Question

Derive an expression for magnetic induction at a point along the axis of bar magnet.

Answer 1

Let us consider that a bar magnet of length 2l and magnetic pole strength m. suppose a point P on the axis of the bar magnet at a distance from from its centre.

so, distance of P from north pole = (d - l)

and distance of P from south pole = (d + l)

now, magnetic field intensity due to north pole at point P is given by, $B_1=\frac{\mu_0}{4\pi}\frac{m}{r^2}$

here, r = (d - l)

so, $B_1=\frac{\mu_0}{4\pi}\frac{m}{(d-l)^2}$, it is directly from the north pole.

again, magnetic field intensity due to South pole at point P is given by,

$B_2=\frac{\mu_0}{4\pi}\frac{m}{(d+l)^2}$ it is directly towards the south pole.

so, magnetic field intensity at point P, B = $B_1+(-B_2)$

=$\frac{\mu_0m}{4\pi}\left[\frac{1}{(d-l)^2}-\frac{1}{(d+l)^2}\right]$

=$\frac{\mu_0m}{4\pi}\left[\frac{4dl}{(d^2-l^2)^2}\right]$

=$\frac{\mu_0}{4\pi}\frac{m(4dl)}{(d^2-l^2)^2}$

we know, M = m × 2l

so, required expression is $\boxed{B=\frac{\mu_0}{4\pi}\frac{2Md}{(d^2-l^2)^2}}$