Question

Derive an expression for the intensity of magnetic field on broad side on position due to a bar magnet.

Answer 1

Let $NS$ be a bar magnet of pole strength $m$ and effective length $2l$. Consider a point $P$ on the neutral axis at a distance $d$ from the centre of the magnet.
Let $PN=PS=x$.
Now, the intensity of field at $P$ due to $N$-pole is,
$B_1=\frac{{\mu }_0}{4\pi }\cdot \frac{m}{x^2}\left(along\overrightarrow{NP}\right)$
Similarly, the intensity of field at $P$ due to $S$ pole is given by
$B_2=\frac{{\mu }_0}{4\pi }\cdot \frac{m}{x^2}\left(along\overrightarrow{PS}\right)$
Since, $B_1=B_2=\frac{{\mu }_0}{4\pi }\cdot \frac{m}{x^2}$
Hence, $B_1=B_2$
Hence, by the law of parallelogram of force the resultant will be given by
$B=\sqrt{B_1^2+B_2^2+2B_1{B}_2\cos 2\theta }$ (Where $2Q$ is the angle between $B_1$ and $B_2$)
But, $B_1=B_2$, hence
$B=\sqrt{B_1^2+B_1^2+2B_1^2\cos 2\theta }$
$=\sqrt{2B_1^2+2B_1^2\cos 2\theta }$
$=\sqrt{2B_1^2(1+\cos 2\theta )}$
$=\sqrt{2B_1^2[1+2{\cos }^2\theta -1]}$
$=\sqrt{2B_1^2\cdot 2{\cos }^2\theta }=2B_1\cos \theta$
or $B=2\cdot \frac{{\mu }_0}{4\pi }\cdot \frac{m}{x^2}\cos \theta$
or $B=\frac{{\mu }_0}{4\pi }\cdot \frac{2m}{x^2}\cdot \frac{1}{x},(\because \cos \theta =\frac{1}{x})$
or $B=\frac{{\mu }_0}{4\pi }\cdot \frac{2ml}{x^3}$
But, $2ml=M$ (magnetic moment)
$\therefore B=\frac{{\mu }_0}{4\pi }\cdot \frac{2M}{x^3}$
But, by the Pythagoras theorem in right angle $\Delta PQS$
$x=\sqrt{d^2+1^2}$
or $x^3={(d^2+1^2)}^{3/2}$
Hence, $B=\frac{{\mu }_0}{4\pi }\cdot \frac{M}{{(d^2+1^2)}^{3/2}}$
If the magnet is small i.e. $d\gg 1$
$\therefore B=\frac{{\mu }_0}{4\pi }\cdot \frac{M}{d^3}$
This is the required expression.
The direction of magnetic field is from north pole to south pole and parallel to magnetic axis of the magnet.