The work done in rotating the magnet from the direction of the uniform field by 600 is W. The potential energy of the magnet at the position where it experiences half the maximum couple position is

W=MB(1−cos60)=MB2τ=MBsinθMB2=MBsinθ1⇒θ1=30And U = -M.B = -MB cos θ1⇒U=−3W

The work done in rotating the magnet from the direction of the uniform field by 600 is W. The potential energy of the magnet at the position where it experiences half the maximum couple position is

W=MB(1−cos60)=MB2τ=MBsinθMB2=MBsinθ1⇒θ1=30And U = -M.B = -MB cos θ1⇒U=−3W

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The work done in rotating the magnet from the direction of the uniform field by 60⁰ is W. The potential energy of the magnet at the position where it experiences half the maximum couple position is

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$- \sqrt{3} W$

$\frac{\sqrt{3}}{2} W$

W/2

answer is D.

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Detailed Solution

$W = MB (1 - \cos 60) = \frac{MB}{2}$
$τ = MBsinθ$
$\frac{MB}{2} = {MBsinθ}_{1} \Rightarrow θ_{1} = 30$
And U = -M.B = -MB cos $θ_{1}$
$\Rightarrow U = - \sqrt{3} W$

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