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A bar magnet is kept on a table in the earth’s magnetic field with its north pole pointing earth’s north. The distance between the null points is 20 cm. If horizontal component of earth’s magnetic field is $4 \times 10^{- 5} T$ , then magnetic moment of the magnet is

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$0.4 A m^{2}$

$0.1 A m^{2}$

$0.2 A m^{2}$

$0.3 A m^{2}$

answer is D.

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Detailed Solution

The null points are formed on the equatorial line. At null point the resultant field is zero.

$∴ B_{e} = \frac{μ_{0}}{4 π} \frac{M}{d^{3}} \Rightarrow 4 \times 10^{- 5} = \frac{4 π \times 10^{- 7}}{4 π}$ $\times \frac{M}{10^{- 3}}$

$\Rightarrow M = \frac{4 \times 10^{- 5} \times 10^{- 3}}{10^{- 7}} = 0.4 A m^{2}$

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