The magnitude of the magnetic field due to a short bar magnet having a magnetic moment of $0.2 Am^2$, at a distance of 50 cm from the centre of the magnet on its axial line will be

$3.2 × 10^{-7} T$

The correct answer is Option (3) → $3.2 × 10^{-7} T$

Given:

Magnetic moment, $M = 0.2\,A·m^2$

Distance, $r = 50\,cm = 0.5\,m$

Formula:

Magnetic field on axial line, $B = \frac{μ_0}{4π} \cdot \frac{2M}{r^3}$

Substituting values:

$B = (10^{-7}) \cdot \frac{2(0.2)}{(0.5)^3}$

$B = 10^{-7} \cdot \frac{0.4}{0.125}$

$B = 10^{-7} \times 3.2$

$B = 3.2 \times 10^{-7}\,T$

Final Answer:$ B = 3.2 \times 10^{-7}\,T$