# (i)Show that the magnetic field $\mathbf{B}$ on the axis

1. Feb 11, 2009

### latentcorpse (i)Show that the magnetic field $\mathbf{B}$ on the axis of a circular current loop of radius a is

$\mathbf{B}=\frac{\mu_0 I a^2}{2(a^2+z^2)^{\frac{3}{2}}} \mathbf{\hat{z}}$
where I is the current and z is the distance along the x axis from the centre of the loop.

I've done this part!!!

{ii} An insulating disc of radius a has uniform surface charge density $\sigma$. It rotates at angular velocity $\omega$ about a perpindicular axis through its centre. What is the surface current density $\mathbf{K(r)}$ at $\mathbf{r}$ from it's centre?
Find the contribution $d \mathbf{B}(z)$ to the magnetic field on the axis of the disc from a ring with radii between r and r+dr, and thus find the magnetic field on the axis of the spinning disc.
Show that as $z \rightarrow \infty$,

$\mathbf{B}(z) ~ \frac{1}{8} \mu_0 \sigma \omega \frac{a^4}{z^3} \mathbf{\hat{z}}$

(iii) What would the corresponding results be for a spinning ring of inner radius a and outer radius b. Recover the result in part (i) by taking the limit $b \rightarrow a$

As I said I've already done (i)

For (ii) I said the surface charge density was given by $\mathbf{K(r)}=\sigma \mathbf{\omega} \wedge \mathbf{r}$

As for the next part about the contribution, I took the perpendicular axis as the z direction and said that since
$\mathbf{I}=I \mathbf{dr}=\mathbf{K} dA$, we get
$\oint I \mathbf{dr}= \int_S \mathbf{K} dA$ and so
$I 2 \pi r = \sigma \mathbf{\omega} \wedge \mathbf{r} dr$

Now I can use the formula from (i) with I as above and a=r giving

$B_z = \frac{\mu_0}{2} \int_{0}^{a} \frac{\sigma \omega r^3 dr}{(r^2+z^2)^{\frac{3}{2}}}$ where essentially we integrate in order to add up all the rings and get a disc.

however I can't integrate this though!!!

2. 3. Feb 11, 2009

### Hootenanny Staff Emeritus
Re: Electromagnetism

I haven't looked at the rest of you work, but I can tell you that this integral may be evaluated with the use of a substitution.

4. Feb 13, 2009

### latentcorpse Re: Electromagnetism

can anybody advise me on this integral? it's very similar to the one i posted about in my other thread "magnetic field integral" except for the extra copmlication that this time we have a r^3 on the numerator....

5. Feb 13, 2009

### Hootenanny Staff Emeritus
Re: Electromagnetism

Have you tried my suggestion?

6. Feb 14, 2009

### xboy Re: Electromagnetism

I am thinking by parts, and the same substitution as in the other problem. But perhaps Hootenanny has something simpler in mind ?

7. Feb 14, 2009

### latentcorpse Re: Electromagnetism

i can't find a suitable substitution that the r^3 term doesnt mess up...

8. Feb 14, 2009

### Hootenanny Staff Emeritus
Re: Electromagnetism

I was thinking exactly the same thing.

latentcorspe: if you use the same substitution as before and then apply integration by parts you should be able to find the anti-derivative, which turns out to be quite nice.

9. Feb 14, 2009

### gabbagabbahey Re: Electromagnetism

Alternatively, you can avoid integration by parts by noticing that $r^3=r(r^2+z^2)-z^2r$ and separating it into two easy integrals.

10. Feb 14, 2009

### xboy Re: Electromagnetism

hey, cool !

11. Feb 16, 2009

### latentcorpse Re: Electromagnetism

ok cheers so i get:

$B_z=\frac{\mu_0 \sigma \omega}{2} $\int_0^a \frac{r dr}{(r^2+z^2)^{\frac{1}{2}}} - z^2 \int_0^a \frac{r dr}{(r^2+z^2)^{\frac{3}{2}}}$$
then if$u=r^+z^2 \Rightarrow r dr=\frac{1}{2} du$ we get,

$B_z=\frac{\mu_0 \sigma \omega}{2} (\frac{1}{2} \int_{z^2}^{a^2+z^2} u^{-\frac{1}{2}}} du - \frac{z^2}{2} \int_{z^2}^{a^2+z^2} u^{-\frac{3}{2}} du) = \frac{\mu_0 \sigma \omega}{2} [\sqrt{a^2+z^2} -z +\frac{z^2}{\sqrt{a^2+z^2}} - z]$

however this doesn't give me what i want when i take the limit as $z \rightarrow \infty$

any ideas?

Last edited: Feb 16, 2009
12. Feb 16, 2009

### xboy Re: Electromagnetism

You'll have to expand the terms under square roots using binomial expansion up to the third term. That's where z^3 will come from.

Note that if you put z tends to infinity first and done the integral then, you'd get the same result more easily.

Last edited: Feb 16, 2009
13. Feb 16, 2009

### latentcorpse Re: Electromagnetism

ok. so using $(1+x)^n=1+x+\frac{n(n-1)}{2}x^2 +...$, we get

$B_z=\frac{\mu_0 \sigma \omega}{2} [(1+\frac{a^2}{2z^2} +\frac{\frac{1}{2}(-\frac{1}{2}}{2} \frac{a^4}{z^4}+...) + z^2(1-\frac{a^2}{2z^2}-\frac{\frac{1}{2}(-\frac{3}{2}}{2} \frac{a^4}{z^4}+...) -2z]$

simplifying to

$B_z=\frac{\mu_0 \sigma \omega}{2} [1+\frac{a^2}{2z^2} +\frac{1}{8} \frac{a^4}{z^4}+z^2-\frac{a^2}{2}+\frac{3}{8} \frac{a^4}{z^2} -2z]$.

that looks like $B_z \rightarrow \infty$ as $z \rightarrow \infty$

???

14. Feb 16, 2009

### xboy Re: Electromagnetism

you seem to have made some error. didn't you take z s outside from the square root terms?

15. Feb 16, 2009

### latentcorpse Re: Electromagnetism

ahhh cheers mate.

I get the right answer but i haven't had to use any limiting property of $z \rightarrow \infty$ have I? or is that why I've been able to ignore higher order terms in my binomial expansion?

Also do you have any ideas for how to start part (iii) as given in my original post?

16. Feb 16, 2009

### victorbomba Re: Electromagnetism

Might try

$$\tan{\theta} = \frac{r}{z}$$

17. Feb 16, 2009

### latentcorpse Re: Electromagnetism

where did you pull that from????

18. Feb 16, 2009

### victorbomba Re: Electromagnetism

Sorry, I wasn't paying attention. That was a suggestion for a substitution to solve the integral
which you had already correctly solved, but I think it was the wrong integral as the answer certainly diverges as z goes to infinity.

Isn't the numerator in the expression $\mathbf{B}=\frac{\mu_0 I a^2}{2(a^2+z^2)^{\frac{3}{2}}} \mathbf{\hat{z}}$ just $$J\mu_0$$ - In which case the numerator in your integral should be r not r cubed.

19. Feb 16, 2009

### latentcorpse Re: Electromagnetism

i dunno. r u referring to part (ii) cos im content with that solution.

any advice for part (iii) with the a and the b stuff?

20. Feb 16, 2009

### xboy Re: Electromagnetism

Yes, it is. Only because a/z is such a small term you could ignore the higher order terms. The idea is to get the first term with 1/z and leave the rest, because for large z they would be too small compared to this term.

If you have done (ii) you can also work out (iii). The field contribution dB from each small ring is still the same. So what's different here?

21. Feb 17, 2009

### latentcorpse Re: Electromagnetism

but to get the right answer I had to take the second order term as well??

i have no idea what (iii) is asking me?