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 $Am^2$ 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. 9.7
• B. 4.9
• C. 19.4
• D. 14.6

Answer 1

Answer: (A)

9.7

Answer 2

$|\bar{B}|$ at an equatorial point
$=\frac{\mu_{o}}{4 \pi} \frac{|\bar{m}|}{r^{3}}=B_{ H }$
$\Rightarrow|\bar{m}|=\frac{B_{ H } r^{3}}{10^{-7}}$
$=\frac{3.6 \times 10^{-5} \times(0.3)^{3}}{10^{-7}}= 9.7$