I wanted to calculate the magnetic field in the figure using the magnetic vector potential.

enter image description here

In case the drawing I made is not too clear, that is a circular wire of radius $a$ with a current $I$ across it. The $z$-axis goes through the center of the circle. For simplicity, I assumed that the circle is on the plane at $z=0$.

Let $|\mathbf{r}-\mathbf{r'}|=R$. So the vector potential is calculated as

$$\mathbf{A}(\mathbf{r})=\frac{\mu _0}{4\pi}\iiint\limits_V\frac{\mathbf{J}(\mathbf{r'})}{R}dV'$$

In this case the wire is infinteliy thin, so the volume integral becomes a line one and using cylindrical coordinates we have that

$$\mathbf{A}=\frac{\mu _0}{4\pi}\oint\limits_C \frac{Ia\mathbf{\hat{\phi}}}{R} d\phi'=\frac{\mu _0 Ia}{4\pi \sqrt{(z^2+a^2)}}\oint\limits_C \mathbf{\hat{\phi}} d\phi'$$

But $\oint\limits_C \mathbf{\hat{\phi}} d\phi'=\mathbf{0}$. So, according to these calculations:

$$\mathbf{A}=\mathbf{0}\implies \mathbf{\nabla}\times\mathbf{A}=\mathbf{B}=\mathbf{0}$$

So I got that $\mathbf{B}=\mathbf{0}$ but I know this is wrong, because the result for this problem is well-known and I have calculated without using $\mathbf{A}$.

However, I want to know what I did wrong during these steps. Can anyone see it?

  • 1
    $\begingroup$ What you've done is equivalent to saying that if $f(x_0) = 0$, then we must have $f'(x_0) = 0$. The curl of $\mathbf{A}$ can be nonzero at a point where $\mathbf{A}$ is zero. $\endgroup$
    – knzhou
    Sep 6, 2016 at 3:32
  • $\begingroup$ @knzhou Yes, I get that, but I don't understand why you say that I did such thing. I mean, I thought I had got that $\mathbf{A}=\mathbf{0}$ for any point in space... Or maybe I didn't? $\endgroup$
    – Tendero
    Sep 6, 2016 at 3:41
  • $\begingroup$ You got it for any point on the $z$-axis. $\endgroup$
    – knzhou
    Sep 6, 2016 at 3:43
  • $\begingroup$ @knzhou Oh, I see it now. So $A_z (z)=0$ but I don't know how is it like for other points, right? $\endgroup$
    – Tendero
    Sep 6, 2016 at 3:44
  • $\begingroup$ Yeah.$\hspace{0mm}$ $\endgroup$
    – knzhou
    Sep 6, 2016 at 3:46

1 Answer 1


Using the line integral, you need to find a general vector function for A.

Say you are trying to find the field at a point (in cylindrical coordinates): ($r_1,\phi_1,z_1$) due to the circular loop.

Let the vector from the origin to this point be $\vec{r_1}$ and the vector from the origin to any point on the loop be $\vec{r'}$. The coordinates of a point on the loop are ($a,\phi ,0$).

Then the distance from a point on the loop to ($r_1,\phi_1,z_1$) $=r=|\vec{r_1}-\vec{r'}|=(a^2+r_1^2-2r_1a*cos(\phi-\phi_1)+z_1^2)^{1/2}$

We have,

$$\mathbf{A}(\mathbf{r})=\frac{\mu _0}{4\pi}\int\frac{\mathbf{I}(\mathbf{r'})} {r}dl'$$

$$\mathbf{A}(\mathbf{r})=\frac{\mu _0I}{4\pi}\int\frac{\mathbf{dl'}} {r}$$

$$\mathbf{A}(\mathbf{r})=\frac{\mu _0I}{4\pi}\int\limits_{0}^{2\pi}\frac{a*d\phi*\hat{\phi}} {(a^2+r_1^2-2r_1a*cos(\phi-\phi_1)+z_1^2)^{1/2}}$$

After evaluating this integral, you can take the curl with respect to $r_1,\phi_1,z_1$, hence generalizing these coordinates and obtaining a function for B.

The integral for A is frustratingly hard to evaluate, because it is exact (and involves elliptical integrals) So, we just use the multipole expansion to find the B field.

If you do want to see how to work through the integral, this treatment is absolutely fantastic. The video finds the field at points in the x-z plane, but if you change the $\phi'$ everywhere to $\phi'-\phi_1$, where $\phi_1$ is the azimuth angle of point P, the solution will be valid for all space.

  • $\begingroup$ So the A field is only in the azimuthal direction? $\endgroup$
    – Parabola
    Sep 6, 2016 at 18:57
  • $\begingroup$ Yes. That indicates that the B-field is only in the $\hat{\rho}$ and $\hat{z}$ directions, which mean it lies entirely in the x-z plane or in planes parallel to the x-z plane. If you watch the video, you will see that in Cartesian coordinates, the A-field is entirely in the y-direction, which indicates that the B-field is in the x and z directions only, agreeing with this. $\endgroup$
    – GeeJay
    Sep 7, 2016 at 13:51

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