Question

Derive the expression for magnetic field induction on the equatorial line of a bar magnet.

Accepted Answer

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

Now, the magnetic induction, $B_{1}$ at point $P$ due to the North Pole along $NP$ is,

$B_{1} = \frac{μ_{0}}{4 π} . \frac{m}{{NP}^{2}} = \frac{μ_{0}}{4 π} . \frac{m}{{(d^{2} + l^{2})}^{}} (as {NP}^{2} = {ON}^{2} + {OP}^{2})$ ........................(1)

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

$B_{2} = \frac{μ_{0}}{4 π} . \frac{m}{{PS}^{2}} = \frac{μ_{0}}{4 π} . \frac{m}{{(d^{2} + l^{2})}^{}} (as {PS}^{2} = {OS}^{2} + {OP}^{2})$ ..................(2)

Therefore, we see that $B_{1} = B_{2} = B (say)$ .
Now, the net magnetic induction at $P$ due to bar magnet= $M_{p} = B_{1} + B_{2} = 2 B$ ...........(3)
Resolving $B_{1}$ and $B_{2}$ into horizontal and vertical components, The vertical components will be cancelled and horizontal will be added together. So, we can write,
$B_{p} = 2 B \cos θ$ .
Or $B_{P} = \frac{2 μ_{0}}{4 π} . \frac{m}{{(d^{2} + l^{2})}^{}} . (\frac{1}{\sqrt{d^{2} + l^{2}}}) as \cos θ = \frac{1}{\sqrt{d^{2} + l^{2}}}$ .

As the magnet is very small, $l < < d$ , so we can write ,

$B_{P} = \frac{2 μ_{0}}{4 π} . \frac{m}{d^{3}} . = \frac{μ_{0}}{4 π} . \frac{M}{d^{3}}$ .

Where $2 ml = M$ , the magnetic moment of the bar magnet.

Derive the expression for magnetic field induction on the equatorial line of a bar magnet.

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