Inner and outer ring, Biot-Savart law So basically I have an inner ring with radius $R_2$ and an outer ring with radius $R_1$, with current $I_0$ flowing through the outer ring. I am asked to evaluate the magnetic field strength at the position of the inner ring using the Biot-Savart law $$\vec{B}(\vec{r}) = \frac{\mu_0 I}{4 \pi} \int \frac{d \vec{s} \times (\vec{r}- \vec{s})}{| \vec{r}-\vec{s}|^3}$$ with the assumption that $R_2 << R_1$. I tried doing that and came to the expression $$\vec{B}(\vec{R}_2) = \frac{\mu_0 I}{4 \pi} \int_0^{2 \pi} \frac{-R_1R_2 + R_1^2 \cos(\varphi - \varphi')}{\sqrt{R_1^2+R_2^2-2R_1R_2 \cos(\varphi-\varphi')}} \vec{e}_z d \varphi'$$ with $\vec{r} =\vec{R}_2 = R_2 \begin{bmatrix} \cos(\varphi) \\ \sin(\varphi) \\ 0 \end{bmatrix}$ and $\vec{s} = R_1 \begin{bmatrix} \cos(\varphi') \\ \sin(\varphi') \\ 0 \end{bmatrix}$, $d \vec{s} = R_1 \begin{bmatrix} -\sin(\varphi') \\ \cos(\varphi') \\ 0 \end{bmatrix} d \varphi$   but I can't continue. Any advice? Because I feel like this might be a bit hard to solve :/
 A: Assuming $\hat{e}_z$,$R_1$ and $R_2$ is constant, these integrals can be put into forms that look like elliptic integrals. $F(x,m) = \int_0^x (1-m\sin^2{x})^{-1/2} dx$ is the elliptic integral of the first kind. $E(x,m) = \int_0^x (1-m\sin^2{x})^{1/2} dx$ is the elliptic integral of the second kind. Using these special functions, Mathematica can give us useful antiderivatives:
$$ \int \frac{dx}{(b-\cos{x})^{3/2}} = 2 \frac{(b-1)^{1/2}}{b^2-1} E(x/2,2/(1-b)) + 2 \frac{\sin{x}}{(-1+b^2)\sqrt{b-\cos{x}}} + C $$
$$ \int \frac{\cos{x} \, dx}{(b-\cos{x})^{3/2}} = \frac{2b\sin{x}}{(b^2-1)\sqrt{b-\cos{x}}} + \frac{2b(b-1)^{1/2}}{b^2-1} E(x/2,2/(1-b)) - \frac{2}{(b-1)^{1/2}} F(x/2,2/(1-b)) + C$$
For more info: https://analyticphysics.com/Special%20Functions/A%20Miscellany%20of%20Elliptic%20Integrals.htm
A: Stop reading and check your numerator again.
PS:
Rather
\begin{align*}
\vec{B}(\vec{R}_2) &= \frac{\mu_0 I}{4 \pi} \int_0^{2 \pi} \frac{R_1^2 - R_1R_2 \cos(\varphi - \varphi^\prime)}{\sqrt{R_1^2+R_2^2-2R_1R_2 \cos(\varphi-\varphi^\prime)}} \vec{e}_z d \varphi^\prime\\
\vec{B}(\vec{R}_2) &= \frac{\mu_0 I}{4 \pi} \int_0^{2 \pi} \frac{R_1^2\left(1 - \frac{R_2}{R_1}\cos(\varphi - \varphi^\prime)\right)}{R_1^3\sqrt{1+\frac{R_2^2}{R_1^2}-2\frac{R_2}{R_1} \cos(\varphi-\varphi^\prime)}} \vec{e}_z d \varphi^\prime\\
\vec{B}(\vec{R}_2) &= \frac{\mu_0 I}{4 \pi R_1} \left(\int_0^{2 \pi}d \varphi^\prime\right)\vec{e}_z&(\because R_1\gg R_2)\\
\vec{B}(\vec{R}_2) &= \frac{\mu_0 I}{2R_1}\vec{e}_z\\
\end{align*}
which is the magnetic field at the center of a current carrying ring with radius $R_1$. :p
