1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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

  1. Feb 11, 2009 #1
    (i)Show that the magnetic field [itex]\mathbf{B}[/itex] on the axis of a circular current loop of radius a is

    [itex]\mathbf{B}=\frac{\mu_0 I a^2}{2(a^2+z^2)^{\frac{3}{2}}} \mathbf{\hat{z}}[/itex]
    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 [itex]\sigma[/itex]. It rotates at angular velocity [itex]\omega[/itex] about a perpindicular axis through its centre. What is the surface current density [itex]\mathbf{K(r)}[/itex] at [itex]\mathbf{r}[/itex] from it's centre?
    Find the contribution [itex]d \mathbf{B}(z)[/itex] 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 [itex]z \rightarrow \infty[/itex],

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

    (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 [itex] b \rightarrow a [/itex]


    As I said I've already done (i)

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

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

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

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

    however I can't integrate this though!!!
     
  2. jcsd
  3. Feb 11, 2009 #2

    Hootenanny

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    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 #3
    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 #4

    Hootenanny

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    Re: Electromagnetism

    Have you tried my suggestion?
     
  6. Feb 14, 2009 #5
    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 #6
    Re: Electromagnetism

    i can't find a suitable substitution that the r^3 term doesnt mess up...
     
  8. Feb 14, 2009 #7

    Hootenanny

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    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 #8

    gabbagabbahey

    User Avatar
    Homework Helper
    Gold Member

    Re: Electromagnetism

    Alternatively, you can avoid integration by parts by noticing that [itex]r^3=r(r^2+z^2)-z^2r[/itex] and separating it into two easy integrals.
     
  10. Feb 14, 2009 #9
    Re: Electromagnetism

    hey, cool !
     
  11. Feb 16, 2009 #10
    Re: Electromagnetism

    ok cheers so i get:

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

    [itex]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] [/itex]

    however this doesn't give me what i want when i take the limit as [itex]z \rightarrow \infty[/itex]

    any ideas?
     
    Last edited: Feb 16, 2009
  12. Feb 16, 2009 #11
    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 #12
    Re: Electromagnetism

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

    [itex]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] [/itex]

    simplifying to

    [itex]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] [/itex].

    that looks like [itex]B_z \rightarrow \infty[/itex] as [itex]z \rightarrow \infty[/itex]

    ???
     
  14. Feb 16, 2009 #13
    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 #14
    Re: Electromagnetism

    ahhh cheers mate.

    I get the right answer but i haven't had to use any limiting property of [itex]z \rightarrow \infty[/itex] 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 #15
    Re: Electromagnetism

    Might try

    [tex]\tan{\theta} = \frac{r}{z}[/tex]
     
  17. Feb 16, 2009 #16
    Re: Electromagnetism

    where did you pull that from????
     
  18. Feb 16, 2009 #17
    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 [itex]
    \mathbf{B}=\frac{\mu_0 I a^2}{2(a^2+z^2)^{\frac{3}{2}}} \mathbf{\hat{z}}
    [/itex] just [tex]J\mu_0[/tex] - In which case the numerator in your integral should be r not r cubed.
     
  19. Feb 16, 2009 #18
    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 #19
    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 #20
    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?
     
  22. Feb 17, 2009 #21
    Re: Electromagnetism

    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.
     
  23. Feb 17, 2009 #22
    Re: Electromagnetism

    so do i just integrate between a and b and then take the limit as b -> a and hope i recover the result?
     
  24. Feb 17, 2009 #23
    Re: Electromagnetism

    Precisely.
     
  25. Feb 17, 2009 #24
    Re: Electromagnetism

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

    [itex]B(z)=\frac{!}{8} \mu_0 \sigma \omega \frac{b^4-a^4}{z^3}[/itex] which doesn't fair too well when I take the limit [itex]b \rightarrow a[/itex]
     
  26. Feb 17, 2009 #25
    Re: Electromagnetism

    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.
     
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook




Loading...