Question

A short bar magnet has a magnetic moment of $0.4 \,J\, T ^{-1}$. The magnitude of the magnetic field (in gauss) produced by the magnet at a distance of $20\, cm$ from the centre of the magnet on the equatorial line of the magnet is

Answer 1

Here, $m=0.4\, J\, T ^{-1}, d=20\, cm =0.2\, m$
On the equatorial line of magnet,
$B =\frac{\mu_{0}}{4 \pi} \cdot \frac{m}{d^{3}}$
$=10^{-7} \times \frac{0.4}{(0.2)^{3}}=\frac{0.4 \times 10^{-7}}{2^{3} \times 10^{-3}}$
$=\frac{0.4}{8} \times 10^{-4} $
$=0.05 \times 10^{-4} T , N - S $ direction
$=0.05 G ; N - S$ direction