Magnetic Properties of Matter: Practice Problem & Solution
18. The work done in turning a magnet of magnetic moment $M$ by an angle of $90^{circ}$ from the meridian, is $n$ times the corresponding work done to turn it through an angle of $60^{circ}$. The value of $n$ is given by (1995)
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 for $90^{circ}$ is $W_1 = MB(1 - cos 90^{circ}) = MB$. Work done for $60^{circ}$ is $W_2 = MB(1 - cos 60^{circ}) = frac{MB}{2}$. Thus, $W_1 = 2 W_2$, which gives $n = 2$.
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