# Magnetic Field of Rotating Circular Ring

• zero1342
In summary, the problem involves finding the magnetic field at a specific position along the rotation axis of a circular ring carrying a uniform linear charge density and rotating with an angular velocity. To solve this, we use Biot-Savart's law and consider the wire element as a rotating linear charge distribution. After calculating the current and displacement vectors, the resulting magnetic field equation is the same as that of a loop of current, indicating the similarities between the two situations.
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.

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?

zero1342 said:
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.

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?

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.

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?

Yeah, that's correct.

## 1. What is a magnetic field?

A magnetic field is a region in space where a magnetic force can be detected. It is created by the movement of electrically charged particles or by the presence of a magnetic material.

## 2. How is the magnetic field of a rotating circular ring generated?

The magnetic field of a rotating circular ring is generated by the movement of electrically charged particles, specifically the flow of electric current through the ring. As the ring rotates, the electric current creates a circular magnetic field around the ring.

## 3. How does the magnetic field of a rotating circular ring affect nearby objects?

The magnetic field of a rotating circular ring can affect nearby objects by exerting a force on them. This force is known as the magnetic force and can cause objects with magnetic properties to move or align in a certain direction.

## 4. What factors affect the strength of the magnetic field of a rotating circular ring?

The strength of the magnetic field of a rotating circular ring is affected by several factors, including the amount of electric current flowing through the ring, the distance from the ring, and the size and shape of the ring itself.

## 5. How can the magnetic field of a rotating circular ring be used in practical applications?

The magnetic field of a rotating circular ring has various practical applications, such as in electric motors and generators. It can also be used in scientific research to study the effects of magnetic fields on different materials and objects.

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