Magnetic field at a point and differential i

[fluidistic]

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 : $$Q=\pi (b^2-a^2)\sigma$$, where $$\sigma$$ is the charge density of the annulus.
Charge of differential annulus : $$q=2\pi r dr \sigma$$. (Though I don't know why it wouldn't be $$2\pi ((r+dr)^2-r^2)$$.)

$$v=\omega r$$,
The current of a differential annulus is $$i=\frac{dq}{dt}=\frac{2\pi r dr \sigma}{dt}$$ but $$\frac{dr}{dt}=v$$. Hence $$i=2\pi r v \sigma$$.
Now I wonder, the i of a differential annulus is the "di" they're talking about?
Oh yes it is!
Well, I replace $$\sigma$$ by $$\frac{Q}{\pi (b^2-a^2)}$$ and I get $$di=\frac{2rvQ}{(b^2-a^2)}=\frac{2r^2 \omega Q}{(b^2-a^2)}$$.
Did I got it right?

2)I enclosed $$= \frac{(b^3-a^3)}{3}\cdot \frac{2 \omega Q}{b^2-a^2}$$ of the entire annulus.
$$B\cdot 2\pi b=\mu _0 I_{\text{enclosed}}$$.
Have I done right this?

 

[cepheid]

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

Well, if, in the annulus, the radius were changing (from inner to outer) by some finite amount $\Delta r$, then it would be true that the area of the annulus was the difference between the areas of the outer and inner circles:

$$A = \pi[ (r + \Delta r)^2 - r^2 ]$$​

However, in this case we're talking about an infinitesimal change in the radius, i.e. it is really really really small. So small, in fact, that it is basically not a change at all. What I mean by this is that when we do the math, we try to have our cake and eat it too. We regard the fact that there are two circles as being evidence that there is some (really really) small area between them, yet we assume that the inner and outer radii are essentially the same. (Yes, this is a contradiction). At this point, the mathematicians cringe, because we have not made this idea of an "infinitesimal" (a.k.a. "differential") quantity very precise (i.e. we have not shown that its properties are mathematically well-defined). My understanding is that a lot more advanced work has to be done in order to come up with a precise definition. That's why calculus is not taught using this concept, instead relying upon the much more precise idea of a limit. The object "dr", we are told, is not meant to be regarded as a number, but rather just a very suggestive symbol/notation. Physicists, however, know that they can get away with treating differentials as though they were actual quantities and still get the right answer. As a result, they have a tendency to abuse the notation. Given that we are doing so, how do we calculate the "infinitesimal" area of the annulus?

$$dA = \pi[ (r + dr)^2 - r^2 ] = \pi [ r^2 + 2rdr + (dr)^2 - r^2]$$

$$dA = \pi[2rdr + (dr)^2]$$​

Now, the reasoning is that if dr is really really small, then (dr)² will be even smaller still, so much so that we can regard it as neglible. To use fancy terminology: we ignore any terms that are of second order in dr. Doing so, the result becomes:

$$dA = 2\pi rdr$$​

Conceptually, the way to think of this approximation is that the inner and outer radii are so similar, that the area of the annulus is the same as it would be if we just had a straight rectangular strip whose length was the same as the circumference and whose width was dr.

Another important thing that you should learn from this is that any time you have an equation involving these infinitesimal quantities, both sides of the equation should be infinitesimal. It doesn't make sense for an infinitesimal to be equated to some ordinary (finite) value. That's how you know that your equation for di is probably wrong. I'll get back to you in a sec explaining where I think you might have gone wrong in calculating it.

 

[cepheid]

The current of a differential annulus is $$i=\frac{dq}{dt}=\frac{2\pi r dr \sigma}{dt}$$

Not quite. Current, is a flow of charge. More specifically, the current at a point in space is defined as follows: if I sit at that point in space and use a "charge-o-meter" (i.e. some sort of fictitious charge measuring device) to measure the total amount of charge that passes by me in one second, that is the current. In the case of the annulus, this is NOT dq, which is the total amount of charge present throughout the entire annulus.

To measure the current in this way, I'd have to sit at a point that is initially located within the annulus, and measure how much charge passes by me. Because the annulus is simply rotating about its centre, a point that is initially within the annulus will always lie within the annulus. Furthermore, if I wait long enough, eventually all of the charge within the annulus will pass by me. The time I have to wait is just the time it takes for the annulus to sweep past me once. The time it takes to sweep past me once is the time it takes to go through one full rotation, which is just the period of the rotation, T. So the current at any point within the annulus due to the rotational motion is just di = (total charge of annulus / rotation period of annulus).

$$di = \frac{dq}{T}$$​

Of course, the rotational period is just the reciprocal of the frequency:

$$T = \frac{1}{\nu} = \frac{2\pi}{2\pi \nu} = \frac{2\pi}{\omega}$$.​

That's a pretty big hint I just gave you.

 

[fluidistic]

Thank you very much cepheid. I completely missed the part of the period...

And to your first reply, it has been extremely helpful to me. I was wondering whether or not saying the parenthesis part "(Though I don't know why it wouldn't be $$2\pi ((r+dr)^2-r^2)$$ .)". I'm glad I did it, you've erased my doubt.
I'm thinking to print out this page :D.
Thanks a million!