Ch 27 Using Ampère’s Law
Problems: 45, 46, 49, 50, 51
45 •
[SSM] A long, straight, thin-walled cylindrical shell of radius R carries a
current I parallel to the central axis of the shell. Find the magnetic field (including
direction) both inside and outside the shell.
Picture the Problem We can apply Ampère’s law to a circle centered on the axis
of the cylinder and evaluate this expression for r < R and r > R to find B inside
and outside the cylinder. We can use the right-hand rule to determine the
direction of the magnetic fields.
Apply Ampère’s law to a circle
centered on the axis of the
cylinder:
Evaluate this expression for r < R:

Evaluate this expression for r > R:
Note that, by symmetry, the field is the
same everywhere on this circle.



Binside  d  0 0  0
Binside  0

C
Solve for Boutside to obtain:
0 C
C
C
Solve for Binside to obtain:

 B  d   I


Boutside  d   B2R   0 I
Boutside 
0 I
2R
The direction of the magnetic field is in the direction of the curled fingers of your
right hand when you grab the cylinder with your right thumb in the direction of
the current.
46 •
In Figure 27-55, one current is 8.0 A into the page, the other current is 8.0 A
 
out of the page, and each curve is a circular path. (a) Find  B  d  for each path,
C
assuming that each integral is to be evaluated in the counterclockwise direction. (b)
Which path, if any, can be used to find the combined magnetic field of these currents?
 
Picture the Problem We can use Ampère’s law,  B  d   0 I C , to find the line
C
 
integral  B  d for each of the three paths.
C
(a) Noting that the angle between


B and d  is 180, evaluate
 
B
  d for C1:
C
 
B
  d    0 8.0 A
C1
The positive tangential direction on C1
is counterclockwise. Therefore, in
accord with convention (a right-hand
rule), the positive normal direction for
the flat surface bounded by C1 is out of
 
the page.  B  d is negative because the
C
current through the surface is in the
negative direction (into the page).
Noting that the net current bounded
 
by C2 is zero, evaluate  B  d :
 
B
  d   0 8.0 A  8.0 A  0
C2
C

Noting that the angle between B and

 
d  is 0, Evaluate  B  d for C3:
 
B
  d    0 8.0 A
C3
C

(b) None of the paths can be used to find B because the current configuration

does not have cylindrical symmetry, which means that B cannot be factored out
of the integral.
49 ••
[SSM] A long cylindrical shell has an inner radius a and an outer radius b
and carries a current I parallel to the central axis. Assume that within the material of the
shell the current density is uniformly distributed. Find an expression for the magnitude of
the magnetic field for (a) 0 < R < a,
(b) a < R < b, and (c) R > b.
Picture the Problem We can use Ampère’s law to calculate B because of the
high degree of symmetry. The current through C depends on whether R is less
than the inner radius a, greater than the inner radius a but less than the outer
radius b, or greater than the outer radius b.
(a) Apply Ampère’s law to a circular
path of radius R < a to obtain:
(b) Use the uniformity of the current
over the cross-section of the
conductor to express the current I
enclosed by a circular path whose
radius satisfies the condition
a < R < b:
Solving for I C  I ' yields:
Substitute in Ampère’s law to
obtain:

C


Br a  d   0 I C   0 0  0
and Br a  0
I'
I

2
2
 R a
 b  a2


2
I C  I'  I

B
C
a  R b


R2  a2
b2  a2

 d   Bar b 2R 
  0 I'   0 I
Solving for Ba<r<b yields:
Ba  r  b 
(c) Express IC for R > b:
IC  I
Substituting in Ampère’s law yields:
B

C
Solve for BR>b to obtain:
R b
B R b 
R2  a2
b2  a2
0 I R 2  a 2
2r b 2  a 2

 d   Br b 2R  0 I
0 I
2R
50 ••
Figure 27-57 shows a solenoid that has n turns per unit length and carries a
current I. Apply Ampère’s law to the rectangular curve shown in the figure to derive an
expression for the magnitude of the magnetic field. Assume that inside the solenoid the
magnetic field is uniform and parallel with the central axis, and that outside the solenoid
there is no magnetic field.
Picture the Problem The number of turns enclosed within the rectangular area is
na. Denote the corners of the rectangle, starting in the lower left-hand corner and
proceeding counterclockwise, as 1, 2, 3, and 4. We can apply Ampère’s law to
 
each side of this rectangle in order to evaluate  B  d .
C
Express the integral around the
closed path C as the sum of the
integrals along the sides of the
rectangle:








 B  d   B  d   B  d   B  d
C
12
23
34
 
  B  d
41
Evaluate


 B  d :
1 2

For the paths 2  3 and 4  1, B is
either zero (outside the solenoid) or

is perpendicular to d  and so:

For the path 3  4, B =0 and:


 B  d   aB
12
 
B
  d 
23
 
B
  d  0
41
 
B
  d  0
34


Substitute in Ampère’s law to
obtain:
 B  d   aB  0  0  0  aB
Solving for B yields:
B  0 nI
C
  0 I C  0 naI
51 •• [SSM] A tightly wound 1000-turn toroid has an inner radius
1.00 cm and an outer radius 2.00 cm, and carries a current of 1.50 A. The toroid is
centered at the origin with the centers of the individual turns in the z = 0 plane. In the z =
0 plane: (a) What is the magnetic field strength at a distance of 1.10 cm from the origin?
(b) What is the magnetic field strength at a distance of 1.50 cm from the origin?
Picture the Problem The magnetic field inside a tightly wound toroid is given by
B  0 NI 2r  , where a < r < b and a and b are the inner and outer radii of the
toroid.
Express the magnetic field of a
toroid:
B
 0 NI
2r
(a) Substitute numerical values and evaluate B(1.10 cm):
4 10
B1.10 cm  

N/A 2 10001.50 A 
 27.3 mT
2 1.10 cm 
7
(b) Substitute numerical values and evaluate B(1.50 cm):
B1.50 cm  
4 10

N/A 2 10001.50 A 
 20.0 mT
2 1.50 cm 
7