A bar magnet of magnetic moment 1.5JT−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)?
Here m=1.5J/T,B=0.22T
(a)
W=−mB(cosθ2−cosθ1)
(i)
θ1=0∘(along the field)
θ2=90∘(perpendicular to the field)
W=−0.33(0−1)J=0.33 J
(ii)
θ1=0∘,θ2=180∘
W=−1.5×0.22(cos180∘−cos0∘)=0.66 J
(b)
Torque =mBsinθ
(i)
θ=90∘
⟹τ=0.33 Nm
(ii)
τ=mBsin180∘
=0Nm