Magnetic Properties of Matter: Practice Problem & Solution
8. A magnetic needle suspended parallel to a magnetic field requires $sqrt{3} text{ J}$ of work to turn it through $60^{circ}$. The torque needed to maintain the needle in this position will be: (2012 Mains)
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 is $W = MB(1 - cos 60^{circ}) = frac{MB}{2} = sqrt{3}$, which implies $MB = 2sqrt{3} text{ J}$. The torque required is $tau = MB sin 60^{circ} = 2sqrt{3} times frac{sqrt{3}}{2} = 3 text{ J}$.
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