# Magnetic Field (Hollow Copper Pipe)

1. Aug 25, 2013

### roam

1. The problem statement, all variables and given/known data

The figure below shows a long, hollow copper pipe.

[Broken]​

The inner radius of the pipe is a and the outer radius is b. A uniform current I flows in the walls of the pipe. You may assume that the permeability of copper is the same as free space, that is, $\mu_0$. Use the integral form of Ampere's law to find:

(i) The magnetic induction B for r<a

(ii) The magnetic induction B for a<r<b

(iii) The magnetic induction B for r>b

2. Relevant equations

Ampere's law in integral form

3. The attempt at a solution

(i) Since I enclosed is 0

$\oint B . dl = B 2 \pi r = \mu_0 I \implies B=0$

(ii) $B 2 \pi r = \mu_0 I$

Here I think the I enclosed is $\frac{I'}{I}= \frac{\pi r^2}{\pi (b-a)^2} \implies I' = \frac{I r^2}{(b-a)^2}$. So

$B 2 \pi r = \mu_0 \frac{I r^2}{(b-a)^2} \implies B= \frac{\mu_0 I r}{2 \pi (b-a)^2}$.

(iii) For the magnetic induction outside

$B 2 \pi r = \mu_0 I \implies B = \frac{\mu_0 I}{2 \pi r}$

Is my working correct? I'm mostly doubtful about (i) and (ii). For (ii) shouldn't B be equal to zero inside the material, since copper is diamagnetic?

Thanks.

Last edited by a moderator: May 6, 2017
2. Aug 26, 2013

### Electric Red

3. Aug 28, 2013

### rude man

(b) is incorrect.

Let r = b, then your expression gives B = μ0Ib/2π(b - a)2
whereas you know the answer should be the same as for part c which you got right.

You need to look at this part carefully. Hint: the difference in cross-sectional area between radii r and a is not π(r - a)2, r > a. What is it?