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A bar magnet takes $\frac{π}{10}$ seconds to complete one oscillation in vibration magnetometer. The moment of inertia of the magnet about the axis of rotation is $1.2 \times 10^{- 4} k g m^{2}$ and the earth’s horizontal magnetic field is $30 μ T$ . The magnetic moment of the magnet is

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$1200 A m^{2}$

$800 A m^{2}$

$1600 A m^{2}$

$1800 A m^{2}$

answer is C.

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Detailed Solution

Time period $T = \frac{π}{10} \sec$

Moment of inertia $I = 1.2 \times 10^{- 4} k g m^{2}$

Magnetic field $B = 30 μ T = 30 \times 10^{- 6} T$

Magnetic moment $M = ?$

$T = 2 π \sqrt{\frac{I}{M B_{H}}}$

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

$M = \frac{1.2 \times 10^{- 4}}{30 \times 10^{- 6}} \times 4 \times 100$

$∴ M = 1600 A m^{2}$

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