Magnetic Effects & Magnetism

JEE Main level, single correct, medium

245 of 276

Question 245 of 276

A bar magnet with a magnetic moment 5.0Am25.0A{m}^{2} is placed in parallel position relative to a magnetic field of 0.4T0.4T. The amount of required work done in turning the magnet form parallel to antiparallel position relative to the field direction is ________.

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Answer: (A) 4J4J

For a bar magnet, the work done in turning from parallel to antiparallel position relative to the field direction is
W=MB(cos⁡θ1−cos⁡θ2)W=MB\left(\cos {\theta }_{1}-\cos {\theta }_{2}\right), here, θ1=0∘andθ1=180∘{\theta }_{1}=0^{\circ}and{\theta }_{1}=180^{\circ}.
W=−MBcos⁡180∘(−mBcos⁡0∘)W=-MB\cos 180^{\circ}\left(-mB\cos 0^{\circ}\right)
W=2MB=2×5×0.4=4JW=2MB=2\times 5\times 0.4=4J

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