Question

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

Answer 1

Answer: 0.05

Here, $m=0.4J{T}^{-1},d=20cm=0.2m$
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
$=0.05G;N-S$ direction