This article is translated from a Chinese article on my Zhihu account. The original article was posted at 2020-02-17 18:49 +0800.

In the vacuum, inside a fixed ring of radius RR with fixed charge QQ uniformly distributed, there is a point charge with charge qq and mass mm moving in the plane of the ring due tue the electrostatic force. It moves in the small region around the center of the ring, and the motion is periodic along a closed curve. The area of the region enclosed by the curve is SS. Denote the distance from the center to the point charge as rr, and r≪Rr\ll R. Find the magnetic induction BB at the center of the ring.

By using the cosine law, we can write the electrical potential in the plane of the ring inside the ring as the integral

U=∫02πdθ2πQ4πε0R2+r2−2Rrcos⁡θ.U=\int_0^{2\pi}\frac{\frac{\mathrm d\theta}{2\pi}Q}{4\pi\varepsilon_0\sqrt{R^2+r^2-2Rr\cos\theta}}.

Note that the elliptic integral of the first kind is

K ⁣(μ)≔∫0π2dφ1−μsin⁡2φ=∫0π2dφ1−μ1−cos⁡2φ2=∫0πdθ4−2μ+2μcos⁡θθ≔2φ=124−2μ∫02πdθ1+μ2−μcos⁡θ.\begin{align*} K\!\left(\mu\right)&\coloneqq\int_0^{\frac\pi2}\frac{\mathrm d\varphi}{\sqrt{1-\mu\sin^2\varphi}}\\ &=\int_0^{\frac\pi2}\frac{\mathrm d\varphi}{\sqrt{1-\mu\frac{1-\cos2\varphi}{2}}}\\ &=\int_0^\pi\frac{\mathrm d\theta}{\sqrt{4-2\mu+2\mu\cos\theta}}&\theta\coloneqq2\varphi\\ &=\frac1{2\sqrt{4-2\mu}}\int_0^{2\pi}\frac{\mathrm d\theta}{1+\frac\mu{2-\mu}{\cos\theta}}.\\ \end{align*}

On the other hand, the potential

U=Q8π2ε0R2+r2∫02πdθ1−2RrR2+r2cos⁡θ.U=\frac{Q}{8\pi^2\varepsilon_0\sqrt{R^2+r^2}}\int_0^{2\pi}\frac{\mathrm d\theta}{1-\frac{2Rr}{R^2+r^2}\cos\theta}.

By comparing the two equations, we are motivated to find μ\mu such that

μ2−μ=−2RrR2+r2,\frac\mu{2-\mu}=-\frac{2Rr}{R^2+r^2},

and we may solve to get

μ=−4Rr(R−r)2.\mu=-\frac{4Rr}{\left(R-r\right)^2}.

Therefore,

U=Q8π2ε0R2+r2⋅4R2+r2R−rK ⁣(−4Rr(R−r)2)=Q2π2ε0(R−r)K ⁣(−4Rr(R−r)2).\begin{align*} U&=\frac{Q}{8\pi^2\varepsilon_0\sqrt{R^2+r^2}}\cdot\frac{4\sqrt{R^2+r^2}}{R-r}K\!\left(-\frac{4Rr}{\left(R-r\right)^2}\right)\\ &=\frac Q{2\pi^2\varepsilon_0\left(R-r\right)}K\!\left(-\frac{4Rr}{\left(R-r\right)^2}\right). \end{align*}

We can expand UU in terms power series of rr (how?), and we get

U=Q4πε0R+Q8πε0R3r2+O ⁣(r4).U=\frac Q{4\pi\varepsilon_0R}+\frac{Q}{8\pi\varepsilon_0R^3}r^2+O\!\left(r^4\right).

Then, the potential energy Ep=qUE_\mathrm p=qU (and omit constant term and higher order terms) is

Ep=qQ8πε0R3r2.E_\mathrm p=\frac{qQ}{8\pi\varepsilon_0R^3}r^2.

To make the trajectory a closed curve, the second derivative of the potential at the equilibrium should be positive, so qQ>0qQ>0, i.e. the ring and point charge have the same sign of charge.

As we all know, for the potential energy Ep=12mω2r2E_\mathrm p=\frac12m\omega^2r^2, the motion is

{x=acos⁡ωt,y=asin⁡ωt,\begin{cases}x=a\cos\omega t,\\y=a\sin\omega t,\end{cases}

where aa and bb are determined by the initial conditions.

Then, we can solve the equation

qQ8πε0R3=12mω2\frac{qQ}{8\pi\varepsilon_0R^3}=\frac12m\omega^2

to get

ω=12RqQπε0mR.\omega=\frac1{2R}\sqrt{\frac{qQ}{\pi\varepsilon_0mR}}.

Because the trajectory is an ellipse, the area is

S=πab.S=\pi ab.

We can take the derivate of the coordinates w.r.t. tt to get the velocity

{vx=−aωsin⁡ωt,vy=aωcos⁡ωt,\begin{cases}v_x=-a\omega\sin\omega t,\\v_y=a\omega\cos\omega t,\end{cases}

By Biot–Savart law, the magnetic induction BB at the center of the ring is

B=μ0q4πr3∣xvxyvy∣=μ0q4πr3ωab(cos⁡2ωt+sin⁡2ωt)=μ0qS4π2r3ω=μ0qS8π2Rr3qQπε0mR.\begin{align*} B&=\frac{\mu_0q}{4\pi r^3}\left|\begin{matrix}x&v_x\\y&v_y\end{matrix}\right|\\ &=\frac{\mu_0q}{4\pi r^3}\omega ab\left(\cos^2\omega t+\sin^2\omega t\right)\\ &=\frac{\mu_0qS}{4\pi^2r^3}\omega\\ &=\frac{\mu_0qS}{8\pi^2Rr^3}\sqrt{\frac{qQ}{\pi\varepsilon_0mR}}. \end{align*}