The radius of a circular coil having 100 turns is 5 cm and a current of `0.5A` is flowing through this coil. If it is placed in a uniform magnetic field of stregth `0.001` T, then what torque will act on the coil, when the plane of the coil is inclined at `30^(@)` with the magnetic field,

Explanation

The work `W_(a)` done by the applied torque would be equal to the change in the coil.s orientation energy due to its change in orientation.
Calculations : From `(W_(a) = U_(f) - U_(i))`, we find
`W_(a) = U(90^(@) - U(0^(@)))`
`= - mu B cos 90^(@) - (-mu B cos 0^(@)) = 0 + mu B`
`= mu B`
Substituting for `mu` from , `(mu = NiA)`, we find that
`W_(a) = (NiA)B`
`= (250) (100 xx 10^(-6)A)(2.52 xx 10^(-4) m^(2)) (0.85 T)`
`= 5.355 xx 10^(-6)J ~~ 5.4 mu J`.
Similarly, we can show that to change the orientation by another `90^(@)`, so that the dipole moment is opposite the field, another `5.4 mu J` is required.