NEET-XII-Physics
36: Permanent Magnets
- #21A 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.Ans : 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, BH = 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
`` \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}``
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