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Faraday's law of induction

  1. Mar 14, 2005 #1
    Faraday's law of induction is

    emf = - (d/dt) ∫S B∙da .

    When the closed loop (serving as the boundary of the surface S) is independent of time, the above relation is equivalent to the Maxwell equation

    curl E = - ∂B/∂t .

    However, when the closed loop C (i.e. the boundary of S) is itself a function of time, the following two questions seem relevant to ask:

    (i) How is the first equation above to be applied?

    (ii) Is this method consistent with Galilean (or Lorentz) invariance?
     
    Last edited: Mar 14, 2005
  2. jcsd
  3. Mar 14, 2005 #2

    reilly

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    Great question. Requires subtle and sophisticated arguments to indicate that Faraday's Law works for moving loops/field sources. One of the best discussions of this issue, I think, is given by Purcell in his Electricity and Magnetism, Vol. 2 of the Berkeley Physics Course. (When I taught this subject, it took one or two lecture sessions to cover the topic, both at the intermediate undergraduate level(Purcell), and the graduate level (Jackson or Panofsky and Phillips). A good part of the issue is calculating the flux through a moving loop. And, again, Purcell's discussion is superb.

    So, I suggest some reading is in order. But, perhaps someone has a better intuitive approach, which can be explained more briefly than several pages of argument.

    Regards,
    Reilly Atkinson
     
  4. Mar 14, 2005 #3
    The answer is straightfoward calculus. If boundary S is a function of time, then the limits of the integrand are functions of time, and you just carry the derivative through according to what is called liebniz's rule.

    (look it up www.mathworld.com)

    If you are worrying about lorentz invariance, setup the integrand in the unprimed frame and differentiate wrt unprimed time, do it again in some other primed frame with respect to that prime time. Then it remains for you to calculate an invariant, transform it, and satisfy your curiosity.

    For the record, the invariants are E*B (dot product) and (E^2)-(B^2).

    It is funny, the title of your post is "Faraday" (the name of the em field tensor) but you phrased your question in terms of maxwells equations. If you are taking a lorentz transformation, or doing anything I suggested, use the field tensor.
     
    Last edited: Mar 14, 2005
  5. Mar 15, 2005 #4
    Yes. Working out the right-hand side of Faraday's law is just straightforward calculus.

    On the right-hand side, we have

    RHS ≡ - (d/dt) ∫S(t) B∙da

    = - ∫S(t)B/∂t da - lim∆t→0(1/∆t)[ ∫S(t+∆t) - ∫S(t) ] B∙da

    = ∫S(t) curl E da - ∫C(t) (B x vC) dl

    = ∫C(t) (E + vC x B) dl ,

    where vC denotes the "velocity" at time t of the line-element dl on C(t).

    Note that, in the above derivation, the following two relations have been employed:

    (i) div B = 0 ;

    (ii) curl E = - ∂B/∂t .

    This is all straightforward enough.
    ______________

    The difficulty, however, lies in the left-hand side of Faraday's law. How is the emf at time t associated with the closed loop C(t) to be defined? According to the usual definition, the emf is given by the work done on a unit charge moved once around the loop. This suggests using the following definition:

    emf(t) ≡ ∫C(t) (E + v x B) dl .

    But what is the v here? It obviously includes a component parallel to dl corresponding to the (hypothetical) motion of the unit charge along the curve C(t), and this component, of course, gives no contribution to the emf. It is only a component of v perpendicular to dl which can give some contribution ... yet, the (hypothetical) motion of the unit charge along C(t) would not include such a component.

    So, perhaps the answer is to simply use the above definition in a 'conceptual way' with v set equal to the component of the velocity of the line-element dl perpendicular to the dl-direction. In that case, we can then just as well set v = vC , because any component of vC parallel to dl makes no contribution anyways. Doing this yields

    emf(t) ≡ ∫C(t) (E + vC x B) dl ,

    which is precisely the expression for the RHS of Faraday's law as it was derived above.
    ______________
    I don't have Purcell (although it does seem worthwhile to attempt to procure a copy).

    You wrote: "A good part of the issue is calculating the flux through a moving loop." This hurdle, I think, I have been able to overcome; and as Crosson points out, it requires no more than straightforward calculus. The result is to be found in what I recorded above, namely,

    (d/dt) ∫S(t) B∙da = ∫S(t)B/∂t da + ∫C(t) (B x vC) dl ,

    in which div B = 0 has been utilized. (Also, remember that vC is the velocity at time t of the line-element dl on C(t); thus, in general, vC will vary from point to point along C(t) depending upon how the curve changes over time at the given point.)

    So, it appears to me that my only real difficulty lies in finding the 'proper' definition of: emf for a time-dependent closed loop.

    Upon achieving that, I would then like to probe the question of where/when/how (?) either of Galilean or Lorentz invariance comes into the picture.

    ... I do have Jackson's book. I'll take a look there and see what it has to offer.
     
  6. Mar 17, 2005 #5
    three questions

    J. D. Jackson, Classical Electrodynamics

    Section 6.1, Faraday's Law of Induction
    ----------------------------------------------
    ... Okay, but why write E' and not just plain E without a prime?

    1) Does the above prescription for defining the emf in terms of E' also apply to a Lorentz invariant scenario?

    2) If so, then just how, specifically?

    3) Then, in terms of the E and B fields in the laboratory frame, what does the emf look like?



    Next, the relevant principle of relativity happens to be:
    And finally, Faraday's law is said to be precisely valid in the context of special relativity:
    ... Are there any comments on and/or answers out there to my three questions above?
     
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