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Magnetic Induction (Non-uniform Current)

  1. Aug 24, 2013 #1
    1. The problem statement, all variables and given/known data

    long wire of radius R0 carries a current density j given by

    [Broken]​

    Find the magnetic induction B inside and outside the wire.

    2. Relevant equations

    Current density: ##J=\frac{I}{A}##

    Ampere's law: ##\oint B.dl = \mu_0 I_{enc}##

    3. The attempt at a solution

    For the magnetic field inside (ρ<R0) using Ampere's law:

    ##\oint B . dl = B 2 \pi \rho = \mu_0 I_{enc}##

    Now I'm not sure what to use as Ienc. If I use the relationship

    ##I_{enc}=JA=j_0 \frac{\rho}{R} \pi \rho^2 = j_0 \frac{\rho^3 \pi}{R}##

    I get

    ##B= \frac{\mu_0 j_0 \rho^2 }{2 R_0}##

    But if I integrate (in cylindrical coordinates) I will get a different value for Ienc:

    ##I_{enc}= \int \frac{j_0 \rho}{R_0} (\rho \ d \rho \ d \phi) = \frac{j_0}{R_0} 2 \pi \int^\rho_0 \rho^2 d \rho = \frac{2 \pi j_0 \rho^3}{3R_0}##

    Therefore I get a different value for B:

    ##B=\frac{\mu_0 j_0 \rho^2}{3R_0}##

    So which method is correct? :confused:

    And for the magnetic field outside, I get ##B=0## since the RHS of Ampere's equation is 0. But shouldn't the magnetic field inside a conducting wire be zero, and non-zero outside it?

    Any explanation would be greatly appreciated.
     
    Last edited by a moderator: May 6, 2017
  2. jcsd
  3. Aug 24, 2013 #2
    Because J is a function of radius. [tex]I=\int { J\cdot dA } [/tex] The second method looks fine.

    For the second question: What is the enclosed current outside? Is it zero?
     
  4. Aug 28, 2013 #3

    rude man

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    You don't need to integrate over phi explicitly. Using r in place of ρ:

    A differential annulus of area dA = 2πr dr contains current di = j(r)dA = 2πr j(r) dr
    so current within radius r = 2πj0/R00r r2 dr
    = 2πj0r3/3R0
    = 2πrB/μ0 etc.
     
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