# Calculating the magnetic field due to a rotating ring using Biot–Savart law

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?

I think that the problem is that you are considering an electric current at distance $r=0$ with the Biot-Savart formula. It's like when you have a wire with a current and you want to find magnetic field on the wire, or electric field on a point-like charge.
In your problem current $J$ is a linear function of distance, but in Biot-Savart you have something like $\frac{1}{\vec{r}^2}$, so in the end you have a $\frac{1}{\vec{r}}$ that give you a divergent integral whenever you integrate it from $0$ to something $\neq0$.
Electric current far from P doesn't matter, this why in the integral the divergent point are $0$ and $2\pi$