A bar magnet with a magnetic moment $5.0 {Am}^{2}$ is placed in parallel position relative to a magnetic field of 0.4 T. The amount of required work done in turning the magnet from parallel to antiparallel position relative to the field direction is _________

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zero

answer is B.

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Detailed Solution

The potential energy stored in a bar magnet is given by

$\begin{array}{l} U = - \vec{M} \cdot \vec{B} \end{array}$

Therefore, work done to alter the position of the bar magnet can be calculated by $w = ΔU = U_{f} - U_{i} = - mBcos π - (- MBcos 0)$

$\Rightarrow w = 2 mB = 2 \times 5 \times 0.4 = 4 J$

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