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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

  1. $A_2=\mu_oA_1$

  2. $A_2=A_1$

  3. $A_1=\mu_oA_2$

  4. $A_2=\mu_o^2A_1$

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.

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I don't understand why is$\oint HdH=0$?

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1 Answer 1

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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.

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  • $\begingroup$ So, this is just using the property of cyclic integral right? $\endgroup$
    – Arsenic
    Commented Jun 18, 2022 at 17:38
  • $\begingroup$ @Arsenic, yes, perhaps. $\endgroup$
    – HEMMI
    Commented Jun 19, 2022 at 9:27
  • $\begingroup$ alright, makes sense to me- Thanks! $\endgroup$
    – Arsenic
    Commented Jun 19, 2022 at 14:43

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