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Relation Between Magnetic Permeability and Magnetic Susceptibility

Relation Between Magnetic Permeability and Magnetic Susceptibility :- When a magnetic material is placed in a current carrying solenoid, then the total magnetic field in the core of the solenoid is :-

$\displaystyle B={{B}_{0}}+{{B}_{m}}$ …..(1)

Here

B₀ = Magnetic field due to free current in the solenoid

B_m = Magnetic field due to bound currents generated by magnetisation of the material

Now

$\displaystyle {{B}_{\text{0}}}={{\mu }_{0}}H$ …..(2)

$\displaystyle {{B}_{m}}={{\mu }_{0}}I$ …..(3)

From equation (1),

$\displaystyle B={{\mu }_{0}}H+{{\mu }_{0}}I$

$\displaystyle \Rightarrow B={{\mu }_{0}}\left( H+I \right)$ …..(4)

Magnetic susceptibility of material,

$\displaystyle {{\chi }_{m}}=\frac{I}{H}$

$\displaystyle \therefore I={{\chi }_{m}}H$

From equation (4),

$\displaystyle B={{\mu }_{0}}\left( H+{{\chi }_{m}}H \right)$

$\displaystyle \Rightarrow B={{\mu }_{0}}H\left( 1+{{\chi }_{m}} \right)$

But $\displaystyle B=\mu H$ , So

$\displaystyle \mu H={{\mu }_{0}}H\left( 1+{{\chi }_{m}} \right)$

$\displaystyle \Rightarrow \mu ={{\mu }_{0}}\left( 1+{{\chi }_{m}} \right)$

$\displaystyle \Rightarrow \frac{\mu }{{{\mu }_{0}}}=1+{{\chi }_{m}}$

$\displaystyle \Rightarrow {{\mu }_{r}}=1+{{\chi }_{m}}$ …..(5)

The above equation (5) is the desired relation between magnetic permeability and magnetic susceptibility.

Example 1.

A rod of ferromagnetic material has dimensions of 10 cm × 0.5 cm × 0.2 cm. When it is placed in a magnetising field of 0.5 × 10⁴ A m⁻¹, a magnetic moment of 5 Am² is induced in it, then the value of magnetic induction will be :-

$(A) 0.54$ T

(B) $0.358$ T

$(C) 6.28$ T

(D) $2.519$ T

Solution:

M = 5 A m²

V = 10 cm × 0.5 cm × 0.2 cm = 10^-6 m³

H = 0.5 × 10⁴ A m⁻¹

$\displaystyle B={{\mu }_{0}}\left( H+I \right)$

$\displaystyle \Rightarrow B={{\mu }_{0}}\left( H+\frac{M}{V} \right)$

$\displaystyle \Rightarrow B=4\pi \times {{10}^{-7}}\times \left( 0.5\times {{10}^{4}}+\frac{5}{{{10}^{-6}}} \right)$

$\displaystyle B=6.28T$

Example 2.

A magnetic substance of volume 30 cm³ is placed in a magnetic field of 5 Oersted. If the magnetic moment produced in the material is 6 A-m², then calculate the magnetic induction.

Solution:

$\displaystyle V=30c{{m}^{3}}=30\times {{10}^{-6}}{{m}^{3}}$

$\displaystyle H=5\text{ }Oe=5\times \left( 80A{{m}^{-1}} \right)=400A{{m}^{-1}}$

$\displaystyle M=6A{{m}^{2}}$

$\displaystyle B={{\mu }_{0}}\left( H+\frac{M}{V} \right)$

$\displaystyle \Rightarrow B=4\pi \times {{10}^{-7}}\times \left( 400+\frac{6}{30\times {{10}^{-6}}} \right)$

$\displaystyle B=0.25T$

Example 3.

A magnetic flux density of 3000 G is produced when a rod of ferromagnetic material of cross sectional area 1 cm² is placed in a magnetic field of 200 Oersted. Calculate the value of magnetisation and magnetic susceptibility of the rod.

Solution:

$\displaystyle B=3000G=3000\times {{10}^{-4}}T$

$\displaystyle H=200\text{ }Oe=200\times \frac{{{10}^{3}}}{4\pi }A{{m}^{-1}}$

Now

$\displaystyle B=\mu H={{\mu }_{0}}{{\mu }_{r}}H$

$\displaystyle \Rightarrow {{\mu }_{r}}=\frac{B}{{{\mu }_{0}}H}=\frac{3000\times {{10}^{-4}}}{4\pi \times {{10}^{-7}}\times \left( 200\times \frac{{{10}^{3}}}{4\pi } \right)}=15$

$\displaystyle {{\chi }_{m}}={{\mu }_{r}}-1=15-1$

$\displaystyle {{\chi }_{m}}=14$

Example 4.

A magnetizing field of 2 × 10³ Am^-1 produces a magnetic field of 8π T in an iron rod. Find the relative magnetic permeability of the rod.

Solution:

$\displaystyle {{\mu }_{r}}=\frac{B}{{{\mu }_{0}}H}=\frac{8\pi }{\left( 4\pi \times {{10}^{-7}} \right)\times \left( 2\times {{10}^{3}} \right)}$

$\displaystyle {{\mu }_{r}}={{10}^{4}}$

Example 5.

Following is the relation between the magnetic permeability(μ) and the magnetising field(H) of an iron sample :-

$\displaystyle \mu =\left( \frac{\text{0}\text{.4}}{H}+12\times {{10}^{-4}} \right)henry/metre$

Here the units of H are A-m. Find the value of H which will produce a magnetic induction of 1.0 Weber/metre².

Solution:

$\displaystyle B=\mu H$

$\displaystyle \Rightarrow 1=\left( \frac{\text{0}\text{.4}}{H}+12\times {{10}^{-4}} \right)H$

$\displaystyle \Rightarrow 1=0.4+12\times {{10}^{-4}}H$

$\displaystyle \Rightarrow 12\times {{10}^{-4}}H=0.6$

$\displaystyle \Rightarrow H=\frac{0.6}{12\times {{10}^{-4}}}$

$\displaystyle \therefore H=500A{{m}^{-1}}$

Example 6.

The magnetic susceptibility of a paramagnetic material is 3.14 × 10^-4 . It is placed in a magnetising field of 6 × 10⁴ Am^-1 . Find :- (a) Magnetising intensity (b) Relative magnetic permeability.

Solution:

$\displaystyle {{\chi }_{m}}=3.14\times {{10}^{-4}}$

$\displaystyle H=6\times {{10}^{4}}A{{m}^{-1}}$

(a) Magnetising intensity

$\displaystyle I={{\chi }_{m}}H=\left( 3.14\times {{10}^{-4}} \right)\times \left( 6\times {{10}^{4}} \right)$

$\displaystyle I=18.84A{{m}^{-1}}$

(b) Relative magnetic permeability

$\displaystyle {{\mu }_{r}}=1+{{\chi }_{m}}$

$\displaystyle {{\mu }_{r}}=1+3.14\times {{10}^{-4}}$

$\displaystyle {{\mu }_{r}}=1.00034$

Example 7.

The magnetic susceptibility of a magnetic material is 30 × 10^-4. Find the absolute and relative magnetic permeability of this material.

Solution:

$\displaystyle {{\chi }_{m}}=30\times {{10}^{-4}}$

Absolute magnetic permeability,

$\displaystyle \mu ={{\mu }_{0}}\left( 1+{{\chi }_{m}} \right)$

$\displaystyle \mu =4\pi \times {{10}^{-7}}\times \left( 1+30\times {{10}^{-4}} \right)$

$\displaystyle \mu =12.6\times {{10}^{-7}}Wb{{A}^{-1}}{{m}^{-1}}$

Relative magnetic permeability,

$\displaystyle {{\mu }_{r}}=1+{{\chi }_{m}}$

$\displaystyle {{\mu }_{r}}=1+30\times {{10}^{-4}}=1+0.003=1.003$

Example 8.

An iron rod of length 0.5 m and 2 mm² cross-sectional area is placed along the axis of a solenoid. If the magnetising field (H) produced in the solenoid is 5000 Am^-1 and the permeability of iron is 16π × 10^-5 WbA^-1m^-1, then find the value of the magnetic moment of the rod.

Solution:

Given

$\displaystyle \begin{array}{l}H=5000A{{m}^{-1}}\\\mu =16\pi \times {{10}^{-5}}Wb{{A}^{-1}}{{m}^{-1}}\\A=2m{{m}^{2}}=2\times {{10}^{-6}}{{m}^{2}}\\l=0.5m\\M=?\end{array}$

Now

$\displaystyle I=\frac{M}{V}$

$\displaystyle \Rightarrow M=VI$

$\displaystyle \Rightarrow M=V\left( {{\chi }_{m}}H \right)$

$\displaystyle \Rightarrow M=VH\left( {{\mu }_{r}}-1 \right)$

$\displaystyle \Rightarrow M=VH\left( \frac{\mu }{{{\mu }_{0}}}-1 \right)$

$\displaystyle \Rightarrow M=AlH\left( \frac{\mu }{{{\mu }_{0}}}-1 \right)$

$\displaystyle \Rightarrow M=\left( 2\times {{10}^{-6}} \right)\times \left( 0.5 \right)\times \left( 5000 \right)\times \left( \frac{16\pi \times {{10}^{-5}}}{4\pi \times {{10}^{-7}}}-1 \right)$

$\displaystyle \therefore M=1.995A{{m}^{2}}$

Example 9.

A material of relative magnetic permeability 400 is filled in the core of a solenoid. The turns of the solenoid are insulated from the core and an electric current of 2 A is flowing in it. If the number of turns on unit length is 1000 then find :-

(a) H

(b) B

(d) Magnetising current

Solution:

Given :-

$\displaystyle {{\mu }_{r}}=400$

$\displaystyle i=2A$

$\displaystyle n=1000{{m}^{-1}}$

(a)

$\displaystyle H=ni=1000\times 2=2000A{{m}^{-1}}$

(b)

$\displaystyle B=\mu H={{\mu }_{0}}{{\mu }_{r}}H$

$\displaystyle B=\left( 4\pi \times {{10}^{-7}} \right)\times 400\times 2000=1.0048T\approx 1T$

(c)

$\displaystyle I={{\chi }_{m}}H=\left( {{\mu }_{r}}-1 \right)H=\left( 400-1 \right)\times 2000$

$\displaystyle I=7.98\times {{10}^{5}}A{{m}^{-1}}$

(d) Magnetising current (i_m) is that additional electric current which when flows through the turns of the solenoid in the absence of a core, produces the same magnetic field B as it is produced in the presence of the core, hence

$\displaystyle B={{\mu }_{\text{0}}}n\left( i+{{i}_{m}} \right)$

$\displaystyle \Rightarrow {{i}_{m}}=\frac{B}{{{\mu }_{\text{0}}}n}-i$

$\displaystyle \therefore {{i}_{m}}=\frac{1}{4\pi \times {{10}^{-7}}\times 1000}-2$

$\displaystyle {{i}_{m}}=796-2=794A$

Example 10.

A rod of magnetic material having cross-sectional area of 0.25 cm² is placed in a magnetising field of 4000 Am^-1. The value of magnetic flux passing through the rod is 25 × 10^-6 Weber. Find the following quantities for the rod :-

(a) Magnetic permeability

(b) Magnetic susceptibility and

Solution:

Given :-

$\displaystyle A=0.25c{{m}^{2}}=0.25\times {{10}^{-4}}{{m}^{2}}$