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
Answer & explanation
Correct answer: option 3
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$