Question

Derive an expression for magnetic field at a point on the axial line of a bar magnet.

Answer 1

Let $NS$ be a bar magnet of magnetic length $2l$ and having each pole of magnetic strength $m$ . O is the center of magnet and $P$ is a point on axial line at a distance $r$ from the center $O$ of magnet , at which magnetic field has to be measured .

The magnetic field $B_1$ at $P$ due to $N$ pole of magnet ,

$B_1=\frac{{\mu }_0}{4\pi }.\frac{m}{N{P}^2}$

or $B_1=\frac{{\mu }_0}{4\pi }.\frac{m}{(r-l{)}^2}$ (along PX) .......................eq1

And , the magnetic field $B_2$ at $P$ due to $S$ pole of magnet ,

$B_2=\frac{{\mu }_0}{4\pi }.\frac{m}{S{P}^2}$

or $B_2=\frac{{\mu }_0}{4\pi }.\frac{m}{(r+l{)}^2}$ (along PS) ......................eq2

Therefore , resultant magnetic field at point $P$ ,

$B=B_1-B_2$ (-ive sign is due to opposite directionS of $B_1$ and $B_2$)

It is clear from eq1 and eq2 that $B_1>B_2$ ,therefore the direction of $B$ will be along $PX$ .

or $B=$$\frac{{\mu }_0}{4\pi }.\frac{m}{(r-l{)}^2}$$-\frac{{\mu }_0}{4\pi }.\frac{m}{(r+l{)}^2}$ (alongPX)

or $B=\frac{{\mu }_0}{4\pi }.\frac{m\left(4rl\right)}{(r^2-l^2{)}^2}$ (alongPX)

Now , $m\left(2l\right)=M$ (magnetic dipole moment of magnet)

Hence , $B=\frac{{\mu }_0}{4\pi }.\frac{2Mr}{(r^2-l^2{)}^2}$ (alongPX)