# Magnetic flux through circular loop due to infinite wire [closed]

I’m trying to calculate magnetic flux that’s going through circular loop with radius $$R$$, due to magnetic field of a infinite wire that is in distance $$d$$ from the center of the loop. $$\vec{B}$$ vector is parallel to $$\vec{dS}$$ vector. I know that magnetic field of that wire is equal to $$B=\frac{\mu_0 I}{2\pi r}$$ and that flux is equal to $$\int_{S} \vec{B}\cdot \vec{dS}=\int_{S} B\cdot dS$$ where $$dS$$ Is surface of the loop, but what I don’t know is how to change that $$dS$$ so is possible to solve.

• How is the loop oriented relative to wire? Apr 17, 2021 at 17:53
• @R.W.Bird surface vector is parallel to B vector Apr 17, 2021 at 19:56
• Is the wire coplanar with the loop, like its forming a chord? Apr 18, 2021 at 10:15
• @lineage yes, I didn't know the word for that, but it is exactly what I mean. I added photo for better understanding. Apr 18, 2021 at 10:20
• @KacperKowerski hint: what exactly do you think is the area that you need to integrate? What direction does the magnetic field of an infinite wire depend upon? Does it have any symmetry that may simplify the calculation? Apr 18, 2021 at 10:35

The convenient infinitesimal surface $$\rm dS$$ is shown in the Figure-01 :

$$$$\mathrm{dS} \boldsymbol{=}\mathrm{hdw}\boldsymbol{=} (2R\sin\theta)( \mathrm d\ell\sin\theta)\boldsymbol{=} (2R\sin\theta)( R\mathrm d\theta\sin\theta) \tag{01}\label{01}$$$$ so $$$$\mathrm{dS} \boldsymbol{=}2R^2\sin^2\theta\mathrm d\theta\boldsymbol{=}R^2(1\boldsymbol{-}\cos2\theta)\mathrm d\theta \tag{02}\label{02}$$$$ We could verify that $$$$\int\limits_{\theta\boldsymbol{=}0}^{\theta\boldsymbol{=}\pi}\!\!\!\!\mathrm{dS}=\pi R^2 \tag{03}\label{03}$$$$

Hence for the magnetic flux through the circle we have $$$$\Phi=\int\!\!\!\int\limits_{\!\!\!\!\!\!\bf circle}\!\!\mathbf{B}\boldsymbol{\cdot}\mathrm{d}\mathbf{S}=\int\!\!\!\int\limits_{\!\!\!\!\!\!\bf circle}\!\!\mathrm{B}\,\mathrm{dS}=\dfrac{\mu_{0}\mathrm{I}R^2}{\pi }\int\limits_{\theta\boldsymbol{=}0}^{\theta\boldsymbol{=}\pi}\!\!\!\!\dfrac{\sin^2\theta\,\rm d\theta}{\mathrm{ L}-R\cos\theta} \tag{04}\label{04}$$$$

In Figure-02 below we see a detail of Figure-01 corresponding to this comment of OP :

Why $$\rm dw=\sin\theta d\ell$$ ? – cover

• You wrote that h=2R$\cos\theta$ and dw=Rd$\theta\cos\theta$, but later dS equals to $\sin^2\theta$, why $\sin\theta \cdot \cos\theta$ equals $\sin^2\theta$? Apr 18, 2021 at 13:20
• How do you make these visuals?? Apr 19, 2021 at 9:04
• @Buraian : I use GeoGebra. It's free. Apr 19, 2021 at 9:08
• Why $dw = \sin\theta dl$? Apr 19, 2021 at 15:06
• @Buraian The images are fantastics....and also the answers :-) May 13, 2021 at 11:30

fig 1: The geometry of the calculation. Note that the distance of the wire from the center has been taken to be $$D$$, not $$d$$.

Let the magnitude of the magnetic field due to an infinite constant current carrying straight wire at a distance $$r$$ from it be $$k/r$$.

In the given geometry, the wire is co-planar with the loop. This makes the problem effectively $$2$$ dimensional. Taking the interior of the circle (in its embedding plane) as the surface of integration of flux calculation, the magnetic field is everywhere (anti)parallel to the surface normal vector, as noted in the question.

Since the magnetic field is only dependent on the distance from the wire, it is constant in the small differential patch of area $$2\,y\, dx$$, colored gray in fig $$1$$. The magnitude of the total flux is then given by

\begin{align} I&=\int_{-R}^{R}\frac{k \,2 y dx}{r}&\\ &=2 k\int_{-R}^{R}\frac{\sqrt{R^2-x^2}dx}{D-x}\tag{1}\\ \end{align}

which equals$$^{A.1}$$ $$I= \begin{cases} 2\pi k (D-\sqrt{D^2-R^2})& D>R\tag{2}\\ \phantom{}\\ 2\pi k D & D\leq R\\ \end{cases}$$

Appendix

1. The integral in the RHS of equation $$1$$ is (via Mathematica)
$$-y+D\tan^{-1}\frac{x}{y}-\sqrt{D^2-R^2}\tan^{-1}\left(\frac{D x -R^2}{y\sqrt{D^2-R^2}}\right)$$
where $$y=\sqrt{R^2-x^2}$$ with limits $$(x,y)$$ going from $$(-R,0)$$ to $$(R,0)$$. One has to be careful with the limits here. While the third term is imaginary for $$D, it cancels out for the limits.

2. First case of Eqn. $$2$$ has the expected asymptotic form, $$k \pi R^2/D$$, for $$D\to\infty$$. Interestingly, if one considers only its real part, it holds for $$R\to\infty$$ too.

3. The second case of Eqn. $$2$$ viz. the case of the wire being inside the loop, is remarkably independent of $$R$$ and depends linearly on $$D$$. This was indeed verified numerically$$(1\sigma)$$.

fig 2: Monte-carlo verification of eqn. 2 for $$R=1,2\,\,(k=1)$$

• What if i want $d$ to be in the center of the loop, because I came to same equation as you but in my case $d$ is in the center of the loop. How should I write integral with my $d$? Apr 18, 2021 at 14:22
• @KacperKowerski give me a sec I am changing it to centered form Apr 18, 2021 at 14:23
• Well, I meant on integral level. Because when I try to calculate integral with your replacement I get much more complicated answer than I should. Apr 18, 2021 at 14:33
• At the integral level, all that happens is in eqn 1, the denominator becomes $D-x$ ...the "level" of susbstitution shouldn't matter as $d,D,R$ are constant during integration. The limits are the same Apr 18, 2021 at 14:38