A magnetic dipole of moment $96 A m^{2}$ is placed horizontally with the N-pole pointing south. The position of neutral point is..........if horizontal components of the earth's field is $24 μ T$ .

10 c m

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8 c m

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20 c m

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20 m m

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Solution

The correct option is D $20 c m$
Horizontal component of earts's magnetic field is $B_{H} = \frac{2 μ_{0} M}{4 π d^{3}}$ or $d^{3} = \frac{2 μ_{0} M}{4 π B_{H}}$
$d^{3} = \frac{2 \times 4 π \times 10^{- 7} \times 0.96}{4 π \times 24 \times 10^{- 6}}$
$d^{3} = 8 \times 10^{- 3}$ or $d = 0.2 m$

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A bar magnet of magnetic moment $1.2 A m^{2}$ is placed in the magnetic meridian with its south pole pointing the north. If a neutral point is found at a distance of $20 c m$ from the centre of the magnet, the value of the horizontal component of the earth's magnetic field is

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A magnetic dipole of moment 96Am2 is placed horizontally with the N-pole pointing south. The position of neutral point is..........if horizontal components of the earth's field is 24μT.

The correct option is D 20cmHorizontal component of earts's magnetic field is BH=2μ0M4πd3 or d3=2μ0M4πBHd3=2×4π×10−7×0.964π×24×10−6d3=8×10−3 or d=0.2m

A magnetic dipole of moment $96 A m^{2}$ is placed horizontally with the N-pole pointing south. The position of neutral point is..........if horizontal components of the earth's field is $24 μ T$ .

The correct option is D $20 c m$
Horizontal component of earts's magnetic field is $B_{H} = \frac{2 μ_{0} M}{4 π d^{3}}$ or $d^{3} = \frac{2 μ_{0} M}{4 π B_{H}}$
$d^{3} = \frac{2 \times 4 π \times 10^{- 7} \times 0.96}{4 π \times 24 \times 10^{- 6}}$
$d^{3} = 8 \times 10^{- 3}$ or $d = 0.2 m$