Question

A short bar magnet is placed in the magnetic meridian of the earth with the north pole pointing north. Neutral points are found at a distance of $30\text{cm}$ from the magnet on the east - west line, drawn through the middle point of the magnet. The magnetic moment of the magnet in $\left({\text{Am}}^2\right)$ is close to
(Given $\frac{{\mu }_0}{4\pi }={10}^{-7}$ in SI units and $B_H$ = Horizontal component of earth's magnetic field = $3.6\times {10}^{-5}\text{T}$)

• A. $4.9$
• B.
  $14.6$
• C. $32.4$
• D. $9.7$

Answer 1

The correct option is D $9.7$


As we know that magnetic field along the equatorial axis of the short bar magnet of magnetic moment $M$ is given by

$B_{magnet}=\frac{{\mu }_0}{4\pi }\frac{M}{r^3}$

At the neutral point,

$B_{magnet}=B_H$

$\frac{{\mu }_0}{4\pi }\frac{M}{r^3}=3.6\times {10}^{-5}$

$\Rightarrow M=\frac{3.6\times {10}^{-5}\times (0.3{)}^3}{{10}^{-7}}$

$\therefore M=9.72{\text{Am}}^2$

<!--td {border: 1px solid #ccc;}br {mso-data-placement:same-cell;}--> Hence, $\left(D\right)$ is the correct answer.