A bar magnet of length 14 cm is placed in the magnetic meridian with its north pole pointing towards the geographic north pole.
A bar magnet of length $14 \mathrm{~cm}$ 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 $18 \mathrm{~cm}$ from the center of the magnet. If $\mathrm{B}_{\mathrm{H}}=0.4 \mathrm{G}$, the
Correct Option: , 3
i.e. $\frac{2 \mu_{0}}{4 \pi} \frac{\mathrm{m}}{\mathrm{r}^{2}} \times \frac{7}{\mathrm{r}}=0.4 \times 10^{-4}$
$\Rightarrow 2 \times 10^{-7} \times \frac{\mathrm{m} \times 7}{\left(7^{2}+18^{2}\right)^{3 / 2}} \times 10^{4}$
$=0.4 \times 10^{-4}$
$m=\frac{4 \times 10^{-2} \times(373)^{3 / 2}}{14}$
$M=m \times 14 \mathrm{~cm}=m \times \frac{14}{100}$
$=\frac{0.04 \times(373)^{3 / 2}}{14} \times \frac{14}{100}$
$=4 \times 10^{-4} \times 7203.82=2.88 \mathrm{~J} / \mathrm{T}$