Question

Define magnetic intensity at a point in a magnetic field. Deduce an expression for the magnetic field intensity at a point in the end-on-position due to bar magnet.

Accepted Answer

Magnetic intensity at a point in a magnetic field is defined as the force experienced by the unit north pole at that point.

$H = \frac{m}{4 {πr}^{2}}$

The magnetic field intensity at a point in the end-on-position due to bar magnet:

A position on the magnetic axis of a magnetic dipole moment is called end-on-position. Let SN be a bar magnet of magnetic length $2 l$ and pole strength $m$ .

Let $P$ be a point in the end-on-position at a distance $d$ from the mid-point of the bar magnet as shown in the above figure.

The magnetic field at a point $P$ due to the North Pole is

$\vec{B_{N}} = \frac{μ_{0}}{4 π} \frac{m}{{(d - l)}^{2}} along \vec{N P}$

The magnetic field at a point $P$ due to the South Pole is

$\vec{B_{S}} = \frac{μ_{0}}{4 π} \frac{m}{{(d + l)}^{2}} along \vec{P N}$

The resultant magnetic field at a point $P$ due to the bar magnet is

$\vec{B_{P}} = \vec{B_{N}} + \vec{B_{S}} = \frac{μ_{0}}{4 π} (\frac{m}{{(d - l)}^{2}} + \frac{(- m)}{{(d + l)}^{2}}) along \vec{N P} = \frac{μ_{0} m}{4 π} (\frac{{(d + l)}^{2} - {(d - l)}^{2}}{{(d^{2} - l^{2})}^{2}}) along \vec{N P} = \frac{μ_{0} m}{4 π} \frac{4 d l}{{(d^{2} - l^{2})}^{2}} along \vec{N P} = \frac{μ_{0}}{4 π} . \frac{2 M d}{{(d^{2} - l^{2})}^{2}} along \vec{N P} Where . M = 2 m l$

We know that,

$\vec{H_{p}} = \frac{\vec{B_{p}}}{μ_{0}} = \frac{1}{4 π} \frac{2 M d}{{(d^{2} - l^{2})}^{2}}$

For a short bar magnet, $l < < d$

The magnetic induction at end-on-position is

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

Exercise 1

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