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Starting from this image, I have two loops $S_1$ and $S_2$ with two batteries where the currents $I_1$ and $I_2$ flows with clockwise verse. The line $\Gamma=\ell$ have always a clockwise verse.

enter image description here

My solution with Ampère's law is $$\oint_{\Gamma} \vec{B}\cdot d\vec{\ell}=\mu_0I_{\text{enc}}$$ where the total current concatenated $I_{\text{enc}}$ with the path is equal to $I_{\text{enc}}=-I_1+2I_2$. If I'm following the right hand rule, shouldn't the grey arrows of the curve be placed in the anti-clockwise direction?

I am not very convinced of the correctness of this image, and I hope to have welcome answers. Thank you very much.

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  • $\begingroup$ The symbol $I$ in there... is it what you call $\ell$ or is it a current $I$? $\endgroup$ May 4, 2020 at 21:29
  • $\begingroup$ @ZeroTheHero It is \Gamma=\ell. $\endgroup$
    – Sebastiano
    May 4, 2020 at 21:30

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You are right. The line integral in the direction shown in the diagram comes to $\mu_0 (I_1-2I_2)$. Alternatively we could reverse the arrows on the grey path, in which case the line integral would come to $\mu_0 (2I_2-I_1)$.

I don't think there's much more to be said. It is indeed conventional to relate the field direction to the current direction by a right hand rule, so the field due to S1 is in the sense of the grey path; the field due to S2 is in the opposite sense.

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