H2 = 875 A/m, and H3 = 1250 A/m, which is impossible, given the
assumption that none of the field intensities is larger than 1000 A/m in Eq.(P5.63).
Since N2I2 = 3N1I1, it is logical to expect that the middle branch in Fig.P5.3
would first reach saturation, so B2 = 1.5 T. Assuming that the other two branches
are in the linear regime (B1 = µaH1 and B3 = µaH3), Eq.(P5.60) becomes
H1 + H3 = 1000 A/m , (P5.65)
and the solution of the new system of three equations is H1 = 250 A/m, H2 =
1500 A/m, and H3 = 750 A/m, where these values are consistent with the new
assumption made about the magnetization stages in the branches. The remaining
two flux densities in the circuit are B1 = 0.375 T and B3 = 1.125 T.
PROBLEM 5.23 Magnetic circuit with a zero flux in one branch. Refer-
ring to Fig.P5.4, the equations for the magnetic circuit are (note that S2 = 2S1 =
2S3):
B1S1 − B2S2 + B3S3 = 0 (B1 = 0) −→ 2B2 = B3 , (P5.66)
−H1l1 + H3l3 = N1I1 (H1 = 0) −→ H3l3 = N1I1 , (P5.67)
H2l2 + H3l3 = N2I2 , (P5.68)
Bi =
{
µaHi for Hi ≤ 1000 A/m
1 T for Hi > 1000 A/m
, i = 1, 2, 3 (µa = 0.001 H/m) . (P5.69)
N2N1
I1
I2
B1=0
B3
B2
N1
C1
C2
Figure P5.4 Finding the current I1 of the winding in the first branch of the mag-
netic circuit in Fig.5.42(a) such that the magnetic flux in that branch is zero.
Assuming, first, that both branches with nonzero fields in Fig.P5.4 are in the
linear regime (B2 = µaH2 and B3 = µaH3), Eq.(P5.66) gives 2H2 = H3, which,
substituted in Eq.(P5.68), results in H2 = N2I2/(l2 + 2l3) = 714.3 A/m and H3 =
2H2 = 1428.6 A/m > 1000 A/m, which is impossible.
Assuming, then, that the third branch is in saturation, B3 = 1 T, we have,
from Eq.(P5.66), that B2 = 0.5 T, meaning that the second branch is in the linear
regime, so, from Eq.(P5.69), H2 = B2/µa = 500 A/m. Eq.(P5.68) then yields
H3 =
N2I2 − H2l2
l3
= 1500 A/m , (P5.70)
© 2011 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be 
obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, 
mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, 
Pearson Education, Inc., Upper Saddle River, NJ 07458.
Document shared on www.docsity.com
https://www.docsity.com/?utm_source=docsity&amp;utm_medium=document&amp;utm_campaign=watermark
P5. Solutions to Problems: Magnetostatic Field in Material Media 153
which is consistent with the new assumption that this branch is in saturation. Fi-
nally, the current of the winding in the first branch in Fig.P5.4 is obtained from
Eq.(P5.67),
I1 =
H3l3
N1
= 0.3 A . (P5.71)
PROBLEM 5.24 Nonlinear magnetic circuit with two air gaps. Equations
for the magnetic circuit, Fig.P5.5, are [see also Eqs.(5.98) and (5.100) for the air
gap]:
−B1S + B2S + B3S = 0 (B3 = 0) −→ B1 = B2 , (P5.72)
H1(l
′
1 + l
′′
1 ) + H0l0 + H2l2 = N1I1 , where H0 =
B0
µ0
=
B1
µ0
, (P5.73)
−H2l2 = N2I2 (H3 = 0) . (P5.74)
Assuming that B1 = B2 = 1 T (saturation) and H1 = H2, Eq.(P5.73) gives
H1 =
N1I1 − B1l0/µ0
l′1 + l
′′
1 + l2
= 2145.8 A/m , (P5.75)
so that the mmf of the second coil in Fig.P5.5 turns, from Eq.(P5.74), out to be
N2I2 = −H2l2 = −H1l2 = −107.3 A turns . (P5.76)
B3=0
N1 N2
l2
I1
l0
l0
I2l1’ l3 ’
l1” l3”
H1 H2
B3
B2C1 C2
B1
H0
H3=0
Figure P5.5 Analysis of the nonlinear magnetic circuit in Fig.5.43.
PROBLEM 5.25 Reverse problem with a nonlinear magnetic circuit.
With reference to the notation in Fig.P5.6, the flux densities in the ferromagnetic
section (B1) and in the air gap (B10) of the first branch of the circuit are
B1 = B10 =
Φ1
S
= 0.5 T . (P5.77)
From the magnetization curve, i.e., by solving the magnetization equation given
analytically, for the corresponding field intensity in the material,
H1 = 250 tanB1 = 137 A/m , (P5.78)
whereas in the gap,
H10 =
B10
µ0
= 398 kA/m . (P5.79)
© 2011 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be 
obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, 
mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, 
Pearson Education, Inc., Upper Saddle River, NJ 07458.
Document shared on www.docsity.com
https://www.docsity.com/?utm_source=docsity&amp;utm_medium=document&amp;utm_campaign=watermark
154 Branislav M. Notaroš: Electromagnetics (Pearson Prentice Hall)
Kirchhoff’s “voltage” law for the closed path C1 in Fig.P5.6 now yields
H2 =
N1I1 − H1(l′1 + l′′1 ) − H10l0
l2
= 16.6 kA/m , (P5.80)
with the corresponding flux density in the second (middle) branch given by
B2 = arctan
H2
250
= 1.56 T . (P5.81)
The flux density in the third branch is next obtained using Kirchhoff’s “current”
law for the node N1 in Fig.P5.6, and from it the associated H value,
B3 = B1 − B2 = −1.06 T −→ H3 = 250 tanB3 = −442 A/m , (P5.82)
whereas the magnetic field intensity in the air gap of the third branch is found as
H30 =
B30
µ0
=
B3
µ0
= −840 kA/m . (P5.83)
Finally, we apply Kirchhoff’s “voltage” law to the closed path C2 and obtain the
mmf we seek:
N2I2 = −H2l2 + H3(l′3 + l′′3 ) + H30l0 = −1438 A turns . (P5.84)
B30
N1 N2
l2
I1
l0
l0
I2l1’ l3’
l1” l3”
B1
B3
B2
C1 C2B10
F1 N1
Figure P5.6 Analysis of the magnetic circuit in Fig.5.43 but with a nonlinear
magnetization curve given analytically.
PROBLEM 5.26 Remanent flux in a circuit with zero mmf. Eq.(5.122),
for the state in the magnetic circuit after the switch K is closed, now, for a modified
set of numerical data, gives B = 1.114 T, and hence the maximum value for the
magnetic flux density and the corresponding field intensity in Fig.5.34(b) are Bm =
1.114 T and Hm = Bm/µa = 1.114 kA/m, as indicated in Fig.P5.7. After the switch
K is opened, the load line in Eq.(5.123) becomes
H + 795.8B = 0 (H in A/m ; B in T) . (P5.85)
The assumption that the intersection of this line with the demagnetization curve
in Fig.P5.7 belongs to the horizontal segment of the curve, i.e., that B = Bm =
1.114 T, results, upon substituting this value in Eq.(P5.85), in H = −886.5 A/m,
which is consistent with the assumption, as H > −Hc = −Hm = −1.114 kA/m, so
the assumption proves correct. The operating point for the circuit in this state is
© 2011 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be 
obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, 
mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, 
Pearson Education, Inc., Upper Saddle River, NJ 07458.
Document shared on www.docsity.com
https://www.docsity.com/?utm_source=docsity&amp;utm_medium=document&amp;utm_campaign=watermark
P5. Solutions to Problems: Magnetostatic Field in Material Media 155
0
H
B
Bm=1.114 T
Hm=1114 A/m
-Hc=-1114 A/m
switch open switch closed
P2
P1
-886.5 A/m
Figure P5.7 Operating points for the magnetic circuit in Fig.5.30(a) in two sta-
tionary states (switch K closed and open, respectively) for a modified set of numer-
ical data.
also shown in Fig.P5.7. The magnetic field intensity in the gap is H0 = B/µ0 =
Bm/µ0 = 886.5 kA/m.
PROBLEM 5.27 Linear magnetic circuit with three branches. (a)-(b)
Fig.P5.8 shows the equivalent electric circuit for the problem, with reluctances,
Eq.(5.97),
R1 = R3 =
l′1 + l
′′
1
µrµ0S
=
l′3 + l
′′
3
µrµ0S
= 7.64× 105 H−1 , R2 =
l2
µrµ0S
= 1.59× 105 H−1 ,
R10 = R30 =
l0
µ0S
= 1.91 × 106 H−1 . (P5.86)
The figure alsoshows the adopted loops for the loop analysis of the (electric) cir-
cuit (of course, there are multiple other ways to solve this electric circuit). The
corresponding loop equations are
(R10 + R1 + R2)Φ1 + (R10 + R1)Φ2 = N1I1 , (P5.87)
(R10 + R1)Φ1 + (R10 + R1 + R3 + R30)Φ2 = N1I1 + N2I2 , (P5.88)
F1 F2
+ N I1 1 +
N I2 2
R1
R2
R3
R10 R30
Figure P5.8 Equivalent electric circuit for the loop analysis of a linear magnetic
circuit with three branches, two air gaps, and two mmf’s (in Fig.5.43).
© 2011 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be 
obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, 
mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, 
Pearson Education, Inc., Upper Saddle River, NJ 07458.
Document shared on www.docsity.com
https://www.docsity.com/?utm_source=docsity&amp;utm_medium=document&amp;utm_campaign=watermark
156 Branislav M. Notaroš: Electromagnetics (Pearson Prentice Hall)
and their solution is Φ1 = 200.5 µWb and Φ2 = 199 µWb. The magnetic field
intensities in the air gaps in Fig.5.43 come out to be
H10 =
B10
µ0
=
Φ1 + Φ2
µ0S
= 1.27 MA/m and H30 =
B30
µ0
=
Φ2
µ0S
= 633 kA/m .
(P5.89)
Section 5.11 Maxwell’s Equations for the
Time-Invariant Electromagnetic Field
PROBLEM 5.28 Continuity equation from Ampère’s law in integral
form. Consider an arbitrary closed surface S in the time-invariant electromagnetic
field and a contour C that splits S into two parts, the upper part S1 and the lower
part S2 (similar to Fig.4.27 for S1 and S3). Let the surfaces S1 and S2, which are
both enclosed by C, be oriented in the same way, that in accordance with the right-
hand rule with respect to the orientation of the contour. Applying Maxwell’s second
equation in integral form, Eqs.(5.70), to the contour C and either S1 or S2 must
give the same result, i.e., the fluxes through S1 and S2 of the current density vector,
J, are the same. This means in turn that [note the similarity with Eq.(4.101)]
∮
S
J · dS = 0 , (P5.90)
which is the time-invariant integral form of the continuity equation, Eqs.(3.59). Note
that this derivation parallels the one given in Example 5.19 in differential notation.
© 2011 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be 
obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, 
mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, 
Pearson Education, Inc., Upper Saddle River, NJ 07458.
Document shared on www.docsity.com
https://www.docsity.com/?utm_source=docsity&amp;utm_medium=document&amp;utm_campaign=watermark