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 $20cm$ from the centre of the magnet on the equatorial line of the magnet is

• A. $0.049G$, $N-S$ direction
• B. $4.95G$, $S-N$ direction
• C. $0.0195G$, $S-N$ direction
• D. $19.5G$, $N-S$ direction

Answer 1

Answer: (A)

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