# Magnetic Field Along the Axis of the Current Ring - Alternative way to compute

This is Example 5.6 in Griffith's Introduction to Electrodynamics (4th Edition):

Find the magnetic field a distance $z$ above the centre of a circular loop of radius $\ R$, which carries a steady current $\ I$. I've attached the Figure from the textbook in the above.

The answer is that along the $z$-axis $\ \mathbf{B}(z)=\frac{\mu_{0}I}{2}\frac{R^{2}}{(R^{2}+z^{2})^{\frac{3}{2}}} \hat{\mathbf{z}}$ .

I understand the way Griffiths computed this result in his textbook (by considering the angle $\ \theta$ and whatnot). However, I would like to ignore this method and calculate this result using the Biot-Savart law ($without$ any symmetry arguments), which I cannot seem to do! I outline my process below:

$\vec{\mathscr{r}}=\mathbf{r}-\mathbf{r}'=z\ \hat{\mathbf{z}}-R\ \hat{\mathbf{s}}$

$\hat{\mathscr{r}}=\frac{z\ \hat{\mathbf{z}}\ -\ R\ \hat{\mathbf{s}}}{\sqrt{R^{2}+z^{2}}}$

The current $\ I$ runs in the $\ \hat{\mathbf{\phi}}$ direction (at radius $\ R$), and so I write $d\mathbf{l}'= R\ d\phi\ \hat{\mathbf{\phi}}$.

Then according to the Biot-Savart law, the magnetic field is given by:

$\mathbf{B}(\mathbf{r}) = \frac{\mu_{0}I}{4\pi} \int \frac{ d\mathbf{l}'\ \times\ \hat{\mathscr{r}} }{\mathscr{r}^{2}}$

And then:

$d\mathbf{l}'\ \times\ \hat{\mathscr{r}}$ = $\frac{IR^{2}\ d\phi}{\sqrt{R^{2}+z^{2}}} \hat{\mathbf{z}} - \frac{IRz\ d\phi}{\sqrt{R^{2}+z^{2}}} \hat{\mathbf{s}}$

I plug this into the Biot-Savart law, integrate all the way around the loop from $\phi = 0$ to $\phi = 2 \pi$, and get:

$\ \mathbf{B}(z)=\frac{\mu_{0}I}{2}\frac{R^{2}}{(R^{2}+z^{2})^{\frac{3}{2}}} \hat{\mathbf{z}} - \frac{\mu_{0}I}{2}\frac{Rz}{(R^{2}+z^{2})^{\frac{3}{2}}} \hat{\mathbf{s}}$

But I can't get rid of the term which points in the $\ \hat{\mathbf{s}}$ direction.....where have I gone wrong?

I hear all the time to "use symmetry arguments" to say that this part cancels to zero - but why isn't the math agreeing with this? Logically I can see the magnetic field only points along the $z$-axis for points on the $z$-axis - but I'd like to see the math behind this.

You have forgotten that the vector $\vec{{S}}$ is actually a function of $\phi$. If you quickly switch into cartesian coordinates $\vec{S}=\cos\left(\phi\right)\vec{i}+\sin\left(\phi\right)\vec{j}$ you can see that when you integrate the $\vec{S}$ component around $2\pi$ the sine and cosine give zero.
In cylindrical coordinates you have to remember that the vector away from the origin in the x-y plane (commonly called $\vec{\rho}$) is given by $\cos\left(\phi\right)\vec{i}+\sin\left(\phi\right)\vec{j}$ and in spherical coordinates the radial vector is given by $\vec{r}=\cos\left(\phi\right)\sin\left(\theta\right)\vec{i}+\sin\left(\phi\right)\sin\left(\theta\right)\vec{j}+\cos\left(\theta\right)\vec{k}$