# Magnetic field at a point and differential i

• fluidistic
In summary: However, if you have two point charges (like electrons) that are orbiting around one another, their current is not just the total amount of charge that flows by them in one second, but is specifically the amount of charge that flows through the point where they are orbiting. The reason that this matters is that when we are talking about the magnetic field around a differential annulus, we are not just talking about the total amount of charge that is within the annulus, we are also talking about the amount of charge that is flowing through the point where the annulus meets the surrounding medium.
fluidistic
Gold Member

## 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.

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?

fluidistic said:
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.

Last edited:
fluidistic said:
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.

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.
Thanks a million!

## 1. What is a magnetic field?

A magnetic field is an invisible force that is created by moving electric charges. It is represented by lines of force that show the direction and strength of the force at any given point.

## 2. How is the magnetic field at a point calculated?

The magnetic field at a point is calculated using the formula B = μ₀I/2πr, where B is the magnetic field, μ₀ is the permeability of free space, I is the current, and r is the distance from the point to the current.

## 3. What is the direction of the magnetic field at a point?

The direction of the magnetic field at a point is perpendicular to the direction of the current and follows the right-hand rule. This means that if you point your thumb in the direction of the current, your fingers will curl in the direction of the magnetic field.

## 4. What is differential i in regards to the magnetic field?

Differential i refers to a small change in current. It is used in the calculation of the magnetic field at a point when the current is not constant, but varies over a small distance.

## 5. How is the magnetic field affected by the distance from the current?

The magnetic field at a point is inversely proportional to the distance from the current. This means that as the distance increases, the magnetic field decreases. This relationship is represented by the formula B ∝ 1/r, where B is the magnetic field and r is the distance from the current.

Replies
6
Views
357
Replies
6
Views
570
Replies
17
Views
893
Replies
5
Views
702
Replies
12
Views
617
Replies
3
Views
563
Replies
14
Views
1K
Replies
10
Views
526
Replies
11
Views
434
Replies
7
Views
1K