# Homework Help: Axial B field from disc of charge

1. Jan 10, 2014

### CAF123

1. The problem statement, all variables and given/known data
An insulating disc of radius a has uniform surface charge density σ and is rotating at constant angular velocity w about a perpendicular axis through it's centre.
a)What is the surface current density $\vec{j}(\underline{r})$ at a position $\underline{r}$ relative to the centre?
b)Consider a part of the disc with radii between $r$ and $r+\delta r$. Find it's contribution $\delta \underline{B}(z)$ to the field on the axis.

2. Relevant equations
Surface current density $\vec{j}(\underline{r}) = \sigma \underline{v} = \frac{d \underline{I}}{dr},$ where $dr$ is the magnitude of the vector perpendicular to the flow of current.

3. The attempt at a solution
a) $|\vec{j}(\underline{r})| = \sigma (\omega r )$
b)My method originally was to find the B field contribution from the imaginary loops at radius r and that at radius r + δr and subtract, but that would mean needing the currents contained in these individual loops. I believe this method is incorrect here because in this case I would be ignoring the width of such loops and since we have associated a current density I should be taking the infinitesimal length into account,yes?

So, then $j = \frac{dI}{δr}$, dI the current in the loop of infinitesimal length (r+δr - r = δr), so dI = σwrδr (This is not a differential equation despite the notation) is the current in that part of the disc between some arbritary r ≤ a - δr and r + δr ≤ a.

I think the definition of surface current density is that it is the current through a length perpendicular to the flow of current. How does this definition hold here? I presume I am not to assume the current is completely in the tangential direction to the loop since I am again given current densities,yes?

The analogous problem of summing up imaginary rings of no finite dimension that constitute the disc is fine. In this case, since there is a current density associated, I think I need to take the width of the rings into account.

Many thanks.

2. Jan 10, 2014

### voko

Why are they "imaginary" and why would you subtract?

What is the radial density of current? What is the current flowing through the loop of radius $r$ and width $\mathrm{d} r$? What field does it generate at the center?

3. Jan 11, 2014

### CAF123

Hi voko,
My choice of words is incorrect, I meant that we consider the disc to be assembled of rings and the summation of all these rings gives a disc. In the limit of a dimensionless ring (I mean of no finite width), then this summation is an integral. My main concern here is that I believe this approach to be wrong because since the question has specified current densities I should be using them and thus considering the infinitesimal width of the rings.

The B field from the ring at radius r axially is $$B = \frac{\mu_o I_1 r^2}{2(r^2 + z^2)^{3/2}},$$where $I_1$ is the current in this ring.

That from the ring at radius r+δr is the same form as above but with $r \rightarrow r + δr$ and $I_1 \rightarrow I_2$. So the contribution from the part of the disc between these two loops, (i.e a ring of width δr) is given by $B_z(r+\delta r) - B_z(r)$, the B field due to the ring at r+δr and that at r. However, this assumes I know the current in a dimensionless ring, which I do not.

Is the term 'radial density of current' analogous to 'surface current density'? I got that to be $|\vec{j(\underline{r})}| = \sigma \omega r$, so it increases linearly with r.

The current in a loop of radius r and width dr (taking δr infinitesimal) is then $\sigma \omega r dr$ using the equations in the OP.

So the contribution is then $$dB_z = \frac{\mu_o \sigma \omega r^3 dr}{2(r^2 + z^2)^{3/2}}$$

4. Jan 11, 2014

### voko

This is the magnetic field of a loop. A loop is something that has radius, but no width. A ring is something that has a width (the extent between the inner and the outer radii), which is probably why you get confused. In a ring of a very small width, the current density is approximately constant, so the field due to such a ring is simply the field of a loop of that radius times the width.

I dislike the term "surface current density". "Current density" is charge per time per area, and what you call "surface current density" is charge per time per length; adding "surface" to something that is "per length" is very confusing.

Other than that, I think you are on the right track now.

5. Jan 11, 2014

### CAF123

Is there anyway to find the total current in each of the loops of radius r and r + dr? (Just so I could use my original method involving subtraction of B fields). I thought at first given the current density σωr, (dimensions of I/m) the total current in the loop would be $σωr(2\pi r)$ for the loop of radius r. But when I checked how the 'surface' current density is defined (I agree that this term is a misnomer, but I am using the convention in my notes and Griffiths), this is now obviously incorrect.

Last edited: Jan 11, 2014
6. Jan 11, 2014

### voko

You have found that the radial current density is $\sigma \omega r$. Total current in the ring between $r$ and $r + dr$ is $\sigma \omega r dr$. Observe that the units are charge per time.

7. Jan 11, 2014

### CAF123

Yes, that made sense, but I was wondering if it is possible (I.e if we have enough information) to deduce the current flowing around a loop of radius r? Then I could use my previous method of subtracting the B field due to loop at r from that due to loop at r + δr to obtain the B field contribution axially from the part of the disc between the two loops.

This method is motivated by the next part of the question which is to consider a ring of inner radius a and outer radius b and then to show that as b → a, we recover the result of the axial magnetic field from a loop of current.

8. Jan 11, 2014

### voko

I do not understand what you are trying to achieve. A loop has no width, it is, figuratively speaking, an edge all over the place. If you have two concentric loops of different radii, subtracting the field due to one from the field due to another does not make any sense to me; even if that means something, I can't see how that could give the field due to the ring between the loops.

Subtraction would make sense if you knew the field of a disk, and the field of a bigger disk, then you could find the field of the ring. But you need to find the field of disk to begin with, so that leads you nowhere.

9. Jan 11, 2014

### CAF123

Thanks, it makes sense to me now.
Consider a ring with inner radius a and outer radius b. Then it has width b-a. The axial B field contribution from the ring is that of a disc of radius b minus that of a disc of radius a. The result is $$B = \frac{\mu_o \sigma \omega}{2} \int_a^b \frac{r^3 dr}{(r^2 + z^2)^{3/2}}$$ In taking the limit $b \rightarrow a$, the axial B field should resemble that of a loop (i.e the ring of previous width b-a is tending to one of no width). I have done the above integral, by parts, and obtained $$B = \frac{\mu_o \sigma \omega}{2}\left[\frac{-b^2}{\sqrt{a^2+z^2}} + \frac{a^2}{\sqrt{a^2+z^2}} +2(\sqrt{b^2 + z^2} + \sqrt{a^2+z^2})\right],$$ where $j/a = \sigma \omega$. As $b \rightarrow a$, the outer radius $b \rightarrow 0$ and the above result can be simplified. But I don't think I can get it into a form that resembles $$B_{\text{loop}} = \frac{\mu_o I a^2}{2(a^2+z^2)^{3/2}}.$$

Last edited: Jan 11, 2014
10. Jan 12, 2014

### voko

No. In taking the limit $b \to a$, everything else being the same, the limit is zero, because integrals are continuous functions of their limits. That, by the way, means that you did not compute the integral properly, because you don't get zero when $a = b$.

I do not understand how the latter condition comes about.

More generally, I do not understand what you are trying to achieve with this method. You have answered part (a). The answer to part (b) is answer to (a) times $\delta r$, if we can make the usual assumption $\delta r$ signifies an infinitesimal increment.

11. Jan 12, 2014

### CAF123

There is a typo in the sign before the last term. I realize my error, I was picturing the situation in my head but not realizing that my lower bound on the integral given previously was a and not 0, so indeed I see why my method would ultimately fail.

This should instead work; $$B = \frac{\mu_o \sigma \omega}{2}\left[\int_0^b \frac{r^3 dr}{(r^2+z^2)^{3/2}} - \int_0^a \frac{r^3 dr}{(r^2+z^2)^{3/2}}\right]$$

As $b \rightarrow a$, we are just left with $$B \rightarrow \frac{\mu_o \sigma \omega}{2}\int_0^b \frac{r^3 dr}{(r^2+z^2)^{3/2}}$$ I believe this expression here should resemble the magnetic field of a loop of radius b. I can reexpress $j/b = \sigma \omega$ and compute the integral and the result is $$\frac{\mu_o j}{2b}\left[\frac{-b^2}{\sqrt{b^2+z^2}} + 2 \sqrt{b^2+z^2} - 2z \right].$$ I can reexpress the bracketed term but I cannot see how to make it into a form that resembles the magnetic field of a loop of radius b.

My intention here is solely to explore the question further.

12. Jan 12, 2014

### voko

I really do not see how you are getting that. Let $b = a + \delta$. Then $$\int_0^b \frac{r^3 dr}{(r^2+z^2)^{3/2}} - \int_0^a \frac{r^3 dr}{(r^2+z^2)^{3/2}} = \int_0^{a + \delta} \frac{r^3 dr}{(r^2+z^2)^{3/2}} - \int_0^a \frac{r^3 dr}{(r^2+z^2)^{3/2}} = \int_a^{a + \delta} \frac{r^3 dr}{(r^2+z^2)^{3/2}}$$ That goes to zero as $\delta \to 0$.

That is done by applying the integral mean value theorem to $$\int_a^{a + \delta} \frac{r^3 dr}{(r^2+z^2)^{3/2}} .$$

I do not think you are exploring a physical problem here; you seem to be dealing with a purely calculus-related question, that of approximating the value of an integral taken over a small interval.

13. Jan 12, 2014

### CAF123

Perhaps I should write out my question again;
'Consider a spinning ring of inner radius a and outer radius b. Show that in the limit b→a, we recover the result of an axial B field due to loop of radius a.'

Could you elaborate on how the integral mean value theorem will help here?

14. Jan 12, 2014

### voko

Applying the theorem, $$B = {\mu_0 \sigma \omega \over 2} \int \limits_a^{a + \delta} { r^3 \over (r^2 + z^2) ^{3/2} } dr = {\mu_0 \sigma \omega \over 2} { \tilde a^3 \over (\tilde a^2 + z^2) ^{3/2} } \delta \approx {\mu_0 \sigma \omega \over 2} {a^3 \over (a^2 + z^2) ^{3/2} } \delta \\ = {\mu_0 \delta I a^2 \over 2 (a^2 + z^2) ^{3/2} } ,$$ where $\tilde a \in [a, a + \delta]$ and $\delta I = \sigma \omega a \delta$.

15. Jan 12, 2014

### CAF123

Very nice voko, what prompted you to use the integral mean value theorem here? After all, the theorem says that for a continuous function f in an interval [a,b], there exists an element c such that f(c) corresponds to the average value of f. Since we were not dealing with averages in the set up, I was just wondering how it came to mind.
Thanks.

16. Jan 12, 2014

### voko

The mean value theorems are the heart of calculus, even though they are rarely appreciated as such.

A derivative of function $f(x)$ is $$\lim_{\delta \to 0} {f(x + \delta) - f(x) \over \delta} .$$ which intuitively means $$f'(x) \approx {f(x + \delta) - f(x) \over \delta}$$ if $\delta$ is small enough. The differential MVT gives that intuitive feel an exact form: $$f'(\tilde x) = {f(x + \delta) - f(x) \over \delta}.$$ The intuitive feeling can be recast differently: if $\delta$ is small, then $$f(x + \delta) - f(x) \approx f'(x) \delta,$$ which in the integral form is $$\int \limits_x^{x + \delta} f'(x) dx \approx f'(x) \delta,$$ to which the integral MVT gives an exact form: $$\int \limits_x^{x + \delta} f'(x) dx = f'(\tilde x) \delta.$$

Perhaps the name "mean value" is a misnomer. To me they are more like "approximate value" theorems.

So in this problem, once we had to approximate an integral over a small interval, the first thing springing to my mind was the intuitive integral approximation, which I mentioned in #4 in physical terms as "In a ring of a very small width, the current density is approximately constant, so the field due to such a ring is simply the field of a loop of that radius times the width." As that had no apparent effect, I then cited what I hoped would be better known: the MVT.

Edit: fixed typos.

Last edited: Jan 12, 2014