# Induced voltage by a rotating ring

Two concentric rings dielectrics and uniformly charged are suspended on the same floor.

The outer ring has a radius R and mass M, while the other has radius r << R and mass M.

The outer ring that has overall charge Q, is made to rotate around the axis passing through the center and perpendicular to the plane of the ring itself, with angular acceleration α.

Calculate the angular acceleration α' of thr inner ring.

My solution is the following:

The outer ring produce a variation of magnetic flux on the inner ring equal to

(μ * Q * π^2 * r^2 ) / ( α * R )  (*)


The inner ring produce a self-induced variation of magnetic flux for the law of Faraday-Lenz equal to

(μ * Q * π^2 * r' ) / α'   (**)


Comparing the ( * ) and ( ** ), I obtain

α' = α * R / r


Is it my solution right?

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Hi user1328847. Welcome to Physics.SE. This site uses an unique TeX markup style called MathJax. This markup is very useful for understanding math equations and parameters. Please have a look here for an intro or our FAQ for more info. For example, $\theta$ results $\theta$, $\omega$ inserts $\omega$, etc. It's quite interesting. You can revise your post if you can ;-) – Waffle's Crazy Peanut Apr 27 '13 at 12:44
thanks @CrazyBuddy for your comment. – user1328847 Apr 27 '13 at 13:04

Assume that radius and charge density for outer ring and inner ring is $R$ , $\lambda$ and $r$ ,$\lambda'$ respectively. Because $r<<R$ you can assume the magnetic field on the whole surface of inner ring to be: $$B=\frac{\mu_0I}{2R}$$ now,$I$ is equal to $\lambda v$ and $v=R\alpha t$, so $$B=\frac{\mu_0\lambda R\alpha t}{2R}$$ and $(r<<R)$ $$\phi=\pi r^2B$$

now we find induced electric field around the first ring .according to faraday's law: $$E(2\pi r)=\frac{d\phi}{dt}$$

so $$E=\frac{\mu_0\lambda r\alpha}{4}$$

now we find total torque exerted on inner ring: $$\tau=r\lambda' E (2\pi r)=\frac{\mu_0\lambda \lambda'\pi r^3\alpha}{2}$$

so you can find the angular acceleration by having it's moment of inertia (for a ring $I_0=Mr^2$):

$$\alpha'=\frac{\tau}{I_0}=\frac{\mu_0\lambda \lambda'\pi r^3\alpha}{2Mr^2}=\frac{\mu_0\lambda \lambda'\pi r\alpha}{2M}$$

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