Question

A bar magnet of magnetic moment $10^{4}\, J / T$ is free to rotate in a horizontal plane. The work done in rotating the magnet slowly from a direction parallel to a horizontal magnetic field of $4 \times 10^{-5}\, T$ to a direction of $60^{\circ}$ from the field will be

• A. $0.2\, J$
• B. $2\, J$
• C. $4.18\, J$
• D. $2\times {10}^{2}\,J$

Answer 1

Answer: (A)

Magnetic moment $=10^{4} J / T$
$B=4 \times 10^{-5} T,\, \theta=60^{\circ}$
Work done in moving the magnet in uniform magnetic field.
$W =M B(1-\cos \theta)$
$=10^{4} \times 4 \times 10^{-5}\left(1-\cos 60^{\circ}\right)$
$=0.2\, J$