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:
$W_{90^{\circ}} = MB(1 - \cos 90^{\circ}) = MB(1 - 0) = MB$.
$W_{60^{\circ}} = MB(1 - \cos 60^{\circ}) = MB(1 - 0.5) = 0.5MB$.
Since $W_{90^{\circ}} = n W_{60^{\circ}}$, we have $MB = n(0.5MB) \Rightarrow n = 2$.
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