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To show that $\int_{C} \vec{B}\cdot \vec{dl}=4\pi I/c$ for this loop

enter image description here

Purcell uses this other path ($C'$)

![enter image description here

He argues that since $C'$ doesn't enclose the wire

$$\begin{align*}\int_{C'}\vec{B}\cdot \vec{dl'}&=0\\ \int_{C_1}\vec{B}\cdot \vec{dl_1}+\int_{C}\vec{B}\cdot \vec{dl}&=0\end{align*}$$

and since $\int_{C_1}\vec{B}\cdot \vec{dl_1}=-4\pi I/c$ then $$\int_{C}\vec{B}\cdot \vec{dl}=4\pi I/c.$$

But I might as well choose to add another circular loop like $C_2$ (assuming $C$ is non planar). Now in this case I would get $$\int_{C}\vec{B}\cdot \vec{dl}=\color{red}{2}\times 4\pi I/c=8\pi I/c$$

enter image description here

Here, $C_2$ is in front of $C_1$. Where does this contradiction arise from?

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  • $\begingroup$ Why is C2 non-planar? and why do you think there is a contradiction? Are the other current loops in the same plane? $\endgroup$
    – jamie1989
    Apr 25 '20 at 16:40
  • $\begingroup$ There is only one wire and it's represented by the dot(the wire is coming at us out of the screen). The loops $C$, $C_1$, $C$ and $C'$ are not wires. $\endgroup$
    – Hilbert
    Apr 25 '20 at 16:44
  • $\begingroup$ sorry, I meant paths, not current loops. $\endgroup$
    – jamie1989
    Apr 25 '20 at 16:46
  • $\begingroup$ 1) I didn't say $C_2$ was non planar, it is planar: it is within the plane that's perpendicular to the wire. 2) The contradiction is that we get two different values of the line integral of $B$ around $C$. $\endgroup$
    – Hilbert
    Apr 25 '20 at 16:50
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    $\begingroup$ I see, but now it looks like your integral around C' being equal to zero isn't correct as the entire path including C2 does enclose the wire? $\endgroup$
    – jamie1989
    Apr 25 '20 at 17:23
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Imagine that you pulled and reshaped the part of the loop you called $C1$ as shown below.

enter image description here

In doing this you have not cut through the current carrying conductor.

You can now see that within your Amperian loop you have the current carrying conductor.

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