Magnetic Field of Rotating Circular Ring

[zero1342]

Homework Statement

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.

Homework Equations

I=q/t
ω=2πf
f=1/period
Biot-Savart Law

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?

 

[ShayanJ]

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.

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.

 

[zero1342]

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?

 

[ShayanJ]

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.

 

[zero1342]

So I originally solved a different problem which was to find the magnetic field for a loop of current and got: $$B=\frac{{μ_0}Ir^2}{2(r^2+z^2)^\frac{3}{2}}$$

Then the problem changed and it was a charged loop rotating and using your help with dl I got: $$B=\frac{{μ_0}qr^3ω}{4πr(r^2+z^2)^\frac{3}{2}}$$ and using λ=q/(2πr) it simplifies to: $$B=\frac{{μ_0}Ir^2}{2(r^2+z^2)^\frac{3}{2}}$$ 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?

 

[ShayanJ]

Yeah, that's correct.