Question

A bar magnet has a magnetic moment of $2.5\,JT^{-1}$ and is placed in a magnetic Held of 0.2T. Work done in turning the magnet from parallel to antiparallel position relative to field direction is

• A. 0.5 J
• B. 1 J
• C. 2 J
• D. 0 J

Answer 1

Answer: (B)

1 J

Answer 2

Work done, W = - MB(cos $\theta_2$ - cos $\theta_1$) = - MB(cos 180$\circ$ - cos 0$\circ$) = - MB(-1 - 1) = - MB( - 2) = 2 MB = 2 $\times$ 2.5 $\times$ 0.2 = 1 J