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Bar magnet in a uniform field – torque and work
Concepts tested here
- magnet-torque-energy
All Questions
2016 Phase II 1 question
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A bar magnet is hung by a thin cotton thread in a uniform horizontal magnetic field and is in equilibrium state. The energy required to rotate it by 60° is W. Now the torque required to keep the magnet in this new position is:
Divide $MB\sin\theta$ by $MB(1-\cos\theta)$.
$\tau = MB\sin60^\circ$ ...(1)
$W = MB(1 - \cos60^\circ)$ ...(2)
From (1) and (2): $\frac{\tau}{W} = \frac{\sqrt{3}/2}{1/2} \Rightarrow \tau = \sqrt{3}W$