# Vector potential of a current carrying circular loop

1. Jul 21, 2014

### ShayanJ

Consider a circular loop of radius R that carrys a uniform current I. We know(by Biot-Savart law) that the magnetic field it produces on its axis is given by $\vec{B}=\frac{\mu_0 I R^2 \hat z }{2(z^2+R^2)^\frac 3 2}$.
But let's calculate its vector potential:
$\vec{A}=\frac{\mu_0}{4\pi} \int \frac{I \vec{dl}}{|\vec{r}-\vec{r}'|}=\frac{\mu_0}{4\pi} \int_0^{2\pi} \frac{I R d\varphi \hat \varphi}{\sqrt{R^2+z^2}}=\frac{\mu_0 I R}{4\pi \sqrt{R^2+z^2}}\int_0^{2\pi} \hat \varphi d\varphi \\$
But we have $\hat \varphi=-\sin\varphi \hat x +\cos\varphi \hat y$ and so $\int_0 ^{2\pi} \hat\varphi d\varphi=0$ which gives $\vec{A}=0$ and so $\vec{B}=0$ !!!
What's wrong?
Thanks

2. Jul 21, 2014

### Staff: Mentor

What is the relationship between A and B? Given that relationship, does A = 0 at a point imply B = 0 at that point?

3. Jul 21, 2014

### ShayanJ

$\vec{B}=\vec\nabla \times \vec A$ which gives B=0 for A=0. Gauge freedom can't help too(By definition).

Last edited: Jul 21, 2014
4. Jul 21, 2014

### Staff: Mentor

No, it doesn't.

Remember, you have evaluated A only along a line. Is that enough information to calculate its curl?

5. Jul 21, 2014

### ShayanJ

Oh God....yeah man!!!
Sorry...thanks

6. Jul 21, 2014

### Delta²

Well if i understand correctly you find the potential only along the z-axis. For a point outside the z-axis $|\vec{r}-\vec{r'}|$ will depend on $\varphi$ as well so the integral wouldnt be always zero.

7. Jul 22, 2014

### vanhees71

Yep, and unfortunately you are led to elliptic integrals :-(.