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# 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)?

Solution
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#### 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

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