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Trouble understanding this derivation of Faraday's law

  1. Oct 2, 2013 #1
    I am having trouble understanding part of the derivation of Faraday's law given in this lecture at around 57:00


    So the first goal is to calculate the change in magnetic flux through a changing closed loop in a dynamic magnetic field, and it's given by the following

    [tex]\phi(t+\Delta t)-\phi(t)=\iint_{S+\Delta S} B(t+\Delta t).dA - \iint_S B(t).dA[/tex]

    where S is surface bounded by the loop at time t, S+ΔS is the surface bounded by the loop at t+Δt

    [tex]\iint_{S+\Delta S} B(t+\Delta t).dA[/tex] is then rewritten as [tex]\iint_S B(t+\Delta t).dA + \iint_{\Delta S} B(t).dA[/tex]

    The part I don't understand is why the surface integral over the surface ΔS now calculated for the field at time t - shouldn't it be B(t+Δt)? In the lecture (at 1:02:15), he says it doesn't matter because ΔS is first order in time, it's proportional to Δt. And the difference* is going to be proportional to Δt so we don't have to worry about the change in time over an infinitesimal surface because it's second order in time.

    What does first order/second order in time mean?

    *I think the difference he refers to is this
    [tex]\iint_{\Delta S} B(t+\Delta t).dA - \iint_{\Delta S} B(t).dA[/tex]

    And in the next part, the equation we arrive at is

    [tex]\Delta \phi=\Delta t \iint_S \frac{\partial B}{\partial t}.dA + \iint_{\Delta S} B.dA[/tex]

    I'm not sure why the Δs in the first integral become derivatives (shouldn't we then have dt rather than Δt in front of the integral?), but I think these are all meant to be infinitesimals.
    And another thing which relates to my first question is why B in the second integral is no longer time-dependent.
     
    Last edited by a moderator: Sep 25, 2014
  2. jcsd
  3. Oct 2, 2013 #2
    The second order term is proportional to the square of Delta t. It is an infinitesimal of higher order, negligible compared to the terms of first order, so it can be dropped.
     
  4. Oct 2, 2013 #3
    Which is the second order term? And why square of delta t?
     
  5. Oct 2, 2013 #4
    The difference between B(t) and B(t+Δt) is an infinitesimal proportional to Δt. The Surface ΔS is also an infinitesimal proportional to Δt. So the difference between the integrals

    ∬ΔS B(t+Δt).dA−∬ΔS B(t).dA

    is an infinitesimal of higher order and is negligible.
     
  6. Oct 3, 2013 #5
    Hmm...is there a 'formal' or more rigorous way of describing this?
     
  7. Oct 3, 2013 #6
    That's standard procedure. Compare for instance with the derivation of the derivative of a product of two functions

    [f*g]'(x)=[f(x+dx)*g(x+dx) - f(x)*g(x)]/dx = [(f(x)+dx f '(x))*(g(x)+dx g'(x)) - f(x)*g(x)]/dx =
    = [dx f '(x)*g(x) + dx f(x)*g'(x) + dx^2 f '(x)*g'(x)]/dx = f '(x)*g(x) + f(x)*g'(x).

    Notice the extra term dx^2 f '(x)*g'(x) was dropped because it was an infinitesimal of higher order.
     
  8. Oct 4, 2013 #7
    Still not sure I quite understand it yet...I'm not sure if I've seen the product rule derived that way before. I'm unfamiliar with the idea of 'infinitesimals of higher order'.

    I just noticed an error in my original post, the difference between the integrals should be ∬S B(t+Δt).dA−∬S B(t).dA, so over the original surface. So the total change in flux is given by

    ∬S B(t+Δt).dA−∬S B(t).dA + ∬ΔS B(t).dA

    and of course B(t) in the last integral is what I can't quite grasp. It gives the flux through the change in surface, so if we want to calculate the total change, why not calculate it at the later time?
    Are we saying that the magnetic field cannot 'that fast'? That is, in the time it takes for the surface to change by ΔS, the magnetic field cannot have changed a lot from what it was at B(t)?
    Seems a bit dodgy to me to be honest :p
     
  9. Oct 6, 2013 #8
    Does this have anything to do with Taylor expansions?
     
  10. Oct 8, 2013 #9
    Are there any textbooks that discuss this (at this level)? I've yet to come across one that does so without using full vector calculus (and on this note, which books presents Faraday's law in the form given in the lecture?)
     
  11. Oct 14, 2013 #10
    Or any webpages...
     
  12. Oct 15, 2013 #11
    Found an error in post #7...
     
  13. Oct 20, 2013 #12
    Anyone? :p
     
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