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Magnetic Field of Rotating Circular Ring

  1. Jul 7, 2015 #1
    1. The problem statement, all variables and given/known data
    Find the magnetic field at position z (z=0 in the plane of the ring) along the rotation axis for a circular ring of radius r, carrying a uniform linear charge density λ, and rotating about its axis with angular velocity ω.

    2. Relevant equations
    Biot-Savart Law

    3. The attempt at a solution
    I can determine the magnetic field when the ring is just a current loop that is not rotating. Once the rotation comes into play I get really confused about how to handle the linear charge density λ and the angular velocity.

    I see that I can solve for time in the equation for current (I=q/t) and end up with: I=(qω)/(2π)

    I think λ=charge/length but should it instead be: λ=dq/dl?
  2. jcsd
  3. Jul 8, 2015 #2


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    You're misunderstanding the problem. Its not a current loop rotating, its a charged loop rotating. So if it wasn't rotating, it was simply a charged loop. But now that its rotating, its setting charged particles in motion in a circular path. So its actually the same as a current loop.
  4. Jul 8, 2015 #3
    I understand that there are many similarities between the magnetic field above a current loop and the magnetic field above a spinning charged loop. The issue is, how do I incorporate the charge density into the problem?

    The differential element of current di = (2πdq)/ω and dq = λdl so di=(λ2πdl)/ω

    Does dl = 2πrdr?
  5. Jul 8, 2015 #4


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    To find the magnetic field, you need to use Biot-Savart's law ## \vec B=\frac{\mu_0}{4\pi} \int \frac{I d \vec l \times \vec r}{r^3} ##.
    Where ## d\vec l ## is the differential length(so its dimension is length not length2 so rdr can't be right) along the wire in the direction of current and ##\vec r ## is the displacement vector from the wire element to the point of observation.
    Here we're talking about a circle. If we use cylindrical coordinates and assume the loop is at z=0, its obvious that the loop is defined by ## \rho=const=R ##. So the differential length should be in the direction of the azimuthal angle ## \phi ##, which means ## d\vec l =Rd\phi \hat \phi##.
    The problem wants the magnetic field along the loops axis, so the point of observation is located on the z axis, ##\vec r_o=z_o \hat z ##. The wire element is located at ## \vec r_e=R\hat \rho ##. So we have ##\vec r=\vec r_o-\vec r_e=z_o\hat z-R\hat \rho ## and ##r=\sqrt{z_o^2+R^2}##.
    The current, as you mentioned before, is ## I=\frac{dq}{dt} ##. Here we have a rotating linear uniform charge distribution. So we have ## I=\frac{\lambda Rd\phi}{dt}=\lambda R \omega ##.
    Now you should put all of the above in the integral and calculate it.
  6. Jul 8, 2015 #5
    So I originally solved a different problem which was to find the magnetic field for a loop of current and got: [tex]B=\frac{{μ_0}Ir^2}{2(r^2+z^2)^\frac{3}{2}}[/tex]

    Then the problem changed and it was a charged loop rotating and using your help with dl I got: [tex]B=\frac{{μ_0}qr^3ω}{4πr(r^2+z^2)^\frac{3}{2}}[/tex] and using λ=q/(2πr) it simplifies to: [tex]B=\frac{{μ_0}Ir^2}{2(r^2+z^2)^\frac{3}{2}}[/tex] which is the same exact result for a loop of current!

    So they're both the same and now your first response makes more sense. Is this all correct?
  7. Jul 8, 2015 #6


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    Yeah, that's correct.
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