Question

A bar magnet has a length of 8 cm. The magnetic field at a point at a distance 3 cm from the centre in the broadside-on position is found to be 4 × 10⁻⁶ T. Find the pole strength of the magnet.

Answer 1

Given:
Length of the magnet, 2l = 8 cm = 8 $\times$ 10^$-2$ m
Distance of the observation point from the centre of the dipole, d = 3 cm
Magnetic field in the broadside-on position, B = 4 × 10⁻⁶ T
The magnetic field due to the dipole on the equatorial point $\left(B\right)$is given by
$B=\frac{{\mathrm{\mu }}_{0}m2l}{4\mathrm{\pi }{\left(d^2+l^2\right)}^{3/2}}$
Here, m is the pole strength of the magnet.
On substituting the respective values, we get
$$\begin{gathered} 4\times {10}^{-6}=\frac{{10}^{-7}m\times 8\times {10}^{-2}}{{\left(9\times {10}^{-4}+16\times {10}^{-4}\right)}^{3/2}} \\ \Rightarrow 4\times {10}^{-6}=\frac{m\times 8\times {10}^{-9}}{{\left(25\right)}^{3/2}\times {\left({10}^{-4}\right)}^{3/2}} \\ \Rightarrow m=\frac{4\times {10}^{-6}\times 125\times {10}^{-6}}{8\times {10}^{-9}} \\ \Rightarrow m=6.25\times {10}^{-2}\mathrm{A}-\mathrm{m} \end{gathered}$$
Thus, the pole strength of the magnet is 6.25$\times$10^$-2$ A-m.