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Magnetic Field (Hollow Copper Pipe)

  1. Aug 25, 2013 #1
    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. jcsd
  3. Aug 26, 2013 #2
    Looks like your answer are correct.
     
  4. Aug 28, 2013 #3

    rude man

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    Homework Helper
    Gold Member

    (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?
     
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