Question

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

• A. $3\times {10}^{-5}T$
• B. $3\times {10}^{-4}T$
• C. $3\times {10}^{3}T$
• D. $3\times {10}^{-2}T$

Answer 1

The correct option is A $3\times {10}^{-5}T$

Given,

$d=20cm=0.2m$

$M=1.2A{m}^2$

The earth magnetic field is given by

$B_H=\frac{{\mu }_0}{4\pi }.\frac{2M}{d^3}$

$B_H={10}^{-7}\times \frac{2\times 1.2}{0.2\times 0.2\times 0.2}=3\times {10}^{-5}T$

The correct option is A.