Question

Derive an expression for magnetic induction at a point on the equitorial line of a bar magnet.

Answer 1

Consider a bar magnet NS of length $2l$ and pole strength $m$. Let, P be the point on the equatorial line at distance d from mid-point of magnet i.e. O as shown in figure 1.

Now, the magnetic induction, $B_1$ at point P due to north pole along NP is,

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

$B_1=\frac{{\mu }_0}{4\pi }\frac{m}{(d^2+l^2{)}^2}$ .........(1) ($\because N{P}^2=O{N}^2+O{P}^2$)

Similarly, the magnetic induction, $B_2$ at point P due to south pole along PS is,

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

$B_2=\frac{{\mu }_0}{4\pi }\frac{m}{(d^2+l^2{)}^2}$ .........(2) ($\because P{S}^2=O{S}^2+O{P}^2$)

From equations (1) and (2), we get

$B_1=B_2=B\left(say\right)$

Now, the net magnetic induction $B_P$ at P due to bar magnet is

$B_P=B_1+B_2=2B$ ........................(3)

Resolving $B_1$ and $B_2$ into horizontal and vertical components, as shown in figure 2. The vertical components will be canceled and horizontal will be added together. So, we can write

$B_P=2B\cos \theta$ .............from(3)

$B_P=2\left(\frac{{\mu }_0}{4\pi }\frac{m}{(d^2+l^2{)}^2}\right)\left(\frac{l}{\sqrt{(}{d}^2+l^2)}\right)$ ............($\because \cos \theta =\frac{l}{\sqrt{(}{d}^2+l^2)}$)

For a short dipole, $l<<d$, therefore

$B_P=\frac{{\mu }_0}{4\pi }\frac{2ml}{d^3}$

$B_P=\frac{{\mu }_0}{4\pi }\frac{M}{d^3}$ ........................($\because M=2ml$)