Current carrying wire produces a magnetic field around it, we all know that.

A circular wire carrying current produces a magnetic field around it due to the flow of electric charge. This phenomenon is described by Ampère's law and the Biot-Savart law in electromagnetism.

The magnetic field produced by a circular wire of radius a carrying current "I" can be calculated using the Biot-Savart law. For a circular loop, the magnetic field at a point on the axis perpendicular to the loop's plane (at a distance r from the centre of the loop) is given by:

$$B = \frac{\mu_o Ia^2}{2(a^2+ r^2)^{3/2}} $$

And at the centre of the wire, r = 0, $B = \frac{\mu_o I}{2a} $, a constant.

But what happens when we want to find the field at the point of the wire itself?

Taking an infinitesimal length on the wire, the magnetic field at that point would be the rest of all the rest of the points on the wire.

If the wire were a straight line, it would have been zero, as $I \times dl$ = 0, but that is not the case here.

For each point, the symmetric points would add up to give me some field as the direction of the current would be opposite as seen sitting at a particular point and integrating it directly could miss some points, I think.

How do you think I should approach this?

Thanks in advance.

  • $\begingroup$ Not clear why you suggest that for a straight wire the field would be zero. It would if you are exactly in the center, but it seems more interesting to compute the field at the surface of the wire, which case exactly is the question about? (And anyway, for the loop you can probably get some complete Elliptical integral as a result if you do the calculation.) $\endgroup$ Commented May 5 at 14:03
  • $\begingroup$ Yes, it will be an elliptical integral, but I don't know how to interpret that. $B = const * \int_0^1 \frac{dm}{m\sqrt{1-m^2}}$,, now according to mathematica, it would never converge, but that is not the case, we should get some value of finite field at the particular point in the wire. 'm' is some substitution of $\theta$ $\endgroup$
    – Gandalf73
    Commented May 5 at 14:48
  • $\begingroup$ Maybe a definition problem? I already asked about that here: math.stackexchange.com/questions/4901060/… $\endgroup$ Commented May 5 at 14:55
  • $\begingroup$ Please try to solve what comes to you. My approach was to take two infinitesimal lengths on the circle, Q and P, as dl, and find the r vector between them, and integrate the whole thing from 0 to $2\pi$. Thanks in advance $\endgroup$
    – Gandalf73
    Commented May 5 at 15:09
  • $\begingroup$ Is $r$ the distance to the plane of the current loop and $a$ its radius? $\endgroup$
    – my2cts
    Commented May 5 at 17:37

1 Answer 1


The fastest way I know (admittedly a bit sloppy) to come to solutions is:

  1. From symmetry we know there is only one vector potential component $A_\varphi$, assuming cylindrical coordinates and having the loop in the $z=0$ plane.
  2. If we look outside the wire we can treat the current as if it flows only on the center-line (even though it is on the surface, or distributed, the same trick as treating the mass of the Earth as being in the center).

Let $R$ be the radius of the loop and let's integrate over the center-line current with the Green function to find $A_\varphi$, which will only depend on $\rho$ and $z$:

$$\begin{align} A_\varphi(\rho,z) =&\ \frac{\mu_0\ I}{4\pi}\int\limits_0^{2\pi}\!\!R\,d\varphi\ \frac{\cos\varphi}{\sqrt{(R-\rho \cos\varphi)^2+(\rho \sin\varphi)^2+z^2}} \\[5pt] =&\ \frac{\mu_0\ I\ R}{4\pi}\int\limits_0^{2\pi}\!\!d\varphi\ \frac{\cos\varphi}{\sqrt{R^2+\rho^2+z^2-2R\rho \cos\varphi}} \\[5pt] =&\ \frac{\mu_0\ I\ R}{4\pi\sqrt{R^2+\rho^2+z^2}}\ \int\limits_0^{2\pi}\!\!d\varphi\ \frac{\cos\varphi}{\sqrt{1-u\, \cos\varphi}} \\[5pt] =&\ \frac{\mu_0\ I\ R}{4\pi\sqrt{R^2+\rho^2+z^2}}\ \Big[\frac{-4\sqrt{1+u}}{u}\ E\big(\frac{2u}{1+u}\big) +\frac{4}{u\,\sqrt{1+u}}\ K\big(\frac{2u}{1+u}\big) \Big] \end{align}$$ where after defining $u=2 R \rho/(R^2+\rho^2+z^2)$, the integral was found with Mathematica. This can further be reduced to: $$\begin{align} A_\varphi(\rho,z) =&\ \frac{\mu_0\, I\,\sqrt{(R+\rho)^2+z^2}}{2\pi\,\rho} \ \Big[-E(m) +\big(1-\frac m2 \big)\ K(m) \Big] \\[6pt] &\text{with}\ \ m=\frac{4R\,\rho}{(R+\rho)^2+z^2} \end{align}$$ In these expressions the complete Elliptic integrals $E(m)$ and $K(m)$ are defined as in Mathematica (see [question 4901060] for the definition ambiguity).

For the magnetic field, ${\bf B}={\bf \nabla}\times{\bf A}$, we use the expressions in cylindrical coordinates from [Jackson]. This leads to: $$\begin{align} {\bf B}_\rho(\rho,z) =&\ \frac{\mu_0\,I\, z\ \left[(r^2+R^2+z^2)\,E(m) -\left((r-R)^2+z^2\right)K(m) \right]}{2 \pi\, r\, \big((r-R)^2+z^2\big) \sqrt{(r+R)^2+z^2}} \\[8pt] {\bf B}_z(\rho,z) =&\ \frac{\mu_0\,I\ \left[\big(r^2 (z^2-R^2)+(R^2+z^2)^2\big)\,E(m) - (R^2+z^2) \big((r-R)^2+z^2\big) K(m) \right]}{2 \pi\, r^2\, \big((r-R)^2+z^2\big) \sqrt{(r+R)^2+z^2}} \end{align} $$ If $r_w$ is the wire diameter, then the field right at the surface of the wire is found as ${\bf B}_\rho(R,r_w)$, the horizontal field just above the wire at heigth $z=r_w$, or also as ${\bf B}_z(R-r_w,0)$, the vertical field at the inner side. So we ask Mathematica for some limits close to the current loop :

Limit[ z Br /.r->R, z->0, Direction->"FromAbove", Assumptions->R>0]
Limit[ w Bz /.{r->R-w,z->0}, w->0, Direction->"FromAbove", Assumptions->R>0]

and in both cases get mu0 I/(2Pi). So this proves that, as expected: $$\begin{align} {\bf B}_\rho(R,r_w) \sim &\ \frac{\mu_0\,I}{2 \pi\, r_w} \\[6pt] {\bf B}_z(R-r_w,0) \sim &\ \frac{\mu_0\,I}{2 \pi\, r_w} \end{align} $$ so the field simply goes inversely proportional to the distance from the wire, the same as for a straight wire. Other limits we can look at:

Limit[Bz, r->0]//Simplify
Limit[Bz, z->0]//Simplify

The first one is the field on the $z$-axis that was already meantioned in the question, the second is the field in the plane of the loop. Results: $$\begin{align} {\bf B}_z(0,z) =&\ \frac{\mu_0\,I\, R^2} {4\, (R^2+z^2)^{3/2}} \\[8pt] {\bf B}_z(\rho,0) =&\ \frac{\mu_0\,I\ R^2}{2 \pi\, \rho^2\, (R^2-\rho^2)} \Big[\ (\rho+R)\ E\Big(\frac{4\,\rho\,R}{(\rho+R)^2}\Big) +(\rho-R)\ K\Big(\frac{4\,\rho\,R}{(\rho+R)^2}\Big) \Big] \end{align} $$ If this last one is plotted for $R=1, I=1$, it looks like this:



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