Question

A bar magnet of length of $0.1m$ has a pole strength of $50Am$. Calculate the magnetic field at a distance of $0.2m$ from its centre on (i) its axial line and (ii) its equatorial line.

Answer 1

$\left(1\right)$Magnetic field on its axial line.

$B_1=\frac{{\mu }_0}{4\pi }\frac{2Mr}{{(r^2-l^2)}^2}{B}_1=\frac{4\pi \times {10}^{-7}}{4\pi }\times \frac{2\times 50\times 0.2}{{({\left(0.2\right)}^2-(0.05{)}^2)}^2}={10}^{-7}\times \frac{20}{1.4\times {10}^{-3}}=1.42mT$

$\left(2\right)$ Field at equatorial line

$B_2=\frac{{\mu }_0}{4\pi }\frac{M}{{(r^2+l^2)}^{3/2}}=\frac{4\pi \times {10}^{-7}}{4\pi }\times \frac{50}{{({\left(0.2\right)}^2+(0.05{)}^2)}^{\frac{3}{2}}}=\frac{50\times {10}^{-7}}{8.76\times {10}^{-3}}=5.7\times {10}^{-4}T$