1. Limited time only! Sign up for a free 30min personal 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!

Faraday's Law, Motional Emfs

  1. Sep 8, 2011 #1
    Hello,

    I am currently trying to really come to grips with electromagnetic induction and Faraday's law.

    The text I'm using is Young and Freedman's University Physics, 12th edition.

    How can the emf for a conducting loop moving in a magnetic field be given by the closed loop integral of (v cross B) dot dl where dl is an infinitesimal loop element, when (the way I understand it) that essentially represents work per unit charge done by the MAGNETIC force? And of course we know that the magnetic force does zero work. This is where I'm confused. Also, when we do work to keep such a loop moving, we move it against a magnetic force, how come (as my book says) the magnetic force does no (negative) work as we move this wire?

    Thanks

    Aubin
     
  2. jcsd
  3. Sep 8, 2011 #2

    Philip Wood

    User Avatar
    Gold Member

    What an interesting question!

    As you say, the magnetic force never does any work on the particle, because:

    dW = q(vxB).ds = q(vxB).vdt = 0

    Here, ds is an element of the particle’s path.

    Now let's consider the particles constrained to move along a conductor. Let the velocity through space in our frame of reference of an element dl of the conductor be vcond, and let the velocity, dl/dt of the particles relative to the conductor be vrel.

    Then our formula becomes

    dW = q{(vcond+vrel)xB}.(vcond+vrel)dt = 0

    The terms involving just vcond in the brackets either side of the dot clearly amount to zero, as do the terms with just vrel. So we are left with the heterogeneous terms

    q{vcondxB}.vreldt + q{vrelxB}.vconddt = 0

    that is

    q{vcondxB}.dl + q{vrelxB}.vconddt = 0

    Note that here we're using dl as originally defined - as a directed element of the conductor, rather than as a displacement of the particle through space (for which we're using ds).

    Now, in the second term, q{vrelxB} is the ‘BIL’ motor effect force contributed by q on the element of the conductor, so - {vrelxB} is the external force needed to keep the element moving at constant speed, so - q{vrelxB}.vconddt is the work an external agency has to do to keep the wire moving at constant speed. So:

    Work done by external agency = - q{vrelxB}.vconddt
    =q {vcondxB}.dl
    = work done pushing q along the conductor

    Hope you like this. It shows that the work done pushing q through the conductor does not come from the magnetic force, but from another, external force.
     
    Last edited: Sep 8, 2011
  4. Sep 8, 2011 #3
    Yes, that makes sense, thank you very much. (Took a while to digest because my vector algebra was a bit rusty from last term). This also shows very clearly why the work done must equal the energy dissipated in the circuit doesn't it?
     
  5. Sep 9, 2011 #4

    vanhees71

    User Avatar
    Science Advisor
    2016 Award

    A very nice pedagogical paper on the subject is

    P. J. Scanlon, R. N. Henriksen, J. R. Allen, Approaches to Electromagnetic Induction, Am. J. Phys. 37, 698 (1969)
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Faraday's Law, Motional Emfs
  1. Faraday's law (Replies: 6)

  2. Faraday's Law (Replies: 8)

  3. Faraday's law (Replies: 12)

Loading...