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

Forever_hard

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

A Copper Ring with Radius [itex]r[/itex] and mass [itex]m[/itex] hangs by a thread and rotates with a period [itex]T[/itex]. Ring's coefficient of self inductance is [itex]L[/itex] . What would be a new rotation period of ring, if it was in a horisontal uniform magnetic [itex]B [/itex] 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 [itex]J[/itex]. Ring has no electrical resistance.

2. Answer:

[itex]T' = \frac{T}{\sqrt{1 + \frac{B^2 r^4 T^2}{4LJ}}}[/itex]

3. The attempt at a solution

Ring has no resistance, then magnetic flux is:

[itex]\Phi = LI + \vec{B}\vec{S} = const[/itex]

initial conditions:

[itex]\vec{B}\vec{S} = 0 ; I = 0[/itex]

[itex]LI = - BScos \angle(B,S) \rightarrow I = -\frac{BScos \angle(B,S)}{L}[/itex]

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:
[itex] p_m - <p_{el}> = <p_m'>[/itex]
[itex] Jw - IS = Jw'[/itex]
[itex] \frac{J2\pi}{T} + \frac{BS^2<cos \angle(B,S)>}{L} = \frac{J2\pi}{T'} [/itex]
[itex]<cos \angle(B,S)> = 0[/itex]

ofc, It's the wrong solution :)
help me, please

Forever_hard

oh, I forgot a picture:
http://s019.radikal.ru/i602/1208/c2/12b4695f3d49.jpg

Forever_hard

I have just solved this problem :)