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. `40 pi rad//s^(2)`
B. `20 pirad//s^(2)`
C. `5pirad//s^(2)`
D. `15 pirad//s^(2)`

Answer 1

Correct Answer - A
Magnetic momentum of loop

`vec(M)=ivec(A)=4xxpi(0.5)^(2)-(-hatk)`
`=-4pi(0.5)^(2)hatk`
`tau=vec(M)xxvec(B)`
`tau=(-pihatk)xx(10 hati)=-10 pihatj`
So, axis of rotation is along `vec(tau)`, i.e, along negative `y`-direction.
Moment of inertia of ring about `y`-axis
`I=1/2MR^(2)=1/2xx2xx(0.5)^(2)=1/4 kgm^(2)`
`tau=Ialpha` so `alpha=(tau)/I=(10pi)/(1//4)=40 pida//s^(2)`
`alpha=40 pi rad//s^(2)`