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 in rotating the magnet from equilibrium is $W = MB(1 - cos 60^{circ}) = frac{MB}{2}$, giving $MB = 2W$. The torque required in this position is $tau = MB sin 60^{circ} = (2W)left(frac{sqrt{3}}{2}right) = sqrt{3}W$.
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