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I Why does a metal ring, slid onto a solenoid, fly off?

  1. Dec 1, 2016 #1
    Our professor did a demonstration today for our physics II class. She was demonstrating electromagnetic induction with a solenoid and a metal ring. When she slid the ring down the solenoid and passed alternating current through the solenoid, the metal ring was flung upward off the solenoid.
    My professor was attempting to demonstrate magnetic induction but she never explained the net upward force. I get that currents are induced and the resulting field opposes the original field, but why in the upward direction? why not downward?
     
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  3. Dec 1, 2016 #2

    vanhees71

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    Let's see. You can take the solenoid as a magnet with a time-dependent magnetic field ##\vec{B}=B_0 \cos(\omega t) \vec{e}_z## (quasistationary approximation, valid for ##\omega/c \ll L##, where ##L## are the typical length scales of the magnet and the ring etc.

    Let ##a## be the radius of the ring, with the surface normal vector taken as ##\vec{e}_z##. Then the magnetic flux through the ring is
    $$\Phi=\pi a^2 B_0 \cos(\omega t),$$
    and the induced EMF
    $$\mathcal{E}=-\frac{1}{c}\dot{\Phi}=\frac{\pi a^2 B_0}{c} \sin(\omega t).$$
    Now this EMF causes a current, which is subject to the equation
    $$L \dot{I}+R I=\frac{\mathcal{E}}{R}=\frac{\pi a^2 B_0}{Rc} \sin(\omega t),$$
    where ##R## is the resistance of the ring and and ##L## its self-inductance, and this gives
    $$I=\frac{\pi a^2 B_0}{R^2+\omega^2 L^2} \left [L \omega \exp \left (-\frac{R t}{L} \right )+ R \cos(\omega t)-L \omega \sin(\omega t) \right].$$
    Now there is a magnetic-field component also in radial direction ##B_{\rho} =B_{0\rho} \cos(\omega t)##, where ##K## is some constant. The ##z##-component of the force on a "current element" ##I \mathrm{d} \vec{\ell} = I a \mathrm{d} \varphi \vec{e}_{\varphi}## is ##\mathrm{d} F_z=-I B_{\rho} a/c \mathrm{d} \varphi##, leading to a total force ##F_z=2 \pi I a B_{\rho}/c##.
    Using the above result you get after some algebra that there is indeed a steady positive ##F_z## plus some contributions oscillating with the frequency of the magnetic field from the solenoid:
    $$F_z=\frac{\pi^2 a^3 B_0 B_{0\rho} \omega L}{(R^2+\omega^2 L^2) c^2}+\text{time-dependent contributions}.$$
    So there's indeed a resulting upward force (don't nail me on all the constants in front of the expression, I'd have to check carefully again ;-)).
     
  4. Dec 1, 2016 #3

    Delta²

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    where does the radial magnetic field comes from? The fact that the solenoid is not infinitely long ?
     
    Last edited: Dec 1, 2016
  5. Dec 1, 2016 #4

    vanhees71

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    Yes, there is both a radial, and azimuthal component, because of the finite length of the solenoid. For a complete treatment (with heavy use of elliptic functions, of course), see the following preprint:

    https://archive.org/details/nasa_techdoc_19980227402
     
  6. Dec 1, 2016 #5
     
  7. Dec 2, 2016 #6

    vanhees71

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    Argh, it must of course be ##c/\omega \gg L##. This comes from the typical length scale over which retardation effects are important, i.e., ##c/\omega=\lambda/2\pi \gg L## means that the typical extensions of your setup is much smaller than the wave length, over which the wave changes significantly, i.e., you can consider the changes over the extensions of the setup as instantaneous along this extension ("near field approximation"="quasitstatic approximation").
     
  8. Dec 2, 2016 #7

    Meir Achuz

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    "I get that currents are induced and the resulting field opposes the original field, but why in the upward direction? why not downward?"

    Think of two coils on the same axis. If the current in the lower coil increases, there will be an increasing flux through the upper coil. By lenz's law,
    There will be an induced current in the upper coil that is opposite to the current in the lower coil. The magnetic force between parallel wires with opposite currents is repulsive, pushing the upper coil up.
     
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