Question

A bar magnet of magnetic moment $1.5J{T}^{-1}$ lies aligned with the direction of a uniform magnetic field of 0.22 T.

(A) What is the amount of work required by an external torque to turn the magnet so as to align its magnetic moment:(i) normal to the field direction , (ii) opposite to the filed direction?

(B) What is the torque on the magnet in case (i) and (ii)

Answer 1

It is given that magnetic moment is

$M=1.55J{T}^{-1}$

$B=0.22T$

(A) Amount of work required by an external torque to turn the magnet so as to align its magnetic moment

(1) Normal to the field direction is

$W_1=-MB[\cos 90-\cos 0]$

$=MB$

$=1.55\times 0.22$

$=0.341N-m$

(2) Amount of work required by an external torque to turn the magnet so as to align its magnetic moment opposite to field direction.

$W_2=-MB[\cos 180-\cos 0]$

$W_2=2MB$

$W_2=0.682N-m$

(B) Torque acting on case (1)

$\tau =MB\sin 90$

$=0.341J$

Torque acting in case (2)

$\tau =MB\sin 180$

$=0J$