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)

$0.2\, J$

Answer 2

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. $W =M B(1-\cos \theta)$ $=10^{4} \times 4 \times 10^{-5}\left(1-\cos 60^{\circ}\right)$ $=0.2\, J$