# Confusion regarding Ampère's law and non-planar loops

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

Purcell uses this other path ($$C'$$)

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 $$\int_{C}\vec{B}\cdot \vec{dl}=\color{red}{2}\times 4\pi I/c=8\pi I/c$$

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

• Why is C2 non-planar? and why do you think there is a contradiction? Are the other current loops in the same plane? Apr 25 '20 at 16:40
• 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. Apr 25 '20 at 16:44
• sorry, I meant paths, not current loops. Apr 25 '20 at 16:46
• 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 $C$. Apr 25 '20 at 16:50
• 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? Apr 25 '20 at 17:23

Imagine that you pulled and reshaped the part of the loop you called $$C1$$ as shown below.

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.