Question

A short bar magnet has a magnetic moment of $0.48\, J\, T^{-1}$. The magnitude and direction of magnetic field produced by the magnet at a distance of $10 \,cm$ from the centre of the magnet on its axis is

• A. $0.48 \times 10^{-4}\, T$ along $N-S$ direction
• B. $0.28 \times 10^{-4}\, T$ along $S-N$ direction
• C. $0.28 \times 10^{-4}\, T$ along $N-S$ direction
• D. $0.96 \times 10^{-4}\, T$ along $S-N$ direction

Answer 1

Answer: (D)

On the axis of the magnet
$B = \frac{\mu_{0}}{4\pi} \cdot \frac{2m}{d^{3}}$
Here, $\frac{\mu _{0}}{4\pi } = 10^{-7} \, A\,m^{-2}$
$m = 0.48 \,J \,T^{-1}, d = 10 \,cm = 0.1 \,m$
Then, $B = \frac{10^{-7}\times 2 \times 0.48 }{\left(0.1\right)^{3}}$
$= 0.96 \times 10^{-4}\,T$, along $S-N$ direction.