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

JEE MainJEE Main 2022Magnetism and Matter

Solution:

Work done $= MB \left(\cos \theta_{1}-\cos \theta_{2}\right)$
$\theta_{1}=0^{\circ}, \theta_{2}=60^{\circ} $
$=2 \times 10^{5} \times 14 \times 10^{-5}(1-1 / 2) $
$=14 \,J$