Magnetic Effects & Magnetism

JEE Main level, single correct, medium

251 of 276

Question 251 of 276

A bar magnet having a magnetic moment of 2.0×105JT−12.0\times {10}^{5}J{T}^{-1}, is placed along the direction of uniform magnetic field of magnitude B=14×10−5TB=14\times {10}^{-5}T. The work done in rotating the magnet slowly through 60∘60^{\circ} from the direction of field is

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Answer: (A) 14J14J

Work done by the external force is equal to the change in the potential energy,
Wext=ΔU=Uf−Ui{W}_{ext}=\Delta U={U}_{f}-{U}_{i}
⇒Wext=−MBcos⁡θ2+MBcos⁡θ1\Rightarrow {W}_{ext}=-MB\cos {\theta }_{2}+MB\cos {\theta }_{1}
⇒Wext=MB(cos⁡θ1−cos⁡θ2)\Rightarrow {W}_{ext}=MB\left(\cos {\theta }_{1}-\cos {\theta }_{2}\right)
⇒Wext=14×10−5×2×105(cos⁡0∘−cos⁡60∘)\Rightarrow {W}_{ext}=14\times {10}^{-5}\times 2\times {10}^{5}\left(\cos 0^{\circ}-\cos 60^{\circ}\right)
⇒Wext=14×2×12=14J\Rightarrow {W}_{ext}=14\times 2\times \dfrac{1}{2}=14J

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