Question

A short bar magnet has a magnetic moment of $0.39\, J\, T^{-1}$. 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. $4.95 \,G$, $S-N$ direction
• C. $0.0195\, G$, $S-N$ direction
• D. $19.5 \,G$, $N-S$ direction

Answer 1

Answer: (A)

Here, $m = 0.39\, J\, T^{-1}$, $d = 20\, cm = 0.2 \,m$
On the equatorial line of magnet
$B = \frac{\mu_{0}}{4\pi} \cdot \frac{m}{d^{3}} = 10^{-7} \times \frac{0.39}{\left(0.2\right)^{3} }$
$ = \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
$= 0.049\, G$; $N-S$ direction