A bar magnet of magnetic moment 2.0 J/T lies aligned with the direction of uniform magnetic field of 0.25 T. What is the amount of work required to turn the magnet so as to align its magnetic moment normal to the field direction ?

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0.5 J

1.5 J

2.0 J

1.0 J

answer is A.

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Detailed Solution

We know that potential energy of a bar magnet with its magnetic moment M inclined at an angle O with magnetic field B is
$\begin{array}{l} U = - MBcos θ \\ ∴ U_{0} = - MBcos 0 = - MB \\ U_{90} = - MBcos 90 = 0 \end{array}$
So, work done
$\begin{array}{l} = U_{90} - U_{0} = 0 - (- MB) = MB \\ = 2 \cdot 0 \times 0 \cdot 25 = 0 \cdot 5 J \end{array}$

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