# What is the value of induced magnetic field?

## 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

\rho_b=-\nabla.{P}

P=(\epsilon-1)\epsilon_0.E

E=\lambda\div({2\pi.\epsilon\epsilon_0.r})

If we placed to first equation we get:

\rho_b=0

\sigma_b=P.n

where is n is unit vector
for outer metalic shell:

\sigma_b(r_2)=P(r_2)=(\epsilon-1)\epsilon_0\lambda\div({2\pi.\epsilon\epsilon_0.r_2})

for inner metalic shell:

\sigma_b(r_1)=-P(r_1)=-(\epsilon-1)\epsilon_0\lambda\div({2\pi.\epsilon\epsilon_0.r_1})

For charge for the inner shell

\sigma_1.2\pi.r_1.h=q_1

For charge for the outer shell

\sigma_2.2\pi.r_2.h=q_2

For current

i=q/T

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

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TSny
Homework Helper
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?

sorry, i will continue:
$\sigma_2$ positive, $\sigma_1$ is negative; so current of outer sphere is anticlockwise, current of inner sphere is clockwise.

i_1=q_1\div{T}=\sigma_1.2.\pi.r_1h\div{2\pi/\omega}=\sigma_1.\omega.r_1.h

i_2=q_2\div{T}=\sigma_2.2.\pi.r_2h\div{2\pi/\omega}=\sigma_2.\omega.r_2.h

The direction of magnetic field is up. now lets consider the r<$r_1$ region. We think this system like selenoid.

B=\mu_0.i\div{h}

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)

B_1=0 for this region.TSny pointed out.

B_2=\mu_0.\sigma_2.\omega.r_2.h\div{h}=\mu_0.\sigma_2.\omega.r_2

And i obtain

B=\mu_0.\omega.\sigma_2.r_2

If i put \sigma_2. value:

B=\mu_0.\omega.(\epsilon-1)\lambda\div({2\epsilon.\pi})

Do you see any error in my calculations?

Last edited:
TSny
Homework Helper
Gold Member
That all looks correct to me.

so thanks

TSny
Homework Helper
Gold Member
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?

ohhh yes you are correct i'll edit now.

TSny
Homework Helper
Gold Member
$B=\mu_0.\omega.(\epsilon-1)\lambda\div({2\epsilon.\pi})$
OK.