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Faraday's Law and induced EMF

  1. Apr 15, 2009 #1
    Faraday's law says

    induced emf = - d(flux)/dt

    If this is applied to a loop where induced emf causes currents, and thus flux itself, do we have to consider that flux (of course we don't if it's constant)?

    If the external flux has a nonzero second derivative, then the induced emf is changing with time, thus the induced flux has a nonzero first derivative. Will this varying induced flux need to be considered when applying Faraday's law?
  2. jcsd
  3. Apr 15, 2009 #2
    Yes. The internal flux must be considered except where it is much smaller than the external flux. Lenz' law describes the internal flux as opposite in direction to the external. Hence the net flux decreases. Faraday's law relates the *net* flux to the emf. Thus the emf is determined by the external flux plus the geometry and resistance of the loop itself.

  4. Apr 15, 2009 #3


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    Staff: Mentor

    This leads to the concept of "self inductance" of a coil or loop of wire.
  5. Apr 15, 2009 #4
    Say we have a circular loop of wire with some area and a uniform magnetic field pointing directly into it (no angle).

    What if the magnitude of B is something like

    B(t) = 100T^5 + 100t^4 + 100T^3 + 100T^2 + 100T + 100

    Then finding an expression for emf in loop of wire will be very hard, correct?

    Because the actual flux through the loop at time t is not just Area*B'(t) , but rather (Area*B'(t) + self_flux'(t) )

    Where self_flux(t) is the flux created by the loop itself.

  6. Apr 15, 2009 #5
    I worked this problem out last month, but it's at home and I'm at work right now. I'll scan it and post it later tonight.

  7. Apr 15, 2009 #6
    I just made that question up to explain the issue I'm having in understanding Faraday's law. It's not a problem from anywhere.

    If you mean that you also "considered" this issue a month ago and worked out some proof where we can ignore the self_flux, then that would be great if you can scan that work.
  8. Apr 16, 2009 #7
    Isn't there a minus sign in the total flux because of Lenz' Law?
  9. Apr 16, 2009 #8
    Here it is. I reuploaded it in a jpg format. I forgot about the psd format being unreadable for most. The emf, or voltage if you prefer, and current, is given by:

    V = -j*omega*phi_e*R / (R + j*omega*L);

    I = j*omega*phi_e / (R + j*omega*L).

    Plugging in all boundary conditions makes perfect sense. If R is quite large, >> omega*L, then V reduces to:

    -j*omega*phi_e, which is Faraday's law w/o considering self inductance.

    Note - R = resistance of loop; L = inductance of loop; phi_e = external flux normal to loop; omega = radian frequency of time changing flux.

    Comments are welcome.


    Attached Files:

    Last edited: Apr 17, 2009
  10. Apr 18, 2009 #9
    Also, phi_i = internal fluz due to loop's own current.

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