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 of $\frac{A_2}{A_2}$ is given by

• A. ${\mu }_0$
• B. $\frac{1}{{\mu }_0}$
• C. $1$
• D. $\sqrt{{\mu }_0}$

Answer 1

The correct option is A ${\mu }_0$

The relation between B and H is given by
$B={\mu }_0(H+I)$
$db={\mu }_{0}dH+{\mu }_{0}dl$
$\oint H\cdot dB={\mu }_0\oint H\cdot dH+{\mu }_0\oint H\cdot dl$
Also $\oint H\cdot dH=0$
$\oint H\cdot dB={\mu }_0\oint H\cdot dl$
Area of B-H loop $={\mu }_0\times areaofl-Hloop$
$A_2={\mu }_0{A}_1$
$\Rightarrow \frac{A_1}{A_2}={\mu }_0$