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

3 \times 10^{- 5} T

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3 \times 10^{- 4} T

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3 \times 10^{3} T

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3 \times 10^{- 3} T

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Solution

The correct option is A $3 \times 10^{- 5} T$
Given : $m = 1.2 A m^{2}$
$d = 20 c m$
Horizontal magnetic field $B = \frac{2 μ_{0} m}{4 π d^{3}}$
$B = \frac{2 (10^{- 7}) (1.2)}{(0.2)^{3}}$
$= 300 \times 10^{- 7}$
$= 3 \times 10^{- 5} T$

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A bar magnet of magnetic moment 1.2 Am2 is placed in the magnetic meridian with its south pole pointing the north. If a neutral point is found at a distance of 20 cm from the centre of the magnet, the value of the horizontal component of the earth's magnetic field is

The correct option is A 3×10−5TGiven : m=1.2Am2d=20cmHorizontal magnetic field B=2μ0m4πd3B=2(10−7)(1.2)(0.2)3=300×10−7=3×10−5T

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

The correct option is A $3 \times 10^{- 5} T$
Given : $m = 1.2 A m^{2}$
$d = 20 c m$
Horizontal magnetic field $B = \frac{2 μ_{0} m}{4 π d^{3}}$
$B = \frac{2 (10^{- 7}) (1.2)}{(0.2)^{3}}$
$= 300 \times 10^{- 7}$
$= 3 \times 10^{- 5} T$