A bar magnet of length 14 cm is placed in the magnetic meridian with its north pole pointing towards the geographic north pole.

i.e. $\frac{2 \mu_{0}}{4 \pi} \frac{\mathrm{m}}{\mathrm{r}^{2}} \times \frac{7}{\mathrm{r}}=0.4 \times 10^{-4}$

$\Rightarrow 2 \times 10^{-7} \times \frac{\mathrm{m} \times 7}{\left(7^{2}+18^{2}\right)^{3 / 2}} \times 10^{4}$

$=0.4 \times 10^{-4}$

$m=\frac{4 \times 10^{-2} \times(373)^{3 / 2}}{14}$

$M=m \times 14 \mathrm{~cm}=m \times \frac{14}{100}$

$=\frac{0.04 \times(373)^{3 / 2}}{14} \times \frac{14}{100}$

$=4 \times 10^{-4} \times 7203.82=2.88 \mathrm{~J} / \mathrm{T}$