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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 $\ I$.

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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.

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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.

You have to be quite careful when using curvelinear coordinates to evaluate integrals remembering that your unit vectors might be functions of the three parameters.

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 $\vec{r}=\cos\left(\phi\right)\sin\left(\theta\right)\vec{i}+\sin\left(\phi\right)\sin\left(\theta\right)\vec{j}+\cos\left(\theta\right)\vec{k}$

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