- #1

fluidistic

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## Homework Statement

Consider a 2 dimensional annulus whose radii are a<b. Its total charge is Q and is uniformly distributed over all its surface. The annulus rotates around its center at an angular velocity w.

This situation is equivalent of having a set of annulus of width dr and radii r (a<r<b), each one with a current di.

1)Calculate di in function of r.

2)Calculate the magnetic field in the center of the annulus.

## Homework Equations

None given.

## The Attempt at a Solution

2) is easy once I have 1). I'd apply Ampere's law once I'd have calculated the total current generated by the rotating annulus.

So I'm stuck on part 1).

What I did : [tex]Q=\pi (b^2-a^2)\sigma[/tex], where [tex]\sigma[/tex] is the charge density of the annulus.

Charge of differential annulus : [tex]q=2\pi r dr \sigma[/tex]. (Though I don't know why it wouldn't be [tex]2\pi ((r+dr)^2-r^2)[/tex].)

[tex]v=\omega r[/tex],

The current of a differential annulus is [tex]i=\frac{dq}{dt}=\frac{2\pi r dr \sigma}{dt}[/tex] but [tex]\frac{dr}{dt}=v[/tex]. Hence [tex]i=2\pi r v \sigma[/tex].

Now I wonder, the i of a differential annulus is the "di" they're talking about?

Oh yes it is!

Well, I replace [tex]\sigma[/tex] by [tex]\frac{Q}{\pi (b^2-a^2)}[/tex] and I get [tex]di=\frac{2rvQ}{(b^2-a^2)}=\frac{2r^2 \omega Q}{(b^2-a^2)}[/tex].

Did I got it right?

2)I enclosed [tex]= \frac{(b^3-a^3)}{3}\cdot \frac{2 \omega Q}{b^2-a^2}[/tex] of the entire annulus.

[tex]B\cdot 2\pi b=\mu _0 I_{\text{enclosed}}[/tex].

Have I done right this?