Question

A conducting ring of mass 2kg and radius `0.5m` is placed on a smooth plane. The ring carries a current of `i=4A`. A horizontal magnetic field `B=10 T` is switched on at time `t=0` as shown in fig The initial angular acceleration of the ring will be
.
A. `40pirad//s^(2)`
B. `20pirad//s^(2)`
C. `5pirad//s^(2)`
D. `15pirad//s^(2)`

Answer 1

Correct Answer - A
Magnetic moment of loop

`vecM=ivecA=4xxpi(0.5)^(2)(-hatk)`
`=-4pi (0.5)^(2)hatk`
`tau=vecMxxvecB`
`tau=(-pihatk)xx(10hati)=-10pihatj`
So, axis of rotation is along `vectau`, i.e, along negative `y`- direction.
Moment of inertia of ring about `y`-axis
`I=(1)/(2)MR^(2)=(1)/(2)xx2xx(0.5)^(2)=(1)/(4)kgm^(2)`
`tau=Ialpha` so `alpha=(tau)/(I)=(10pi)/(1//4)=40pirad//s^(2)`
`alpha=40pirad//s^(2)`