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 field direction?
(b) What is the torque on the magnet in cases (i) and (ii)?

Answer 1

Here $m=1.5J/T,B=0.22T$

(a)

$W=-mB(cos{\theta }_2-cos{\theta }_1)$

(i)

${\theta }_1=0^{\circ }$(along the field)

${\theta }_2={90}^{\circ }$(perpendicular to the field)

$W=-0.33(0-1)J=0.33J$

(ii)

${\theta }_1=0^{\circ },{\theta }_2={180}^{\circ }$

$W=-1.5\times 0.22(cos{180}^{\circ }-cos{0}^{\circ })=0.66J$

(b)

Torque $=mBsin\theta$

(i)

$\theta ={90}^{\circ }$

$⟹\tau =0.33Nm$

(ii)

$\tau =mBsin{180}^{\circ }$

$=0Nm$