A bar magnet having a magnetic moment of 2 × 104 J T ⁻¹ is free to rotate in a horizontal plane. A horizontal magnetic field B = 6 × 10 - 4 T exists in the space. The work done in taking the magnet slowly from a direction parallel to the field to a direction 60° from the field is (a) 12 J (b) 6 J (c) 2 J (d) 0.6 J

Answer: (b). W = MB(cos θ₁ - cos θ₂) = MB(1 - cos 60°) W = 2 × 10⁴ × 6 × 10⁻⁴ × 0.5 = 6 J So the answer is (b) 6 J.

A bar magnet having a magnetic moment of 2 × 104 J T ⁻¹ is free to… | NEET 2009 (AIPMT)

Q28NEET 2009 · AIPMTEasy

Field of a Magnet & Magnet in a Uniform Field

A bar magnet having a magnetic moment of 2 $\times$ 104 J T ${}^{-1}$ is free to rotate in a horizontal plane. A horizontal magnetic field B = 6 $\times$ 10 $-$ 4 T exists in the space. The work done in taking the magnet slowly from a direction parallel to the field to a direction 60° from the field is

Show answer and solution

Answer: (b)

$W = MB(\cos\theta_1 - \cos\theta_2) = MB(1 - \cos 60^\circ)$
$W = 2 \times 10^{4} \times 6 \times 10^{-4} \times 0.5 = 6\ \text{J}$
So the answer is (b) 6 J.

Report a mistake

Field of a Magnet & Magnet in a Uniform Field question from NEET 2009 (AIPMT)

More questions on Field of a Magnet & Magnet in a Uniform Field