Question

A short bar magnet is placed in the magnetic meridian of earth with its north pole pointing towards North. Neutral points are found at a distance of $30cm$ 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 approximately equal 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. $14.6$
• B. $19.4$
• C. $9.7$
• D. $4.9$

Answer 1

The correct option is C

$9.7$

Since the magnet is short and neutral point is on the equatorial line(East-West)
Earths Horizontal component of Magnetic field $B_H=\frac{{\mu }_0}{4\pi }\frac{m}{r^3}$

$3.6\times {10}^{-5}={10}^{-7}\times \frac{m}{(0.30{)}^3}$

$m=\frac{3.6\times {10}^{-5}\times (0.30{)}^3}{{10}^{-7}}=9.7A{m}^2$