Question

Derive expression for the intensity of magnetic field at a point in end on position due to a short bar magnet.

Answer 1

Consider a magnet $NS$ of effective length $2l$. The strength of each pole of this magnet is $m$. The magnetic field intensity at a point $P$ on the axis of the magnet has to be determined. The distance of point $P$ from the centre $O$ of the magnet $NS$ is $d$.

The intensity at point $P$ due to the north pole $\left(N\right)$ of the magnet.

$B_1=\frac{{\mu }_0}{4\pi }\frac{m}{{\left(NP\right)}^2}=\frac{{\mu }_0}{4\pi }\frac{m}{{(d-l)}^2}$ ........(1)

where $NP=(d-l)$

The intensity at point $P$ due to the south $\left(S\right)$ Pole

$B_2=-\frac{{\mu }_0}{4\pi }\frac{m}{{\left(SP\right)}^2}=-\frac{{\mu }_0}{4\pi }\frac{m}{{(d+l)}^2}$ .......(2)

Here $SP=(d+l)$

From Resultant intensity $B=B_1+B_2$ .......(3)

Substituting the values from equations (1) and (2) in equation (3)

$B=\frac{{\mu }_0}{4\pi }\frac{m}{{(d-l)}^2}-\frac{{\mu }_0}{4\pi }\frac{m}{{(d+l)}^2}$

$\therefore$ $B=\frac{{\mu }_{0}m}{4\pi }[\frac{1}{{(d-l)}^2}-\frac{1}{{(d+l)}^2}]$

Or $B=\frac{{\mu }_{0}m}{4\pi }\left[\frac{{(d+l)}^2-{(d-l)}^2}{{(d-l)}^2{(d+l)}^2}\right]$

Or $B=\frac{{\mu }_{0}m}{4\pi }\left[\frac{d^2+l^2+2dl-(d^2+l^2-2dl)}{{\left\{(d-l)(d+l)\right\}}^2}\right]$

Or $B=\frac{{\mu }_0}{4\pi }m\left[\frac{4dl}{{(d^2-l^2)}^2}\right]$ $[\because (d-l)(d+l)=d^2-l^2]$

Or $B=\frac{{\mu }_0}{4\pi }\frac{2Md}{{(d^2-l^2)}^2}$ [$\because 2ml=M$ magnetic moment]

If $d\gg l$, for a small magnet

$B=\frac{{\mu }_0}{4\pi }\frac{2M}{d^3}$ Tesla

This is the required expression.