Magnetic Properties of Matter: Practice Problem & Solution
7. 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 $60^{\circ}$ is $W$. Now the torque required to keep the magnet in this new position is: (2016 - II)
Solution Explained:
To solve this problem, we apply the core principles of Magnetic Properties of Matter. Understanding the underlying formula is key to arriving at the correct answer below:
Work done, $W = MB(1 - \cos 60^{\circ}) = \frac{MB}{2} \Rightarrow MB = 2W$.
Torque required, $\tau = MB \sin 60^{\circ} = 2W \times \frac{\sqrt{3}}{2} = \sqrt{3}W$.
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