A bar magnet having a magnetic moment of 2.0 × 105 JT ⁻¹, is placed along the direction of uniform magnetic field of magnitude B = 14 × 10 - 5 T. The work done in rotating the magnet slowly through 60 ° from the direction of field is: (a) 14 J (b) 8.4 J (c) 4 J (d) 1.4 J

Answer: (a). W = MB(1 - cos 60°) W = 2.0 × 10⁵ × 14 × 10⁻⁵ × 0.5 = 14 J So the answer is (a) 14 J.

A bar magnet having a magnetic moment of 2.0 × 105 JT ⁻¹, is placed… | JEE Main 2022 (26 Jun, Shift 2)

Q11JEE Main 2022 · 26 JUN · SHIFT 2Easy

Field of a Magnet & Magnet in a Uniform Field

A bar magnet having a magnetic moment of 2.0 $\times$ 105 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:

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Answer: (a)

$W = MB(1 - \cos 60^\circ)$
$W = 2.0 \times 10^{5} \times 14 \times 10^{-5} \times 0.5 = 14\ \text{J}$
So the answer is (a) 14 J.

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