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

  1. Jul 30, 2008 #1
    1. The problem statement, all variables and given/known data
    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 [itex]B_0[/itex], 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?

    2. Relevant equations
    [tex]\phi = \int B \cdot da [/tex]
    [tex] emf= \delta_t \phi_B [/tex]

    3. 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.

  2. jcsd
  3. Jul 31, 2008 #2
    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.
  4. Aug 1, 2011 #3
    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 [itex]\vec{\nabla} \text{x} \vec{E} = -\frac{\partial{\vec{B}}}{\partial{t}}[/itex]), the induced electric field due to a change in magnetic flux [itex]\Phi = \vec{B} \cdot \vec{A}[/itex] (where [itex]\vec{A}[/itex] is the area vector normal to the surface through which the magnetic field is "fluxing") is given by

    (1) [itex]\oint{\vec{E} \cdot d\vec{l}}=-\frac{d\Phi}{dt}[/itex]

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

    (2) [itex]2\pi a E= \pi a^2 \frac{dB}{dt}[/itex]

    which simplifies to

    (2') [itex]E= \frac{a}{2}\frac{dB}{dt}[/itex]

    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) [itex]\vec{\tau}= \frac{d\vec{L}}{dt}[/itex]

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

    (4) [itex]d\tau=a d F=a \lambda E dl[/itex]

    where [itex]\lambda[/itex] is the linear charge density; i.e., [itex]\lambda=\frac{q}{2\pi a}[/itex]. 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 [itex]\hat{z}[/itex]-direction. Now the total torque is obtained by integrating these infinitesimal torques around the ring, giving

    (5) [itex]\tau=2\pi a^2 \lambda E[/itex]

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

    (6) [itex]\frac{dL}{dt}=\lambda \pi a^3 \frac{dB}{dt}[/itex]

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

    (7) [itex]L=\lambda \pi a^3 B[/itex]

    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.
    Last edited: Aug 1, 2011
  5. Aug 1, 2011 #4

    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" [Broken] 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.
    Last edited by a moderator: May 5, 2017
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook