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In the mit ocw lecture by Prof. Lewin on EMI, He quotes a few statements from here to couple of seconds of the lecture.

I am confused why the sign of $L\frac{dI}{dt}$ changes, when we go around and evaluate the closed loop integral in the opposite direction of the current.

By faraday's law , we have

$$\oint_{+\partial \Sigma} \vec{E} \cdot d\vec{\ell} = - \frac{d}{dt} \int_{+\Sigma} \vec{B}\cdot d\vec{A}$$

This seemingly suggest that the negative sign would be only affected by the dot product $\vec{B}\cdot d\vec{A}$ , and indeed, the direction of the $d\vec{A}$(Area vector) is only affected by the sense of path chosen for $d\vec{\ell}$ .

Then, what might be the role of current (not the time derivative of current) in deciding the sign of $L\frac{dI}{dt}$ ?

If I had made any wrong statement here, please do correct it. Thank you..

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Sorry for answering my own question, i just noticed a really simple solution for my question. The answer is not about the physics itself, but absence of mind , The current indeed decides the direction of magnetic field

I still have a doubt on the conclusion that whether the Area vector convention is strictly necessary for the thing which prof. Lewin mentions, as far as i tried, it is indeed necessary..

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