Question 3 calculate current enclosed by the amperian

Question 3: Calculate current enclosed by the Amperian loop. Be careful about signs!
Remember that

J is non-uniform so you will need to compute the integral
I
enc
=

J ⋅ ˆ n dA
∫∫
.
Question 4: Your answer to Question 3 should be zero when r = a
.
When r = b , the total
current is I in the wire. Find an expression for I
.
Question 5: What is

B ⋅ d

s
 ∫
? (That is, evaluate the integral, the left hand side of
Ampere’s law)
Question 6: Now apply Ampere’s Law. Write down a vector expression for the magnetic
field in the region a < r < b . Be sure to draw any necessary unit vectors on your above
figure showing your Amperian loop.


B ⋅ d

s
 ∫
= μ
0
I
enc
⇒ B (2 π r ) =
μ
0
2 πα
3
( r
3
− a
3
)
Therefore

B = B
ˆ
ϕ = μ
0
α
3
( r 3 − a 3 )
1
r
ˆ
ϕ ; a < r < b .
Question 7: Repeat the steps above to find a vector expression for the magnetic field in
the region r < a .
0 .
Question 8: Repeat the steps above to find a vector expression for the magnetic field in
the region r > b .
Solving 7-4

The magnetic field for the region r > b is given by

B = B
ˆ
ϕ = μ
0
α
3
( b 3 − a 3 )
1
r
ˆ
ϕ
Question 9: Make a plot of the magnitude of the magnetic field vs. distance from the
central axis, i.e. B vs. r . Clearly label any special values.
Solving 7-5

Problem 2: Current Slabs
The figure below shows two slabs of current. Both slabs of current are infinite in the x
and
z
-directions, and have thickness d in the y -direction. The top slab of current is
located in the region 0 < y < d and has a constant current density

J
out
= J ˆ z , directed out
of the plane of the figure below. The bottom slab of current is located in the region
− d < y < 0 and has a constant current density ˆ
in
J = − J z

, directed into the plane of the
figure below. There is a discontinuity in the current density at y = 0
(a) What is the direction and magnitude of the magnetic field for y d > ? Briefly
explain your reasoning. Clearly indicate your choice of unit vectors.
-
.
Solving 7-6