I wish to calculate the magnetic field due to a charged ring rotating about an axis perpendicular to the plane of the ring and passing through a point on its circumference $(P)$. I arrive at something unusual.
Here's what i did
Let the radius of the ring be $R$, let it rotate with an angular velocity $\omega$.
According to Biot-Savart law, the magnetic field due to a moving charge is $\dfrac{\mu_o}{4\pi}\dfrac{dq(\vec{v} \times \hat{r})}{|\vec{r}|^2}$. Let us consider a charge $dq$. Let $\angle POQ$ be $\theta$ and let the angle subtended by the small element $dq$ be $d\theta$. Through some geometry, the distance $PQ$ (let us call it $x$) is found to be $2R\sin \left(\dfrac{\theta}{2}\right)$
We now see that all such elements move in circles with radius $x=2R\sin \left(\dfrac{\theta}{2}\right)$.
So the magnetic field due to all such elements will be pointing out of the plane. We also note that the velocity and $x$ are always perpendicular.
Therefore, the magnetic field due to that element at point $P$ $$|\vec{dB}|=\dfrac{\mu_o}{4\pi}\dfrac{dq \ v}{x^2}=\dfrac{\mu_o}{4\pi}\dfrac{\omega \ dq}{x}$$
Substituting $x=2R\sin \left(\dfrac{\theta}{2}\right)$ and $dq=\dfrac{Qd\theta}{2\pi}$, $$\color{blue}{|\vec{B}|=\dfrac{\mu_oQ\omega}{16\pi^2R}\displaystyle\int_{0}^{2\pi} \csc \left(\dfrac{\theta}{2}\right) \ d\theta}$$
But i don't think the last integral converges.
Note that this integral can be evaluated if the values of the integrating limits are anything other than $0$ and $2\pi$, i.e., if it is not a full ring.
Why is this happening? What does this physically mean?