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Permanent Magnets — Physics STD 11 Science — Question

Gujarat BoardEnglish MediumSTD 11 SciencePhysicsPermanent Magnets4 Marks
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 $1.2 \times 10^{-4} \mathrm{~kg}-\mathrm{m}^2$ and the earth's horizontal magnetic field is $30\mu \text{T}.$ Find the magnetic moment of the magnet.
✓

Answer

We know: $\text{v}=\frac{1}{2\pi}\sqrt{\frac}{\text{mB}_\text{H}}{\text{I}}$For like poles tied together
$M=M_1-M_2$
For unlike poles $\mathrm{M}^{\prime}=\mathrm{M}_1+\mathrm{M}_2$
$\frac{\text{v}_1}{\text{v}_2}=\sqrt{\frac{\text{M}_1-\text{M}_2}{\text{M} _1-\text{M}_2}}\Rightarrow\Big(\frac{10}{2}\Big)^2$
$=\frac{\text{M}_1-\text{M}_2}{\text{M}_1+\text{M}_2}\Rightarrow25=\frac{\text{M}_1-\text{M}_2}{\text{M}_1+\text{M}_2}$
$\Rightarrow\frac{26}{24}=\frac{2\text{M}_1}{2\text{M}_2}\Rightarrow\frac{\text{M}_1}{\text{M}_2}=\frac{13}{12}$

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