A very small magnet is placed in the magnetic meridian with its south pole pointing north. The null point is obtained 20 cm away from the centre of the magnet. If the earth's magnetic field (horizontal component) at this point be 0.3 gauss, the magnetic moment of the magnet i

8.0 \times 10^{2} e . m . u

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1.2 \times 10^{3} e . m . u

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2.4 \times 10^{3} e . m . u

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3.6 \times 10^{3} e . m . u

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Solution

The correct option is B $1.2 \times 10^{3} e . m . u$

At neutral point
$| B | = | B_{H} | \Rightarrow \frac{2 M}{(20)^{3}} = 0.3 \Rightarrow M = 1.2 \times 10^{3} e m u$

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A very small magnet is placed in the magnetic meridian with its south pole pointing north. The null point is obtained 20 cm away from the centre of the magnet. If the earth's magnetic field (horizontal component) at this point be 0.3 gauss, the magnetic moment of the magnet i

The correct option is B 1.2×103 e.m.u At neutral point |B|=|BH|⇒2M(20)3=0.3⇒M=1.2×103 emu

The correct option is B $1.2 \times 10^{3} e . m . u$

At neutral point
$| B | = | B_{H} | \Rightarrow \frac{2 M}{(20)^{3}} = 0.3 \Rightarrow M = 1.2 \times 10^{3} e m u$