Homework #1
P.6-21 A very large slab of material of thickness d lies perpendicularly to a uniform magnetic field


of intensity H 0  az H 0 . Ignoring edge effect, determine the magnetic field intensity in the slab:
a) if the slab material has a permeability ,


b) if the slab is a permanent magnet having a magnetization vector M i  az M i .
Solution:


a) From constitute relation we have B2  2 H 2
and B.C. requires B2 z  B1z .
Thus,  H 2  0 H 0


 
H 2  az H 2  az 0 H 0 .




b) Given B2  0 H 2  M i ,


and B.C. requires B2 z  B1z .


Thus, 0  H 2  M i   0 H 0 H 2  az  H 0  M i  .
P.6-24 Do the following by using Eq. (6-224):
a) Determine the scalar magnetic potential at a point on the axis of a circular loop having a
radius b and carrying a current I.

b) Obtain the magnetic flux density B from  0Vm , and compare the result with Eq. (6-38).
Solution:
a) From Eq. (6-224) in P.6-23, we have
 
I ds  aR
I
Vm 

.
2

4
4
R
Referring the right figure,
 
ds  aR  (cos  )  d  d 
z
z  2
2
 d  d .
R  z2   2 .
Vm 
I
4
2
b
0
0
 
z
z
2
 2 
3/2
I
2 
 d  d   1 

.
z 2   2 
z

Vm 
0 Ib 2

 az
b) B   0Vm  az 0
, which is the same as Eq. (6-38).
3/2
z
2  z 2  b2 
P.6-26 A ferromagnetic sphere of radius b is magnetized uniformly with a magnetization M = azM0.
a) Determine the equivalent magnetization current densities Jm and Jms.
b) Determine the magnetic flux density at the center of the sphere.
Solution:
z

b

M  aˆ z M 0




a) J m    M  0; J ms  M  aˆ n  aˆ R cos   aˆ sin  M 0  aˆ R  aˆ M 0 sin  .
b) Apply Eq. (6-38) to a loop of radius bsin carrying a current Jmsbd

 ( J bd )(b sin  ) 2
 M
 aˆ z 0 0 sin 3 d
dB  aˆ z 0 ms 2 3 / 2
2
2(b )



 M 
2
2
B   dB  aˆ z 0 0  sin 3 d  aˆ z  0 M 0   0 M .
2 0
3
3
P.6-27 A toroidal iron core of relative permeability 3000 has a mean radius R = 80 (mm) and a
circular cross section with radius b = 25 (mm). An air gap  g  3 (mm) exists, and a current I flows
in a 500-turn winding to produce a magnetic flux of 10–5 (Wb). (See Fig. 6-44.) Neglecting flux
leakage and using mean path length, find
a) the reluctances of the air gap and of the iron core,
b) Bg and Hg in the air gap, and Bc and Hc in the iron core,
c) the required current I.
Solution:
a)  g 
g
0 S

3 103
 1.21106 (H 1 ),
7
2
4  10  (  0.025)
c
2  0.08  3  103

 6.75  104 (H 1 ).
7
2
 S 3000  (4 10 )  (  0.025)

 
105

b) Bg  Bc  a
 a 5.09 103 (T),
2
  0.025

1 
 5.09  103 
 a 4.05 103 (A/m),
Hg 
Bg  a
7
4  10
0

1   4.05 103 
 a 1.35 (A/m).
Hc 
B a
0 r c  3000
c 
c) NI    c   g  , I 

 c  g   0.0256 (A) = 25.6 (mA).
N
P.6-32 Consider a plane boundary ( y = 0) between air (region 1,  r1 = 1) and iron (region 2,  r2 =
5000).
a) Assuming B1 = ax0.5 – ay10 (mT), find B2 and the angle that B2 makes with the interface.
b) Assuming B2 = ax10 + ay0.5 (mT), find B1 and the angle that B1 makes with the normal to
the interface.
Solution:
 
 


a) B1  ax 0.5  a y 10 (mT), B2  ax B2 x  a y B2 y .
B2 x
0.5
 H1 x 
 B2 x  2500 (mT).
50000
0
 

B2 y  B1 y  10 (mT), thus B2  ax 2500  a y 10 (mT).
From B.C.: H 2 x 
B 
2
tan 1  5000  1x   250   2  tan 1 250  89.8o ,  2  0.2o.
 B1 y 
1


tan  2 
 
 


b) B2  ax 10  a y 0.5 (mT), B1  ax B1x  a y B1 y .
From B.C.: H1x 
B1x
0
 H2x 
10
5000 0
 B1x  0.002 (mT).
 

B1 y  B2 y  0.5 (mT), thus B1  ax 0.002  a y 0.5 (mT).
1
0.002
 0.004   2  tan 1 0.004  0.23o.
tan  2 
2
0.5
tan 1 
P.6-35 Determine the self-inductance of a toroidal coil of N turns of wire wound on an air frame
with mean radius ro and a circular cross section of radius b. Obtain an approximate expression
assuming b << ro.
Solution:
 
  NI
B  a B  a 0 ,
2 r
in which, r  ro   cos  .

0 NI
2
b
2
0
0



 d d 
 0 NI ro  ro2  b 2 .
ro   cos 


N
 0 N 2 ro  ro2  b 2 .
I
 NI
If ro  b , B  0 (constant).
2 ro
L 
  B S  B  b
2

0 Nb 2 I
2ro

L
0 N 2b 2
2ro
註: 此處用到的積分請查看網頁留的說明。
.
P.6-37 Calculate the mutual inductance per unit length between two parallel two-wire transmission
lines A  A and B  B  separated by a distance D, as shown in Fig. 6-47. Assume the wire
radius to be much smaller than D and the wire spacing d.
Solution:

B at a distance r from an infinitely long line carrying a current I:

 I
B  aˆ 0
2r
For a unit length the flux due to I in line A that links with the second
line pair B - B' is: (We select the surface formed by sub-surfaces B – C
and
C - B'. Since the normal direction of surface C - B' is perpendicular to
the field B, thus the surface integral of this part equals zero and we
only need to consider the surface integral over the surface B – C.)
dr  0 I AC
ln

B r
2
AB
Similarly, that due to I in line A' is:
 A 
A  
0 I
2

C
0 I AC
ln
2 AB
Total flux linkage per unit length
  A  A  
12
L12 
0 I AC AC 0 I AB AB 0 I D 2  d 2
ln

ln

ln
2 AB AB 2 AB AB 2
D2
0  d 2 
ln 1 

2  D 2 
P.6-40 Find the mutual inductance between two coplanar rectangular loops with parallel sides, as
shown in Fig. 6-50. Assume that h1 >> h2 (h2 > w2 > d).
Solution:
Approximate the magnetic flux due to the long loop linking with
the small loop by that due to two infinitely long wires carrying
equal and opposite current I.

 1
1
 h I w d
w1  d 
.
dx  0 2 ln 2



2
w1  w2  d 
 d
 d  x w1  d  x 
12 
0 h2 I
2
L12 
12 0 h2 w2  d w1  d 
ln
.

2
I
d w1  w2  d 

w2
0
Fig. 6-50
P.6-42 Calculate the force per unit length on each of three equidistance, infinitely long, parallel
wires 0.15 (m) apart, each carrying a current of 25 (A) in the same direction. Specify the direction
of the force.
Solution:
I1 = I2 = I3 = 25 (A); d = 0.15 (m).

3 0 I
.
B2  aˆ x 2 B12 cos 30 o  aˆ x
2d
Force per unit length on wire 2:

f 2  aˆ y IB2  aˆ y
3 0 I 2
 aˆ y 1150  0   aˆ y 1.44  10 3 (N/m).
2d
Forces on all three wires are of equal magnitude and toward the
center of the triangle.
P.6-46 The bar AA’ in Fig. 6-53 serves as a conducting path (such as the blade of a circuit breaker)
for the current I in two very long parallel lines. The lines have a radius b and are spaced at a
distance d apart. Find the direction and the magnitude of the magnetic force.
Solution:


1 
 I1
, d   aˆ y dy
Bat y  aˆ z 0  
4  y d  y 

 
1 
 I2  1
dy
dF  Id   B   aˆ x 0  
4  y d  y 

1 
 I 2 d b  1
 I2  d 
dy  aˆ x 0 ln  1.
F   aˆ x 0   
4 b  y d  y 
2  b 
P. 6-47 A d-c current I = 10 (A) flows in an equilateral triangle loop in the xy-plane as in Fig. 6.-54.
Assuming a uniform magnetic flux density B = ay0.5 (T) in the region, find the forces and torque on
the loop. The dimensions are in (cm).
y
B
A (0, 20)
I
B
C
(10,0)O (10,0)
x
FIGURE 6-54
A triangular loop in a uniform
Magnetic field (Problem P.6-47).
Solution:
Assign point names A, B, and C as above, then we can obtain

FAB  B  I ( AB)  (a y 0.5)  10(a x 0.1  a y 0.2)  a z 0.5 (N)

FCA  B  I (CA)  (a y 0.5) 10(a x 0.1  a y 0.2)  a z 0.5 (N)

FBC  B  I ( BC )  (a y 0.5) 10(a x 0.2)  a z 1.0 (N)
Ftotal  0 (N)
T  m  B  (a z IS )  (a y 0.5)  (a z 10  12  0.2  0.2)  (a y 0.5)  a x 0.1 (N  m)
P.6-53 A current I flows in a long solenoid with n closely wound coil-turns per unit length. The
cross-sectional area of its iron core, which has permeability , is S. Determine the force acting on
the core if it is withdrawn to the position shown in Fig. 6-55.
Fig. 6-55
Solution:
Wm 
1
H 2 dv

2
Assume a virtual displacement, x, of the iron core.
Wm ( x  x)  Wm ( x) 
( FI ) x 
1
1
(   0 ) H 2 dv  Wm ( x)  0 (  r  1)n 2 I 2 Sx.

S
x


2
2
Wm 0

(  r  1)n 2 I 2 S , in the direction of increasing x.
x
2