Question

A short bar magnet is placed in the magnetic meridian of the earth with north pole pointing north. Neutral points are found at a distance of 30 cm from the magnet on the East - West line, drawn through the middle point of the magnet. The magnetic moment of the magnet in $A{m}^2$ is close to (Given $\frac{{\mu }_0}{4\pi }={10}^{-7}$ in SI units and $B_H=$ Horizontal component of earth's magnetic field $=3.6\times {10}^{-5}Tesla.)$

• A. 4.9
• B. 14.6
• C. 19.4
• D. 9.7

Answer 1

The correct option is D 9.7

Magnetic field due to bar magnet of pole strength of $m$ is given as at equatorial position is given as

$\overrightarrow{B}=\frac{{\mu }_{\circ }}{4\pi }\frac{m}{r^3}$

r is distance from middle point of magnet

r= 30 cm= .3 m; $\frac{{\mu }_{\circ }}{4\pi }={10}^{-7}$

At neutral point the magnetic field of earth and bar magnet will be equal and opposite, the net magnetic field will be zero at neutral points.

${\overrightarrow{B}}_{magnet}={\overrightarrow{B}}_{earth}={10}^{-7}\frac{m}{{.3}^3}$

${\overrightarrow{B}}_{earth}=B_H$ from diagram

$\Rightarrow m=\frac{3.6\times {10}^{-}5\times {.3}^3}{{10}^{-}7}$

$m=9.72$ A-m