Question

A current is spread uniformly over the cross sectional area of a hollow infinite cylindrical shell with inner radius a and outer radius b. Find the magnitude of the magnetic field in the regions a)r<a (5 pts) b) a<r<b (15 pts) c) b<r (5 pts) I а b

Answer 1

Solution :

According to Ampere's Law :

$\oint B\cdot dl=\mu _0I_{enc}$

Here, Since the current I is spread uniformly.

Thus, The current density in the wire is given by :

$J=\frac{I}{A}=\frac{I}{\pi b^2-\pi a^2}=\frac{I}{\pi (b^2-a^2)}$

.

Part (a) Solution :

for, r < a

Here, I_enc = 0 A

Therefore :

$\oint B\cdot dl=\mu _0I_{enc}=\mu _0\ (0\ A)=0$

Thus, B = 0 T

.

Part (b) Solution :

for, a < r < b

Here,

$I_{enc}=JA=\left [ \frac{I}{\pi (b^2-a^2)} \right ]\left [ \pi (r^2-a^2) \right ]= \frac{I}{(b^2-a^2)}\cdot (r^2-a^2)$

Therefore, From Ampere's Law :

$B(2\pi r)= \mu _0\cdot \frac{I}{(b^2-a^2)}\cdot (r^2-a^2)$

$B= \frac{\mu _0I}{(2\pi r)}\cdot \frac{ (r^2-a^2)}{(b^2-a^2)}$

.

Part (c) Solution :

for, b < r

Here, I_enc = I

Thus :

$B\cdot (2\pi r)=\mu _0I_{enc}=\mu _0\ (I)$

$B=\frac{\mu _0I}{2\pi r}$