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If the area under the I-H hysteresis loop and B-H hysteresis loop are denoted by $A_1$and $A_2$ (The symbols have usual meaning as set in electromagnetics), then
$A_2=\mu_oA_1$
$A_2=A_1$
$A_1=\mu_oA_2$
$A_2=\mu_o^2A_1$
The correct answer for this problem is option 1) $\boxed{A_2=\mu_oA_1}$.
My doubt lies in the official solution provided to us.
Since $\int H dH = \frac{1}{2}H^2$, we get
$$
\oint HdH=\left[\frac{1}{2}H^2\right]_{H_1}^{H_2}+\left[\frac{1}{2}H^2\right]_{H_2}^{H_1}=0
$$
Am I right? I welcome anyone's criticism.