Question

A bar magnet takes $\frac{\pi }{10}$second to complete one oscillation in an oscillation magnetometer. The moment of inertia of the magnet about the axis of rotation is $1.2\times {10}^{-4}$ kg $m^3$ and the earth's horizontal magnetic field is $30\mu T$. Find the magnetic moment of the magnet.

Answer 1

According to oscillation magnetometer,

T= $2\pi \sqrt{\frac{1}{M{B}_H}}$

$\Rightarrow \frac{\pi }{10}=2\pi \sqrt{\frac{1.2\times {10}^{-4}}{M\times 30\times {10}^{-6}}}$

${\left(\frac{1}{20}\right)}^2=\frac{1.2\times {10}^{-4}}{M\times 30\times {10}^{-6}}$

M = $\frac{1.2\times {10}^{-4}\times 400}{30\times {10}^{-6}}$

= $16\times {10}^{2}A-m^2$

= 1600 A-$m^2$