Question

A short bar magnet has a magnetic moment of $0.39J{T}^{-1}$. The magnitude and direction of the magnetic field produced by the magnet at a distance of $20$cm from the centre of the magnet on the equatorial line of the magnet is

• A. $0.049$ G, N-S direction
• B. $4.9$ G, S-N direction
• C. $0.0195$ G, S-N direction
• D. $19.5$ G, N-S direction

Answer 1

The correct option is A $0.049$ G, N-S direction

Here, $m=0.39J{T}^{-1},d=20cm=0.2m$.
On the equatorial line of magnet
$B=\frac{{\mu }_0}{4\pi }.\frac{m}{d^3}={10}^{-7}\times \frac{0.39}{(0.2{)}^3}=\frac{0.39\times {10}^{-7}}{2^3\times {10}^{-3}}=\frac{0.39}{8}\times {10}^{-4}$
= $0.049\times {10}^{-4}T$, N-S direction
= $0.049$ G; N-S direction