Question

A bar magnet of length $14\text{cm}$ is placed in the magnetic meridian with its north pole pointing towards the geographic north pole. A neutral point is obtained at a distance of $18\text{cm}$ from the center of the magnet. If $B_H=0.4G$, the magnetic moment of the magnet is $(1G={10}^{-4}\text{T})$

• A. $28.80{\text{J T}}^{-1}$
• B. $2.880{\text{J T}}^{-1}$
• C. $2.880\times {10}^3{\text{J T}}^{-1}$
• D. $2.880\times {10}^2{\text{J T}}^{-1}$

Answer 1

The correct option is B $2.880{\text{J T}}^{-1}$

A neutral point is obtained when the magnetic field due to bar magnet balances the horizontal component of earth's magnetic field ($B_H$).

$B_H$ points towards north along the magnetic meridian.

From the figure, we can see that net magnetic field at the point due to bar magnet will be
$B=2B_0\sin \theta$
$B=2\frac{{\mu }_0}{4\pi }\frac{m}{r^2}\times \frac{7}{r}$
where
$B_0\to$ magnetic flux density

$m\to$ power of magnetic pole
$\theta \to$ angle made by $B_0$ with the horizontal

Given, $B_H=0.4G=0.4\times {10}^{-4}\text{T}$

For neutral point,
$B=B_H$
$0.4\times {10}^{-4}=2\times {10}^{-7}\times \frac{m\times 7}{{(7^2+{18}^2)}^{\frac{3}{2}}}\times {10}^4$
$\Rightarrow m=\frac{4\times {10}^{-2}\times {\left(373\right)}^{\frac{3}{2}}}{14}$


$\therefore$ Magnetic moment of the bar magnet,
$M=m\times 14\text{cm}=m\times \frac{14}{100}$
$\Rightarrow M=\frac{0.04\times {\left(373\right)}^{\frac{3}{2}}}{14}\times \frac{14}{100}$
$\Rightarrow M\approx 2.88\text{J/T}$

Hence, option (b) is correct.