1
$\begingroup$

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.

Ring

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?

$\endgroup$
2
$\begingroup$

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 $2\pi$

$\endgroup$
  • $\begingroup$ So, then the magnetic field inside the wire (at the center) cannot be found out? $\endgroup$ – pkwssis Aug 14 '14 at 15:32
  • $\begingroup$ The wire itself is not something without dimensions, you have to consider that has thickness, you are using a mathematical approximation. You can find the magnetic field inside a wire with thickness with Gauss Theorem I think, and some simmetry consideration. It's a typical ploblem that I have seen sometimes. $\endgroup$ – Karozo Aug 14 '14 at 15:35

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.