Name: Anti-ferromagnetic and Ferrimagnetic Substances
Uploaded: 2019-07-19
Duration: 2 min 19 s
Description: Ferromagnetic materials align parallel whereas antiferromagnetic materials align antiparallel to the applied magnetic fields. Watch the video & learn more ab...

Question

What do you mean by end-on and broad-side on position of a bar magnet? Deduce an expression for the magnetic intensity at a point in the broad-side-on position due to a bar magnet.

Accepted Answer

The point that is on the axis of the bar magnet is called end-on-position. The point that is the perpendicular bisector of the bar magnet is called broad-end-on position.

An expression for magnetic intensity at a point on the broad-side-on position due to a bar magnet:

A line that is the perpendicular bisector of the bar magnet is known as an equatorial line. Let's consider $Q$ be the point on the broad-side-on position of the bar magnet SN, at a distance $d$ from its mid point $' Q'$ . Let the pole strength of the magnet be $m$ and its magnetic length $2 l$ .

The magnetic field at point $Q$ due to the north pole of the magnet is

$\vec{B_{N}} = \frac{μ_{0}}{4 π} \frac{m \vec{N Q}}{{(N Q)}^{3}}$

The magnetic field at point $Q$ due to the south pole of the magnet is

$\vec{B_{S}} = \frac{μ_{0}}{4 π} \frac{m \vec{Q S}}{{(Q S)}^{3}}$

But, $N Q = Q S = \sqrt{d^{2} + l^{2}}$

The resultant magnetic field at the equatorial point is

$\vec{B_{Q}} = \vec{B_{N}} + \vec{B_{S}} = \frac{μ_{0}}{4 π} \frac{m}{{(d^{2} + l^{2})}^{\frac{3}{2}}} (\vec{N Q} + \vec{S Q}) According to the traingle rule, \vec{NQ} + \vec{SQ} = \vec{NS} = 2 l \vec{B_{Q}} = \frac{μ_{0}}{4 π} \frac{m \times 2 l}{{(d^{2} + l^{2})}^{\frac{3}{2}}} \vec{B_{Q}} = \frac{μ_{0}}{4 π} \frac{- M}{{(d^{2} + l^{2})}^{\frac{3}{2}}}$

The negative sign in the above expression implies the magnetic field is antiparallel to the magnetic moment.

The magnitude of magnetic intensity is

$\vec{H_{Q}} = \frac{\vec{B_{Q}}}{μ_{0}} \vec{H_{Q}} = \frac{1}{4 π} \frac{M}{{(d^{2} + l^{2})}^{\frac{3}{2}}}$

For the bar magnet, $d > > l$ ,

$B = \frac{μ_{0}}{4 π} \frac{M}{d^{3}}$

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