Question

A bar magnet takes π/10 second the complete one oscillation in an oscillation magnetometer. The moment of inertia of the magnet about the axis of rotation is 1.2 × 10⁻⁴ kg m² and the earth's horizontal magnetic field is 30 μT. Find the magnetic moment of the magnet.

Answer 1

Given:
Time taken by the bar magnet to complete one oscillation, T = $\frac{\mathrm{\pi }}{10}\mathrm{s}$
Moment of inertia of the magnet about the axis of rotation, I = 1.2 × 10⁻⁴ kgm²
Horizontal component of Earth's magnetic field, B_H = 30 μT
Time period of oscillating magnetometer $\left(T\right)$ is given by
$T=2\mathrm{\pi }\sqrt{\frac{I}{M{B}_H}}$
Here,
M = Magnetic moment of the magnet
On substituting the respective values, we get
$$\begin{gathered} \frac{\mathrm{\pi }}{10}=2\mathrm{\pi }\sqrt{\frac{1.2\times {10}^{-4}}{\mathrm{M}\times 30\times {10}^{-6}}} \\ \Rightarrow {\left(\frac{1}{20}\right)}^2=\frac{1.2\times {10}^{-4}}{M\times 30\times {10}^{-6}} \\ \Rightarrow M=\frac{1.2\times {10}^{-4}\times 400}{30\times {10}^{-6}} \\ \Rightarrow M=16\times {10}^2\mathrm{A}-{\mathrm{m}}^2 \\ \Rightarrow M=1600\mathrm{A}-{\mathrm{m}}^2 \end{gathered}$$