`L` is a smooth conducting loop of radius `l=1.0 m` & fixed in a horizontal plane.A conducting rod of mass `m=1.0 kg` and length slightly greater than `l` higed at the centre of the loop can rotate in the horizontal plane such that the free end slides on the rim of the loop.There is a uniform magnetic field of strength `B=1.0 T` directed vertically downward.The rod is rotated with angular velocity `omega_(0)=1.0 rad//s` and left.The fixed end of the rod and the rim of the loop are connected through a battery of `e.m.f` `E` a resistance `R=1 Omega`, and initially uncharged capacitor of capacitance `C=1.0 F` in series.Find
(i)the time dependence of `e.m.f` `M` such that the current `I_(0)=1.0 A` in the circuit is contant.
(ii)energy supplied by the battery by the time rod stops.

Answer

A, B, C, D

Explanation

(i)The magnetic equivalent of the loop-rod system.
`epsilon=1/2omegal^(2)`
where `omega`=angular velocity of rod at any time `t` then `omega=omega_(0)-alphat`
`because tau=Ialpha rArr (I_(0)lB)l/2=(ml^(2))/3alpha`
`rArr alpha=(3I_(0)B)/2=3/2rad//s^(2)`
applying loop law, `epsilon+E-I_(0)R-q/C=0`
`1/2Bl^(2)(omega_(0)-alphat)+E-I_(0)R-(I_(0)t)/C=0 rArrE=I_(0)(R+t/C)-1/2Bl^(2)(omega_(0)-alphat)=1/2+(7t)/4`
(ii)`omega=0=omega_(0)-3/2t_(0)` so `t_(0)=(2omega_(0))/3=2/3sec`.
Also energy supplied by battery `=underset(0)overset(t_(@))int EI_(0)di==underset(0)overset(2/3)int(1/2+(7t)/4)I_(0)dt=t/2+(7t^(2))/8| _(0) ^(2/3)=13/18J`