Question

The areas under the $I-H$ hysteresis loop and $B-H$ hysteresis loop are denoted by ${{A}_{1}}$ and ${{A}_{2}}$, then the ratio $\dfrac{{{A}_{2}}}{{{A}_{1}}}$ is given by
A) ${{\mu }_{0}}$
B) $\dfrac{1}{{{\mu }_{0}}}$
C) $1$
D) $\sqrt{{{\mu }_{0}}}$

Answer 1

Hint: The magnetic permeability of a magnetic material gives an idea about the number of magnetic field lines per unit area, the magnetic material permits when placed in a magnetic field. The magnetic susceptibility of a magnetic material is defined as the ratio of the intensity of magnetisation to the magnetising force. The relationship between magnetic permeability and magnetic susceptibility can be used to find the ratio of areas under the $B-H$ and the $I-H$ loops.
The area under the hysteresis curve is equal to the integral loop $\oint{HdB}$ or $\oint{HdI}$.

Complete answer:
When a magnetic field is placed in a magnetising field of magnetising intensity (or magnetising force) $H$, the material gets magnetised. Hysteresis curve represents the relation between the magnetic induction $B$ and the magnetic intensity $H$. Hysteresis curve can also be drawn in terms of the intensity of magnetisation $I$ and the magnetic intensity $H$. Both the shapes of $B-H$ curve and $I-H$ look identical for a particular magnetic material. The shape and size of the hysteresis curve depend upon the nature of the magnetic material. By studying the hysteresis loop of magnetic materials, one can study the differences in the material’s properties such as permeability, retentivity, energy loss, etc. For example, the area under the hysteresis curve can be used to determine the energy loss in the process of magnetisation of the material.

Let us assume that a magnetic material gets magnetised with the help of a magnetising field of magnetic intensity $H$. The total magnetic induction $B$in the material is equal to the sum of magnetic induction ${{B}_{0}}$in vacuum produced by the magnetic intensity and the magnetic induction ${{B}_{m}}$,due to magnetisation of the material.
$B={{B}_{0}}+{{B}_{m}}$
Here,
${{B}_{o}}={{\mu }_{0}}H$ and ${{B}_{m}}={{\mu }_{0}}I$ ,where ${{\mu }_{0}}$ is the magnetic permeability of free space.
Therefore,
$B={{B}_{0}}+{{B}_{m}}={{\mu }_{0}}H+{{\mu }_{0}}I={{\mu }_{0}}(H+I)$ (equation 1)

Now, the area under $B-H$ hysteresis loop is given by the loop integral $\oint{HdB}$
And the area under the $I-H$ hysteresis loop is given by the loop integral $\oint{HdI}$. From the question, it is clearly mentioned that these are ${{A}_{2}}$ and ${{A}_{1}}$ respectively.
From equation 1, we have
$B={{\mu }_{0}}(H+I)={{\mu }_{0}}H+{{\mu }_{0}}I$
On differentiation, we get
$dB={{\mu }_{0}}dH+{{\mu }_{0}}dI$
Let us multiply the above equation by $H$on both sides to obtain the area under the hysteresis curves.
After multiplication, we have
$HdB={{\mu }_{0}}HdH+{{\mu }_{0}}HdI$
Integrating the above equation to get the loop integrals or the area under the curves, we have
$\oint{HdB}={{\mu }_{0}}\oint{HdH}+{{\mu }_{0}}\oint{HdI}$ (equation 2)
It is clear that ${{\mu }_{0}}\oint{HdH}=0$ because the area under this loop cannot be determined.
Therefore, equation 2 becomes

$\oint{HdB}={{\mu }_{0}}\oint{HdI}$
OR
${{A}_{2}}={{\mu }_{0}}{{A}_{1}}$
OR
$\dfrac{{{A}_{1}}}{{{A}_{2}}}=\dfrac{1}{{{\mu }_{0}}}$
OR
$\dfrac{{{A}_{2}}}{{{A}_{1}}}={{\mu }_{0}}$

So, the correct answer is “Option A”.

Note:
Students need not get confused with magnetic intensity $H$and the intensity of magnetisation $I$. Magnetic intensity$(H)$is the magnetising force acting on the magnetic material when placed in a magnetic field. Intensity of magnetisation$(I)$ is the induced magnetisation in the magnetic material when placed in a magnetic field.