Question

A magnetizing field of \[1600\dfrac{A}{m}\] Produces a magnetic flux $2.4 \times {10^{ - 5}}Wb$ in a bar of iron of cross-section $0.2c{m^2}$. Calculate permeability and susceptibility of the bar:

Answer 1

Hint: In this question, we start with the equation $B = \dfrac{\phi }{a}$ then we use this equation to get the permeability as $\mu = \dfrac{B}{H} = 7.5 \times {10^{ - 4}}Weber/Am$ . After that, we use ${\mu _0}$ and $\mu $ where,${\mu _0}$ the permeability of free space and equal to $4\pi \times {10^{ - 7}}Weber/Am$and get the relative permeability ${\mu _r} = \dfrac{\mu }{{{\mu _0}}}$.At last, we use the expression ${\mu _r} = 1 + {\chi _m}$ to get the value of magnetic susceptibility.

Complete step-by-step answer:
Given that
 Magnetic field intensity is $H = 1600A/m$
Flux is $\phi = 2.4 \times {10^{ - 5}}Weber$
Cross-section area is $a = 0.2c{m^2} = 0.2 \times {10^{ - 4}}{m^2}$
We know that permeability is the measure of the ability of a material to support the formation of the magnetic field.
Now we need to find the permeability $\mu $ so we use the expression

$\mu = \dfrac{B}{H}$--------------------------------- (1)
Here B is the Magnetic flux density. B can be calculated as
$B = \dfrac{\phi }{a}$
$ \Rightarrow B = \dfrac{{2.4 \times {{10}^{ - 5}}Weber}}{{0.2 \times {{10}^{ - 4}}{m^2}}}$
\[ \Rightarrow B = 1.2Weber/{m^2}\]------------------------ (2)

Substitute equation (2) in equation (1) we will get the permeability as:

$\mu = \dfrac{{1.2Weber/{m^2}}}{{1600A/m}}$
$ \Rightarrow \mu = 7.5 \times {10^{ - 4}}Weber/Am$
Hence the permeability is $7.5 \times {10^{ - 4}}Weber/Am$
Now we need to find the relative permeability ${\mu _r}$ that is

${\mu _r} = \dfrac{\mu }{{{\mu _0}}}$

Here ${\mu _0}$ is the permeability of free space that is equal to $4\pi \times {10^{ - 7}}Weber/Am$
So we will get relative permeability ${\mu _r}$ after substituting ${\mu _0}$and $\mu $ as.

 ${\mu _r} = \dfrac{{7.5 \times {{10}^{ - 4}}Weber/Am}}{{4\pi \times {{10}^{ - 7}}Weber/Am}}$
$ \Rightarrow {\mu _r} = 0.5968 \times {10^3} = 596.8$------------------------------- (3)

We know magnetic susceptibility denoted by ${\chi _m}$ is a measure of up to what extent a material will get magnetized in an applied magnetic field. Now we write the expression of ${\chi _m}$ as

${\mu _r} = 1 + {\chi _m}$
$ \Rightarrow {\mu _r} - 1 = {\chi _m}$

Using equation (3) we get

$ \Rightarrow {\chi _m} = 596.8 - 1 = 595.8$
Hence the magnetic susceptibility is $595.8$
Therefore, we get permeability and magnetic susceptibility as $7.5 \times {10^{ - 4}}Weber/Am$ and $595.8$ respectively.

Note: For these types of questions we need to understand different terms related to magnetization that are magnetic field intensity H, magnetic flux density B, magnetic flux $\phi $, permeability, relative permeability, magnetic susceptibility, magnetic dipole, magnetic hysteresis. We also need to know the expression of each term and how to relate them to each other.