Magnetic lines of force are observed in a bar magnet, a solenoid, an inductor, and even in Earth. The Earth itself is a giant magnet with North and South poles. The magnetic lines of force start from the North pole and end in the south pole–never intersecting each other. But do you think the magnetic lines of force that permeate around a magnet will be the same in every medium? The answer is no. Air has a different permeability in comparison to steel. Additionally, permeability is used to measure how much magnetism a material is capable of retaining inside it when an external magnetising field is applied. In this article, we will explore permeability in detail.
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When an external magnetising field (H) is applied, a magnetic field is established inside the material(B). The ratio of the magnetic induction produced to the magnetising field is called magnetic permeability. Mathematically,
$B=\mu H$
The SI unit of magnetic permeability is henry/ metre ($\frac{H}{m}$) .Its dimensional formula is $[M{L{T}^{-2}{A}^{-2}].}^{}$.Higher the magnetic permeability, higher is the material’s ability to allow magnetic lines of force to pass through it. In other words, it helps us to determine how much magnetic flux it allows to pass through.
Soft iron core has high magnetic permeability, hence it is used to reduce flux losses in solenoids.
The permeability of free space is defined as the number of magnetic lines of force that air/ vacuum would allow to pass through it. It’s value is given by ${\mu}_{0}=4\pi \times {10}^{-7}\frac{H}{m}.$.
Relative permeability(${\mu}_{r}$) is defined as the ratio of the permeability of the medium/ material to the permeability of vacuum. It is a measure of how many magnetic field lines it allows in comparison to vacuum.
${\mu}_{r}=\frac{{\mu}_{}}{{\mu}_{0}}$
It has no unit.
Fig showing magnetic lines of force around the earth
The value of magnetic permeability depends upon –
1)The nature of the material
2)Humidity
3)Position of the material in the medium
Note:
Magnetic permeability is always a positive value.
The magnetic induction of a material when an external magnetising field I is applied would be,
$B={\mu}_{0}\left(H+I\right)$
$I={\mathcal{X}}_{m}H$is called the intensity of magnetisation. It is defined as the magnetic moment per unit volume.
$B={\mu}_{0}\left(H+{\mathcal{X}}_{m}H\right)={\mu}_{0}H\left(1+{\mathcal{X}}_{m}\right)$
But $B={\mu H={\mu}_{0}{\mu}_{r}H}_{}$
Comparing the above two equations, we get,
${\mu}_{r}=1+{\mathcal{X}}_{m}$
${\mathcal{X}}_{m}-$ susceptibility of the material.
Diamagnetic materials are those which are weakly repelled when placed in an external magnetic field. Examples would be bismuth, gold and antimony.
The susceptibility is a measure of how responsive a material is to the applied magnetic field. Since diamagnetic substances align themselves opposite to the direction of the applied field, their susceptibility is slightly less than zero.
Since, ${\mu}_{r}=1+{\mathcal{X}}_{m},$ ${\mu}_{r}<1$ for a diamagnetic material.
Paramagnetic materials are those which are weakly attracted when an external magnetic field is applied. They acquire a feeble magnetisation in the direction of the field. Examples would be water, chromium and manganese.
The value of ${\mu}_{r}$ is slightly greater than one for paramagnetic materials.
When temperature is increased, the susceptibility of paramagnetic material reduces. This is because the thermal vibration of the molecules reduces the susceptibility.
${\mathcal{X}}_{m}\propto $$\frac{1}{T}$
Ferromagnetic materials align themselves along the direction of the magnetic field since they are strongly attracted towards it. Examples would be nickel, steel and cobalt.
The value of ${\mu}_{r}$ is much greater than one for ferromagnetic materials.
When temperature is increased above the Curie temperature(T_{c}), a ferromagnetic material becomes paramagnetic. This is called Curie Weiss’ law.
Note:
Permanent magnets should have high permeability, high retentivity and high coercivity.
Electromagnets should have permeability, low retentivity and low coercivity.
Q1.An external magnetising field of 20 CGS unit causes a magnetic flux of 2400 CGS in a bar of iron of cross section 0.2 cm^{2}. Find (i)permeability (ii)susceptibility of the bar
Answer. Given, $\varphi =2400CGS$, $A=0.2\times {10}^{-4}{m}^{2}$
$B=\frac{\varphi}{A}=\frac{2400\times {10}^{-8}}{0.2\times {10}^{-4}}=1.2\frac{Wb}{{m}^{2}}$
(i)Permeability, $\mu =\frac{B}{H}=\frac{1.2}{20\frac{}{4\pi}\times {10}^{-4}}=7.54\times \frac{{10}^{-4}H}{m}$
(ii)Susceptibility
$\mu ={\mu}_{0}{\mu}_{r}={\mu}_{0}\left(1+{\mathcal{X}}_{m}\right),{\mathcal{X}}_{m}=\frac{\mu}{{\mu}_{0}}-1=\frac{7.54\times {10}^{-4}}{4\pi \times {10}^{-7}}-1=599$
Q2.A solenoid having 5000 turns/m is supplied with a current of 2 A. If the magnetisation I is $2\times \frac{{10}^{-2}A}{m}$, calculate its susceptibility.
Answer.
i=2 A, n=5000 turns/m
$H=ni=5000\times 2=\frac{{10}^{4}A}{m}.$
$I={\mathcal{X}}_{m}H$
${\mathcal{X}}_{m}=\frac{I}{H}=\frac{2\times {10}^{-2}}{{10}^{4}}=2\times {10}^{-6}$
Q3.The solenoid having 2000 turns /m, carrying a current of 5 A is fitted with a core with a relative permeability 220. Calculate the magnetic field and the magnetic intensity.
Answer.
n=2000 turns/m, ${\mu}_{r}=220,i=5A$
Magnetic intensity, $H=ni=2000\times 5=\mathrm{10,000}\frac{A}{m}.$
Magnetic field, $B=\mu H=4\pi \times {10}^{-7}\times 220\times \mathrm{10,000}=88\pi \times {10}^{-2}T$
Q.4The susceptibility of a paramagnetic material at 200 K is found to be 0.0075. Find out its value at 100 K.
Answer.
${\mathcal{X}}_{m1}=0.0075,{T}_{1}=200K$
${\mathcal{X}}_{m2}=?,{T}_{2}=100K$
For a paramagnetic material,
$\frac{{\mathcal{X}}_{m2}}{{\mathcal{X}}_{m1}}=\frac{{T}_{1}}{{T}_{2}}$
$\frac{{\mathcal{X}}_{m2}}{{\mathcal{X}}_{m1}}=\frac{200}{100}\Rightarrow {\mathcal{X}}_{m2}=2\times 0.0075=0.015$
Q1.Why is magnetic permeability important?
Answer.The permeability is a measure of how much magnetic lines of force a medium or material allows to pass through it. Higher the permeability of a magnet, the better would be its performance as an electromagnet/permanent magnet.
Q2.Where is magnetic permeability used?
Answer.Magnetic permeability finds diverse applications in electromagnets, transformers and inductors. Inductors store magnetic energy in the form of magnetic fields. Higher the permeability, higher the magnetic field in the inductor. Inserting a soft iron core inside the permeability increases the magnetic field strength and also reduces flux losses.
Q3.Which material has highest magnetic permeability?
Answer.Iron,being a ferromagnetic material, has a relative permeability around 6000. Ferromagnetic materials get strongly magnetised when an external magnetic field is applied, since their molecules have a permanent dipole moment.
Q4.Why does the susceptibility of a diamagnetic material not vary with temperature?
Answer.Diamagnetic materials have molecules that do not have their own magnetic moment. When subjected to an external magnetic field, a negative magnetisation is produced, and the susceptibility is found to be negative.