Question

The area of I−H curve and area of B−H curve are related as:
A) Area of I−H curve < area of B−H curve
B) Area of I−H curve > area of B−H curve
C) Area of I−H curve = area of B−H curve
D) None of these

Answer 1

This problem talks about the hysteresis loop created when we plot magnetic field or magnetization vector field against magnetic field strength. Area under these curves depend on the dependence of the magnetic field on magnetic field strength and the magnetization vector.

Formula used:
The relationship between magnetic field, magnetic field strength and magnetization vector is given by:
$\vec B = {\mu _0}(\vec H + \vec I)$...........(1)
Where,
$\vec B$ represent the magnetic field vector,
${\mu _0}$ is the permeability of free space,
$\vec H$ represent the magnetic field strength,
$\vec I$ represent the magnetization vector.

Complete step by step answer:
Step 1:
Take the total derivative of each term of eq.(1) and multiply $\vec H$ with each term. Then take integral along the closed loop to get the relation:

d\vec B = {\mu _0}d\vec H + {\mu _0}d\vec I \\\ \Rightarrow \vec H.d\vec B = {\mu _0}\vec H.d\vec H + {\mu _0}\vec H.d\vec I \\\ \therefore \oint {\vec H.d\vec B = {\mu _0}\oint {\vec H.d\vec H} + } {\mu _0}\oint {\vec H.d\vec I} \\\ $$........(2) Step 2: In the above integral relation of eq.(2) the term $\oint {\vec H.d\vec B} $ represents an area confined by the hysteresis loop in the B-H curve. And similarly, the term $\oint {\vec H.d\vec I} $ represents an area confined by the hysteresis loop in the I-H curve. Now, $\vec H.d\vec H$ is an odd function. So, integrating it over a symmetric limit around 0 yields $\oint {\vec H.d\vec H} = 0$. Hence, the above relation turns out to be: $\oint {\vec H.d\vec B = } {\mu _0}\oint {\vec H.d\vec I}$. From this relationship it is evident that the area under the B-H curve is $${\mu _0}$$times the area under I-H curve. Since, $${\mu _0} = 4\pi \times {10^{ - 7}} \ll 1$$, the area under the I-H curve is significantly larger than that of the B-H curve. **The correct answer is (B), area of I−H curve > area of B−H curve.** **Note:** While finding the area under the closed hysteresis loop for both I-H and B-H curves, just taking normal integration will not give an accurate result. A student must take a path integral around the closed loop to get the area covered by it.