# Does $\int_C B \cdot dr$ have any physical meaning when $C$ is *not* a closed loop?

Question. Does the line integral $$\int_C \textbf{B} \cdot d\textbf{r}$$ have any physical meaning when $$C$$ is not a closed loop? That is, I know that Ampere's law asserts that in the case of a static electric field, the line integral of the magnetic field around a closed loop is proportional to the electric current flowing through the loop. But I'm wondering if the line integral has any meaning when the path is just an arbitrary path through the magnetic field, and specifically not a closed loop?

Context. I am a mathematics grad student, TA'ing a vector calculus course. My professor has assigned a recitation activity where we are to calculate several such line integrals with several paths through the magnetic field generated by a wire. So I'm just wondering if there is any physical meaning to such things, as that will help illustrate the purpose of these integrals!

Thanks!

• If the exercise is done correctly, it can be a good way to demonstrate that that line integral depends on the path and not on the endpoints. That might be why the instructor chose to set those tasks. Oct 24, 2021 at 10:48
• @EmilioPisanty I think that's exactly the point yes. Thanks for your input! Oct 24, 2021 at 19:02

I mean, you can force some artificial meaning onto it: for example, if you need to calculate an integral over a stick perpendicular to the current carrying wire, $$I_1$$, you can write something like that: $$I_1 + I_2 = \mu_0 J$$, where $$J$$ is current. $$I_2$$ is very easy to calculate, it is $$\frac{2\pi-\theta}{2\pi}\mu_0 I$$, therefore you find $$I_1 = \frac{\theta}{2\pi} \mu_0$$.