Chapter 28--Examples
1
Problem
Two long, parallel transmission lines, 40
cm apart, carry 25 A and 75 A currents.
Find all locations where the net magnetic
field of these wires is zero if these
currents are a) in the same direction and
b) in opposite directions
2
Parallel Directions

OO
OO
OO
XO
XO
xX O

75A
XX
XX
XX
X
X
X
Based on the figure, the only
place that the fields can be
zero is between wires.
Assume that the zero point
is a distance x from the 75 A
wire. So now set B75=B25
0 I
 0 I 75
 0 I 25
B


2r
2x
2 0.4  x 
75 * (0.4  x)  25 x  30  100 x
x  0.3m
3
Opposite Directions Directions

Based on the figure, there are two
regions outside the wires.

X
X
O
X
O
X
X
X
O
X
X
x
X
O
X
X
X
X
X
O
X
O
25A

Since one current is smaller than
the other, we must choose the
region closest to the smaller current.
Assume that the zero point is a
distance x from the 25 A wire. So now
set B75=B25
75A
0 I
 0 I 25
 0 I 75
B


2r
2x
2 0.4  x 
25 * (0.4  x)  75 x  10  50 x
x  0.2m
4
Problem
Consider the figure below. The curved segments are
arcs of circles of radii a and b. The straight segments
are along radii. Find the magnetic field at point P (the
center of the curvature) assuming i as the current.
a
i
q
b
i
5
Biot-Savart
 o i ds  rˆ
dB 
4 r 2
Based on Biot-Savart, the
 o i ds  rˆ q o i rdq
dB 

current segments along the
2
2
4

r
4

r
0
radial vector do not
q
contribute since they are
o i dq o i q
B

parallel with r.


 4
0
o
4
at r  b

Bb  o
4
Ba 
a
q
b
r
4 r
at r  a
Now we must integrate
across each arc. The
integration variable is q.
ds=rd q
i

iq
a
iq
b
i
6
Direction
By the RH Rule, the current
from b is out of the page at
Point P.
The current from a is into the
page.
a
i
q
b
i
We will choose out of the page
as positive and the magnetic
field at P is
o i q o i q
B  Bb  Ba 

4 b 4 a
oi q  1 1 
B
  
4  b a 
7
Problem
Four long copper wires are parallel to each other, their
cross-sections forming the corners of a square with
sides a=20 cm. A 20 A current exists in each wire in
the direction shown in the figure below. What are the
magnitude and direction of B at the center of the
square.
a
a
X
a
a
X
8
RH Rule
a
a
a
X
a
The lower right and
upper left corners each
point to the upper right.
X
The fields from the lower
left and upper right point
to the upper left.
Thus the total field point
up the page
9
The Magnitude


The distance to the center from any corner is
a/sqrt(2).
All of the horizontal components have added out.
Only the four vertical components sum to
 oi
oi
2 2 oi
B  4*
cos 45  4 *

a
a
2a 2
2
0
2
2( 4  10 7 )( 20)
B
 8  10 5 T
 ( 0 .2 )
10
Problem
A solid conductor with radius a is supported by
insulating material on the axis of a conducting
tube with inner radius b and outer radius c (see
figure below). The central conductor and the
outer conductor carry an equal current i in
opposite directions. However, the magnitude of
current density in the inner conductor is
expressed by J=kr where k is an appropriate
constant.
Derive an expression for the magnetic field at all
points (r<a, a<r<b, b<r<c, r>c)
Derive an expression for current density in the
outer conductor.
Show that each boundary is matched.
11
First, r<=a



At r=a, total current is i
so i=J*A
J=ka and A=a2 so
i=ka3 or k=i/a3
Using RH rule, with our
thumb pointing in
direction of current, our
fingers are matching
direction of integration
so this is a positive
direction
 
 B  ds  0ienclosed
 
 B  ds  B2r 
i
r3
3
ienclosed  JA  kr * r  3 r  i 3
a
a
r3
B2r    0i 3
a
 0i r 2
B
2 a 3
at r  a
2
 0i a 2  0i
B

3
2 a
2a
12
Next, a<=r<b
 
 B  ds  0ienclosed
 
 B  ds  B2r 
ienclosed  i
 0i
B
2r
at r  a
 0i
B
2a
at r  b
 0i
B
2b
13
b<=r<=c
J’=-i/A
 A=(c2)-(b2)
 J’=-i/[(c2)-(b2)]

B
0 
i
2
2 





i


r


b

2r  c 2   b 2 

at r  b
B
 
B
  ds  0ienclosed
 
B
  ds  B2r 
0 
 0i
i
2
2 







b


b

i 

2b  c 2   b 2 
 2b
at r  b
B
ienclosed  i  i '  i  J ' A
0 
i
2
2 




  0

c


b
i 
2c  c 2   b 2 

   
A  r 2  b 2



r   b 
   


 
i


B
r   b 
i 
2r  c   b 


i
B2r    0i   0i '   0 i 
2
c  b 2

2
0
2
2
2
2
2
14
r>c
 
B

d
s


i
0
enclosed

 


B

d
s

B
2

r

ienclosed  i  i  0
B0
Recall for a coaxial
cable with equal static
charges, E=0 outside
as well
15