1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Terminal velocity of a falling conducting ring

  1. Nov 29, 2009 #1
    1. The problem statement, all variables and given/known data

    A conducting ring with radius a and mass m is placed in a magnetic field at a height H above the origin of the reference frame. The plane of the ring is parallel to the ground, that is, the normal is directed along the z-axis. The electrical resistance per unit length of the ring is R/(2Pia), so that the resistance of the loop is R. The magnetic field has spatial dependence:

    [tex]\vec{B} = \frac{B_0}{L}\left(-\frac{\rho}{2} \hat{\rho} + z \hat{z} \right)[/tex]

    where [tex] \hat{\rho}, \hat{z} [/tex] are unit vectors in the usual cylindrical coordinate system.

    At a certain time, the ring is dropped and falls due to gravity. The plane of the ring remains horizontal as the ring falls. Find the terminal velocity of the ring.

    2. Relevant equations

    We've got Maxwell's equations, F=ma, Lorentz force F=qv X B, etc.

    3. The attempt at a solution

    Here's my attempt so far:

    [tex] \nabla \times \vec{E} = -\frac{\partial \vec{B}}{\partial t} \Rightarrow \oint{\vec{E} \cdot d\vec{l}} = -\frac{d}{dt} \int{\vec{B} \cdot \hat{n} \hspace{1mm} da} \Rightarrow E(2 \pi a) = -\frac{d}{dt} \int{B_z \hspace{1mm}da} \Rightarrow E(2 \pi a) = -\frac{d}{dt} \int_0^{2\pi}{d\phi}\int_0^a{B_z \rho\hspace{1mm} d\rho} = - \frac{B_0 a^2 \pi}{L}\frac{dz}{dt} [/tex]

    But dz/dt is just the velocity we're looking for!!

    Here is where I am now stuck. I have a nice expression for the electric field, E, in terms of velocity, dz/dt =v:

    [tex] \vec{E} = - \frac{B_0 \pi a v_z}{2L} \hat{\phi} [/tex]

    But the electric field points in the circumferential direction, not the z direction. In terms of finding terminal velocity, I want something to the effect of

    [tex] \sum F_z = -mg + F_? = ma [/tex], where F? is the net force in the z direction due to my induced electric field. From this point it is easy to set ma =0 and solve for v-terminal. But I can't just say something like F=qE, because E points in the wrong direction.

    Furthermore, I haven't at all used the fact that the ring is at a height H or that it has resistance R - are those unnecessary bits of information, or am I approaching this entirely wrong? I almost could solve this using conservation of energy, but it's the terminal velocity I want and not clear to me how to find that in an energy formulation of the problem (might be possible, though). So - can anyone help? o:)
     
  2. jcsd
  3. Nov 29, 2009 #2
    Oh, one more note. I also thought of this:

    [tex] -\frac{d}{dt} \int B \cdot da = -\frac{d \mathcal{E}}{dt} [/tex]

    , i.e. the change in magnetic flux. And then we could use the fact that

    [tex] \vec{F}_{mag} = I \int d\vec{l} \times \vec{B} = \frac{\mathcal{E}}{R} \int d\vec{l} \times \vec{B} [/tex]

    but then when I do out the cross product I get for the z-component of magnetic force

    [tex] \vec{F}_{mag} = \frac{B_0 \rho \pi a}{L} \hat{z} [/tex]

    which has no expression for velocity in it. So, how would I go about finding v-terminal from this?
     
  4. Nov 30, 2009 #3

    gabbagabbahey

    User Avatar
    Homework Helper
    Gold Member

    You seem to be assuming (without any justification) that [itex]\oint\textbf{E}\cdot d\textbf{l}=2\pi aE[/itex]... is that really true for any old electric field?

    Isn't there also an electric force on the loop?
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Terminal velocity of a falling conducting ring
Loading...