Angular momentum of a charged insulator ring in a decreasing magnetic field

[SonOfOle]

Homework Statement

Consider a thin ring of mass m that has a radius a and negligible width. The ring lies in a horizontal plan. The ring is an insulator and carries a fixed charge q that is uniformly distributed around its circumference. The ring is located in a magnetic field of strength $B_0$, the field is parallel to the vertical axis through the center of the ring. The ring is also supported so that it can rotate freely about this central vertical axis.

If the magnetic field is switched off,

a) how much angular momentum will the ring acquire?

Homework Equations

$$\phi = \int B \cdot da$$
$$emf= \delta_t \phi_B$$

The Attempt at a Solution

I'm stuck on what the force on each electron is in this process. The ring will begin to spin to try and create a field to compensate for the external, decreasing B field, but some of the force goes into making the ring spin too, as the charged are not free.

I want to think there is some conservation of momentum going on here, but unsure on that.

Ideas?

 

[badphysicist]

You're on the right track with your equations, but you might want to look for what the emf can be expressed as. Maybe try looking at Maxwell's equations in their integral forms.

 

[tstin]

This was an old qual problem I was just looking at. Let me describe how to solve it, and then ask another question I had that my solution doesn't answer.

According to Lenz's Law, (or really to Maxwell's equation $\vec{\nabla} \text{x} \vec{E} = -\frac{\partial{\vec{B}}}{\partial{t}}$), the induced electric field due to a change in magnetic flux $\Phi = \vec{B} \cdot \vec{A}$ (where $\vec{A}$ is the area vector normal to the surface through which the magnetic field is "fluxing") is given by

(1) $\oint{\vec{E} \cdot d\vec{l}}=-\frac{d\Phi}{dt}$

In this case (a ring of radius a), $\Phi = \pi a^2 \vec{B}$, and due to symmetry $\vec{E}$ is the same magnitude along the ring, and due to Lenz's Law $\vec{E}$ will point in the direction such that it opposes the change in flux through the ring, or in the $\hat{\phi}$-direction (parallel to $d\vec{l}$). Using this information, (1) becomes

(2) $2\pi a E= \pi a^2 \frac{dB}{dt}$

which simplifies to

(2') $E= \frac{a}{2}\frac{dB}{dt}$

So it might seem that we can't calculate this time derivative, but we don't need to! Recall Newton's second law in rotational form:

(3) $\vec{\tau}= \frac{d\vec{L}}{dt}$

where $\vec{\tau}$ is the total torque on the object, and $\vec{L}$ is its angular momentum. Now, the torque on the ring is due to the induced electric field, and the infinitesimal torque $d\tau$ on an infinitesimal length of the ring $dl$ is

(4) $d\tau=a d F=a \lambda E dl$

where $\lambda$ is the linear charge density; i.e., $\lambda=\frac{q}{2\pi a}$. I'm dropping vector signs here because I'm lazy, but we know that the force is in the direction of the electric field, and the torque is in the $\hat{z}$-direction. Now the total torque is obtained by integrating these infinitesimal torques around the ring, giving

(5) $\tau=2\pi a^2 \lambda E$

Finally, we substitute in (3) and (2'), leaving us with

(6) $\frac{dL}{dt}=\lambda \pi a^3 \frac{dB}{dt}$

Now we integrate over time, and we get the final angular momentum of the ring,

(7) $L=\lambda \pi a^3 B$


So that's fine, but the question I have is, is there a way to get this result using only the initial momentum and final momentum of the electromagnetic fields alone? The problem I have is, this angular momentum acquired by the ring comes from the magnetic field. But I can't figure out what the initial angular momentum of the field is using Poynting's vector, because there's no electric field inside the ring (right? I tried using a Gaussian surface to see this), and therefore the cross-product of the E and B-fields inside the ring is 0. But that can't be right... the static magnetic field must have some angular momentum, or else there wouldn't be any way for the ring to acquire angular momentum. What am I missing here? Any ideas are appreciated.

 

[tstin]

there's no electric field inside the ring (right? I tried using a Gaussian surface to see this), and therefore the cross-product of the E and B-fields inside the ring is 0.

This is wrong, of course. There's no nice Gaussian surface one can draw to calculate the electric field due to a ring of charge. We can only analytically calculate the field along the ring's axis. Here's an http://www.physics.buffalo.edu/~sen/documents/field_by_charged_ring.pdf" trying to develop an intuitive understanding of the field due to a charged ring.

Anyways, I guess it's not trivial to find the electric field and thus the initial angular momentum. of the static field configuration.

 

[paranormal]

The angular momentum of the ring can be calculated using the equation L = Iω, where I is the moment of inertia and ω is the angular velocity. In this case, the moment of inertia of the ring can be approximated as I = mr^2, where r is the radius of the ring.

When the magnetic field is switched off, there will be a change in magnetic flux through the ring, which will induce an emf (electromotive force) according to Faraday's law. This emf will cause a current to flow in the ring, and this current will produce a magnetic field that opposes the external field. This will result in a torque on the ring, causing it to rotate.

The force on each electron in the ring can be calculated using the Lorentz force law, which states that F = qv x B, where q is the charge, v is the velocity, and B is the magnetic field. As the ring rotates, the velocity of the electrons will change, resulting in a change in force. However, as the ring is an insulator, the electrons cannot move freely and will experience a resistance to this force, leading to a decrease in the torque and a decrease in the angular momentum of the ring.

In terms of conservation of momentum, we can consider the system of the ring and the external magnetic field. When the field is switched off, the ring will experience a torque and rotate, while the external field will decrease in strength. This decrease in the field's momentum will be equal to the increase in the ring's angular momentum, resulting in conservation of momentum in the system.

In summary, the ring will acquire angular momentum due to the torque produced by the induced current, and this can be calculated using the equation L = Iω. The force on each electron can be calculated using the Lorentz force law, and the decrease in the external magnetic field's momentum will be equal to the increase in the ring's angular momentum, resulting in conservation of momentum in the system.