A bar magnet of length 14cm is placed in the magnetic meridian with...

i.e.& $\dfrac{2{μ}_{0}}{4\pi}\dfrac{m}{r2} \times \dfrac{7}{r} = 0 . 4 \times {10}^{- 4}$

$⇒ 2 \times {10}^{- 7} \times \dfrac{m \times 7}{\left({7}^{2} + {18}^{2}\right)3 / 2} \times {10}^{4}= 0 . 4 \times {10}^{- 4}$

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

$⇒ M = m \times 14cm = m \times \dfrac{14}{100}= \dfrac{0 . 04 \times \left(373\right)3 / 2}{14} \times \dfrac{14}{100}$

$⇒ M = 4 \times {10}^{- 4} \times 7203 . 82 = 2 . 88JT^{- 1}$.