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The work done in turning a magnet of magnetic moment M by an angle of 90º from the meridian, is n times the corresponding work done to turn it through an angle of 60º. The value of n is given by

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$\frac{1}{2}$

$\frac{1}{4}$

answer is C.

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Detailed Solution

Angle of magnet $(θ) = 90 º$ and $60 º$

Work done in turning the magnet through 90º

$(W_{1}) = M B (\cos 0 º - \cos 90 º) = M B (1 - 0) = M B$

Similarly

$W_{2} = M B (\cos 0 º - \cos 60 º) = M B$ $(1 - \frac{1}{2}) = \frac{M B}{2}$

Therefore $W_{1} = 2 W_{2}$ or n = 2

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