# Find Period of rotation of the copper ring in a Magnetic Field

1. Aug 3, 2012

### Forever_hard

1. The problem statement, all variables and given/known data

A Copper Ring with Radius $r$ and mass $m$ hangs by a thread and rotates with a period $T$. Ring's coefficient of self inductance is $L$ . What would be a new rotation period of ring, if it was in a horisontal uniform magnetic $B$ field, which is parallel to Ring's plane on a picture? Ring's moment of inertia(axis goes through the centre of mass of Ring) is $J$. Ring has no electrical resistance.

$T' = \frac{T}{\sqrt{1 + \frac{B^2 r^4 T^2}{4LJ}}}$

3. The attempt at a solution

Ring has no resistance, then magnetic flux is:

$\Phi = LI + \vec{B}\vec{S} = const$

initial conditions:

$\vec{B}\vec{S} = 0 ; I = 0$

$LI = - BScos \angle(B,S) \rightarrow I = -\frac{BScos \angle(B,S)}{L}$

And after this stage I have big troubles :(
I think, I have to use the Magnetic and angular moments but I don't know how :(

The only one idea I have is following:
$p_m - <p_{el}> = <p_m'>$
$Jw - IS = Jw'$
$\frac{J2\pi}{T} + \frac{BS^2<cos \angle(B,S)>}{L} = \frac{J2\pi}{T'}$
$<cos \angle(B,S)> = 0$

ofc, It's the wrong solution :)