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What is the value of induced magnetic field?

  • #1

Homework Statement


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

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
 

Answers and Replies

  • #2
TSny
Homework Helper
Gold Member
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Your work looks good, but you didn't show how you are calculating the B field. Are you getting that B = 0 everywhere?
 
  • #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:
  • #4
TSny
Homework Helper
Gold Member
12,321
2,808
That all looks correct to me.
 
  • #5
so thanks
 
  • #6
TSny
Homework Helper
Gold Member
12,321
2,808
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?
 
  • #7
ohhh yes you are correct i'll edit now.
 
  • #8
TSny
Homework Helper
Gold Member
12,321
2,808

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