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What is the value of induced magnetic field?

  1. Jul 14, 2016 #1
    1. The problem statement, all variables and given/known data
    Two coencenteric metalic shell has inner radius $r_1$ outer radius $r_2$. We place along axis infinity wire has $\lambda$ charge in per unit length. The inner region of metalic shells inserted with relative permabilitty coefficent $\epsilon$. This system rotates with $\omega$ angular velocity. What is the value of induced magnetic field?
    [figure:http://i.stack.imgur.com/ywe4p.jpg

    3. The attempt at a solution
    \begin{equation}
    \rho_b=-\nabla.{P}
    \end{equation}
    \begin{equation}
    P=(\epsilon-1)\epsilon_0.E
    \end{equation}
    \begin{equation}
    E=\lambda\div({2\pi.\epsilon\epsilon_0.r})
    \end{equation}
    If we placed to first equation we get:
    \begin{equation}
    \rho_b=0
    \end{equation}
    \begin{equation}
    \sigma_b=P.n
    \end{equation}
    where is n is unit vector
    for outer metalic shell:
    \begin{equation}
    \sigma_b(r_2)=P(r_2)=(\epsilon-1)\epsilon_0\lambda\div({2\pi.\epsilon\epsilon_0.r_2})
    \end{equation}
    for inner metalic shell:
    \begin{equation}
    \sigma_b(r_1)=-P(r_1)=-(\epsilon-1)\epsilon_0\lambda\div({2\pi.\epsilon\epsilon_0.r_1})
    \end{equation}
    For charge for the inner shell
    \begin{equation}
    \sigma_1.2\pi.r_1.h=q_1
    \end{equation}
    For charge for the outer shell
    \begin{equation}
    \sigma_2.2\pi.r_2.h=q_2
    \end{equation}
    For current
    \begin{equation}
    i=q/T
    \end{equation}
    when we calculate current inner and outer's effect of magnetic field canceled. Where is the mistake if there is? Or what variables cause to magnetic field? HELP PLEASE
     
  2. jcsd
  3. Jul 14, 2016 #2

    TSny

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    Gold Member

    Your work looks good, but you didn't show how you are calculating the B field. Are you getting that B = 0 everywhere?
     
  4. Jul 15, 2016 #3
    sorry, i will continue:
    $\sigma_2$ positive, $\sigma_1$ is negative; so current of outer sphere is anticlockwise, current of inner sphere is clockwise.
    \begin{equation}
    i_1=q_1\div{T}=\sigma_1.2.\pi.r_1h\div{2\pi/\omega}=\sigma_1.\omega.r_1.h
    \end{equation}
    \begin{equation}
    i_2=q_2\div{T}=\sigma_2.2.\pi.r_2h\div{2\pi/\omega}=\sigma_2.\omega.r_2.h
    \end{equation}
    The direction of magnetic field is up. now lets consider the r<$r_1$ region. We think this system like selenoid.
    \begin{equation}
    B=\mu_0.i\div{h}
    \end{equation}
    In the this region magnetic field directions are opposite and canceled. Also in the region $r>r_2$ canceled. But the magnetic field in the region$r_2>r>r$ they don't canceled. (I found my mistake)
    \begin{equation}
    B_1=0 for this region.TSny pointed out.
    \end{equation}
    \begin{equation}
    B_2=\mu_0.\sigma_2.\omega.r_2.h\div{h}=\mu_0.\sigma_2.\omega.r_2
    \end{equation}
    And i obtain
    \begin{equation}
    B=\mu_0.\omega.\sigma_2.r_2
    \end{equation}
    If i put \sigma_2. value:
    \begin{equation}
    B=\mu_0.\omega.(\epsilon-1)\lambda\div({2\epsilon.\pi})
    \end{equation}
    Do you see any error in my calculations?
     
    Last edited: Jul 15, 2016
  5. Jul 15, 2016 #4

    TSny

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    That all looks correct to me.
     
  6. Jul 15, 2016 #5
    so thanks
     
  7. Jul 15, 2016 #6

    TSny

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    Oh wait. Sorry. I think you are right that B = 0 for ##r< r_1## and ##r > r_2##. But I don't think you have the correct expression for ##r_1 < r < r_2.##.

    For an ideal solenoid, what is B outside the solenoid?
     
  8. Jul 15, 2016 #7
    ohhh yes you are correct i'll edit now.
     
  9. Jul 15, 2016 #8

    TSny

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    Gold Member

    OK.
     
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