Question

A bar magnet has a length $8\text{cm}$. The magnetic field at a point at a distance $3\text{cm}$ from the centre in the broad- side (or equatorial) on position is found to be $4\times {10}^{-6}\text{T}$. Find the pole strength of the magnet.

• A. $8\times {10}^{-2}\text{Am}$
• B. $1.25\times {10}^{-2}\text{Am}$
• C. $6.25\times {10}^{-2}\text{Am}$
• D. $3\times {10}^{-2}\text{Am}$

Answer 1

The correct option is C $6.25\times {10}^{-2}\text{Am}$

The magnetic field due to a bar magnet in the broadside- on position is given by
$$B=\frac{{\mu }_0}{4\pi }\frac{M}{{[r^2+\frac{l^2}{4}]}^{3/2}}$$

Here, dipole moment, $M=ml$

$m$ is the pole strength of the magnet.

$\Rightarrow 4\times {10}^{-6}={10}^{-7}\times \frac{m\times 8\times {10}^{-2}}{{[(3\times {10}^{-2}{)}^2+\frac{{(8\times {10}^{-2})}^2}{4}]}^{3/2}}$

$40=\frac{m\times 8\times {10}^{-2}}{(5\times {10}^{-2}{)}^3}$

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

Hence, option $\left(C\right)$ is the correct answer.