Name: Hysteresis: Property of Ferromagnetic Substances
Uploaded: 2019-07-19
Duration: 5 min 6 s
Description: The behaviour of ferromagnets is fascinating. The magnetic field due to a ferromagnet exhibits an interesting phenomenon known as hysteresis. Let’s explore.

Question

Two short magnets of magnetic moments $a$ and $b$ of magnetic moments $32 A m^{2}$ and $27 A m^{2}$ are placed on a table, as shown in figure. Find the magnitude and direction of the magnetic field produced.

These magnets at point $P$ situated at the equatorial lines of both the magnets at distances $40 cm$ and $30 cm$ respectively from the centres of the two magnets.

Accepted Answer

The magnetic moment of the magnet $a, m_{1} = 32 {Am}^{2}$

The magnetic moment of the magnet $b, m_{2} = 27 {Am}^{2}$

The distance of point $P$ from the magnet $a, r_{1} = 40 cm = 0 .4 m$

The distance of point $P$ from the magnet $b, r_{2} = 30 cm = 0 .3 m$

As shown in the figure, the point $P$ lies on the equatorial line of both the magnets $a$ and $b$ .

We know the magnetic field due to bar magnet along its equatorial line is
$B_{e q u a} = \frac{μ_{0}}{4 π} . \frac{m}{r^{3}}$

$∴ B_{a} = \frac{μ_{0}}{4 π} . \frac{m_{1}}{r_{1}^{3}} = \frac{10^{- 7} \times 32}{{(0 .40)}^{3}} = 0 .5 \times 10^{- 4} T$

$B_{b} = \frac{μ_{0}}{4 π} . \frac{m_{2}}{r_{2}^{3}} = \frac{10^{- 7} \times 27}{{(0 .30)}^{3}} = 10^{- 4} T$

The two fields are perpendicular to each other as shown in the figure.

So the resultant field at point $P$ is

$B = \sqrt{B_{a}^{2} + B_{b}^{2}} = 10^{- 4} \sqrt{{(0 .5)}^{2} + 1^{2}} = 1 .12 \times 10^{- 4} T$

If the field $B$ makes an angle $θ$ with the direction of $B_{b}$ , then

$\tan θ = \frac{B_{a}}{B_{b}} = \frac{0 .5 \times 10^{- 4}}{10^{- 4}} = 0 .5$

or $θ = 26 {.57}^{o}$

Important Questions on Magnetism

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