- #1
Forever_hard
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Homework Statement
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]
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