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Forever_hard
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Homework Statement
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.
2. Answer:
T' = \frac{T}{\sqrt{1 + \frac{B^2 r^4 T^2}{4LJ}}}
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 :)
help me, please
