(i)Show that the magnetic field [latex]\mathbf{B}[/latex] on the axis

[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!

 

[Hootenanny]

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

however I can't integrate this though!

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.

 

[latentcorpse]

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...

 

[Hootenanny]

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...

Have you tried my suggestion?

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.

 

[xboy]

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

 

[latentcorpse]

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

 

[Hootenanny]

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

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.

 

[gabbagabbahey]

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.

 

[xboy]

hey, cool !

 

[latentcorpse]

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?

 

[xboy]

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.

 

[latentcorpse]

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 $z \rightarrow \infty$

?

 

[xboy]

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

 

[latentcorpse]

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?

 

[victorbomba]

Might try

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

 

[latentcorpse]

Might try

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

where did you pull that from?

 

[victorbomba]

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.

 

[latentcorpse]

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

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

 

[xboy]

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?

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?

 

[latentcorpse]

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?

 

[xboy]

I didn't mean the first term as in first term of the binomial expansion. Rather the first surviving term in the whole expression. If you had expanded further there would be other surviving terms, but we are saying that they for large z they are far too small compared to the z term we already have.

(iii) is asking you to calculate the field for a disc with a hole in its center. You can still break it into small rings as you did here. Then you can integrate the contributions of the small rings, as you did here.

 

[latentcorpse]

so do i just integrate between a and b and then take the limit as b -> a and hope i recover the result?

 

[xboy]

Precisely.

 

[latentcorpse]

this didn't work out too well for me - I ended up with

$B(z)=\frac{!}{8} \mu_0 \sigma \omega \frac{b^4-a^4}{z^3}$ which doesn't fair too well when I take the limit $b \rightarrow a$

 

[xboy]

Suppose that the ring has a very small radius dr . Then b-a = dr. Can you manipulate the above equation (factorising it, for example) and get something in terms of dr.

 

[latentcorpse]

$b^4-a^4=(b^2+a^2)(b+a)dr$

still not sure where this is going?

 

[xboy]

It's going alright. Use b-a = dr again and expand and ignore the drs compared to bigger terms, or just make b=a=r and see what it comes to.

 

[latentcorpse]

i don't have any b-a terms left to make into dr

 

[xboy]

Put b = a + dr then. Or rather save yourself the trouble and put b=a= r, you'll get the same results.

 

[latentcorpse]

ok so with b=a=r, i get 2r^2 . 2r . dr = 4r^3 . dr
but how does that help recover what I'm looking for?

 

[xboy]

Substitute that in your original expression and see. Incidentally you won't get back the original term ( the first expression you wrote) but the modified term for z tending to infinity. This is because you've calculated the field using that condition.

 

[latentcorpse]

$B(z)=\frac{1}{2} \mu_0 \sigma \omega \frac{r^3 dr}{z^3}$ as b-> a dr->0 though?

 

[xboy]

That's what you should get.

 

[latentcorpse]

hmmm...surely if dr->0, then B(z)->0 which doesn't make much sense though?

 

[xboy]

You should try to work out an expression for the current through this ring.