A bar magnet having a magnetic movement of 2 × 1 0 4 JT − 1 is free to rotate in a horizontal plane. A horizontal magnetic field B = 6 × 1 0 − 4 T exists in the space. The work done in taking the magn

Question

A bar magnet having a magnetic movement of $2 \times 1 0^{4} JT^{- 1}$ is free to rotate in a horizontal plane. A horizontal magnetic field $B = 6 \times 1 0^{- 4}$ $T$ exists in the space. The work done in taking the magnet slowly from a direction parallel to the field to a direction $6 0^{\circ}$ from the field is :NEET 2009 (Mains)

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Answer:

D

Solution:

$W = MB (cos θ_{1} - cos θ_{2}) = 2 \times 1 0^{4} \times 6 \times 1 0^{- 4} (cos θ - cos 6 0^{\circ}) = 12 \times \frac{1}{2} = 6 J$

TestHub · Accepted Answer

W ​ = MB ( cos θ 1 ​ − cos θ 2 ​ ) = 2 × 1 0 4 × 6 × 1 0 − 4 ( cos θ − cos 6 0 ∘ ) = 12 × 2 1 ​ = 6 J ​

Physics - Magnetism Question with Solution | TestHub

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