A bar magnet is 8 cm long. At a distance of 3 cm from its centre in...

Given:
Length of the magnet, $2l = 8 \text{ cm} \Rightarrow l = 4 \text{ cm} = 0.04 \text{ m}$
Distance from the centre, $d = 3 \text{ cm} = 0.03 \text{ m}$
Magnetic field, $B = 4 \times 10^{-6} \text{ T}$

The magnetic field at a distance $d$ on the equatorial line (broadside-on position) of a bar magnet is given by:
$B = \dfrac{\mu_0}{4\pi} \dfrac{M}{(d^2 + l^2)^{3/2}}$
where $M = m \times 2l$ is the magnetic moment and $m$ is the pole strength.

Substituting the given values:
$4 \times 10^{-6} = 10^{-7} \times \dfrac{m \times 0.08}{((0.03)^2 + (0.04)^2)^{3/2}}$

$4 \times 10^{-6} = 10^{-7} \times \dfrac{m \times 0.08}{(0.0009 + 0.0016)^{3/2}}$

$4 \times 10^{-6} = 10^{-7} \times \dfrac{m \times 0.08}{(0.0025)^{3/2}}$

$4 \times 10^{-6} = 10^{-7} \times \dfrac{m \times 0.08}{(0.05)^3}$

$4 \times 10^{-6} = 10^{-7} \times \dfrac{m \times 0.08}{1.25 \times 10^{-4}}$

$m \times 0.08 \times 10^{-7} = 4 \times 10^{-6} \times 1.25 \times 10^{-4}$

$m \times 8 \times 10^{-9} = 5 \times 10^{-10}$

$m = \dfrac{5 \times 10^{-10}}{8 \times 10^{-9}} = \dfrac{5}{80} = 0.0625 \text{ A m}$

$m = 6.25 \times 10^{-2} \text{ A m}$