Relation Between Magnetic Permeability and Susceptibility - Class 12 Physics

Permeability is the measure of the ability of a material to support the formation of a magnetic field within itself. In other words, it is the degree of magnetization that a material obtains in response to an applied magnetic field. The magnetic susceptibility is a dimensionless proportionality constant that indicates the degree of magnetization of a material in response to an applied magnetic field. Relative permeability, sometimes denoted by the symbol μr, is the ratio of the permeability of a specific medium to the permeability of free space. Here we will study about the Relation Between Relative Magnetic Permeability and Susceptibility:

Relation Between Relative Magnetic Permeability and Susceptibility

When a magnetic material is kept in a magnetising field (H). Then total number of magnetic lines of force inside the material = magnetic lines of force due to magnetising field + magnetic lines of force due to magnetisation of specimen

i.e. Magnetic induction (B) = $B_{0}$ (no. of lines of force due to H) + $\mu_{0} I$ (no. of lines due to magnetisation of specimen0029

or $B = B _{0}+\mu_{0} I =\mu_{0} H +\mu_{0} I =\mu_{0}( H + I )$

$B =\mu_{0}( H + I )=\mu_{0} H \left(1+\chi_{ m }\right)$ as $\chi_{ m }=\frac{ I }{ H }$

$B =\mu H =\mu_{0} H \left(1+\chi_{ m }\right)$ or $\mu=\mu_{0}\left(1+\chi_{m}\right)$

or $\mu_{ r }=1+\chi_{ m }$.

Ex. A tungsten rod of length 10 cm and area of cross-section $0.25 cm ^{2}$ is placed in a magnetising field of 314 oersted, with its length parallel to the field. The magnetic susceptibility of tungsten is $6.8 \times 10^{-5}$. Calculate the (i) intensity of magnetisation (ii) magnetic moment and (iii) absolute permeability. Sol. Given H = 314 oersted = $\frac{314 \times 10^{3}}{4 \pi}$ amp. $/ m .=\frac{10^{5}}{4}$ amp. $/ m$

1. Intensity of magnetisation $$I = {{{\chi _m}} \over H} = {{6.8 \times {{10}^{ - 5}} \times {{10}^5}} \over 4} = 1.7\,amp./1m$$
2. Magnetic moment M = $\mathbf{I}$V = $1.7 \times 0.1 \times 0.25 \times 10^{-4}=4.25 \times 10^{-6}$ amp./m.
3. Absolute permeability $\mu=\mu_{ r } \mu_{0}=\mu_{0}\left(1+\chi_{ m }\right)=4 \pi \times 10^{-7}\left[1+6.8 \times 10^{-5}\right]=12.56 \times 10^{-7}$Wb/A-m

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Ex. A solenoid of 500 turns/m is carrying a current of 3A. Its core is made of iron which has a relative permeability of 5000. Determine the magnitude of magnetic intensity, magnetisation and magnetic field inside the core. Sol. Magnetic intensity H = ni = 500 × 3 = 1500 A/m $\mu_{ r }=1+\chi_{ m }$ so $\chi_{ m }=\mu_{ r }-1=4999 \approx 5000$ Intensity of magnetisation $I =\chi H =5000 \times 1500=7.5 \times 10^{6} A / m$ Magnetic field $B =\mu_{ r } \mu_{0} H =5000 \times 4 \pi \times 10^{-7} \times 1500=9.4$ tesla.

Ex. An electron in an atom revolves around the nucleus in an orbit of radius 0.53 A. Calculate the equivalent magnetic moment if the frequency of revolution is $6.8 \times 10^{9} MHz$ Sol. Magnetic moment $M=I A=\frac{e}{T} \times \pi r^{2}=e v \pi r^{2}$ So M = $1.6 \times 10^{-19} \times 6.8 \times 10^{9} \times 3.14 \times\left(0.53 \times 10^{-10}\right)^{2}=9.6 \times 10^{-24} Am ^{2}$

Ex. An iron rod 0.2 m long $10^{-2}$ m in diameter and of permeability 1000 is placed inside a long solenoid wound with 300 turns per meter. If a current of 0.5 A is passed through the rod find magnetic moment of rod. Sol. $B =\mu_{0}( H + I )$ so $I =\frac{ B }{\mu_{0}}- H =\frac{\mu H }{\mu_{0}}- H =\left(\mu_{ r }-1\right) H$ For a solenoid $B =\mu_{0} ni$ So $H =\frac{ B }{\mu_{0}}= ni$ so $I=\left(\mu_{r}-1\right) n i$ $I =(1000-1) 300 \times 0.5=999 \times 150 Am ^{2}$ Magnetic moment of rod $M = I \times V = I \pi r ^{2} \ell=999 \times 150 \times 3.14 \times\left(0.5 \times 10^{-2}\right) \times 0.2$ $=0.2325 JT ^{-1}$

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