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Question

A bar magnet of magnetic moment 1.5JT1 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)?

Solution
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Here m=1.5J/T,B=0.22T
(a)
W=mB(cosθ2cosθ1)
(i)
θ1=0(along the field)
θ2=90(perpendicular to the field)
W=0.33(01)J=0.33 J
(ii)
θ1=0,θ2=180
W=1.5×0.22(cos180cos0)=0.66 J

(b)
Torque =mBsinθ
(i)
θ=90
τ=0.33 Nm
(ii)
τ=mBsin180
=0Nm

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