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Question on Faraday's Law

  1. Sep 23, 2012 #1
    For a coil,

    [tex]e=N\frac {d\Phi}{dt}[/tex]

    Where [itex]e\;[/itex] is the instantaneous voltage driving the coil and [itex] \Phi\;[/itex] is the flux generated through the coil with N turns.

    For a coil

    [tex]\oint \vec B \cdot d\vec l =\mu N I \Rightarrow B=\mu N I \Rightarrow \Phi = BS=\mu N I S[/tex]

    In the book Handbook of Transformer Design & Application by Flanagan, page 1.7, it gives

    [tex]e=N\frac{d\Phi}{dt}\times 10^{-8}[/tex]

    It said the multiplier factor depends on the system units. I have no idea how that [itex]10^{-8}\;[/itex] comes from. Please help.


  2. jcsd
  3. Sep 23, 2012 #2
    I have never seen Flanagan, is it an old book?

    The conversion factors between the old cgs system and MKS are

    Magnetic flux density : 1 volt-second/metre2 = 104 emu (gauss)

    Magnetic flux : 1 volt-second (weber) = 10 8 emu (maxwell)

    Inductance : 1 henry = 109 emu

    EMF : 1 volt = 108 emu
  4. Sep 23, 2012 #3
    Thanks for the reply. I still have question:

    [tex]\Phi = BS=\mu N I S[/tex]

    It [itex]\Phi\;[/itex] is in H/m X N X coulomb/sec X m^2. [itex]\frac{d\Phi}{dt}\;[/itex] is in (H/m X N X coulomb/sec X m^2)/sec

    How does this become maxwell/sec in the equation? I am confused with the units. Please help.
  5. Sep 23, 2012 #4
    In the old cgs system one line of induction was called a maxwell, and magnetic induction expressed in maxwells per sq cm.

    One maxwell per sq cm was called a gauss.

    In MKS

    1 weber per sq m = 104 gauss.

    since 1 meter squared = 104 cm2 it follows that

    1 weber = 108 maxwells
  6. Sep 23, 2012 #5
    Yes, I actually studied they since you replied. My question is how to make the two side to be equal units as I posted in #3

  7. Sep 23, 2012 #6
    I am only guessing about your reference.

    You really need to supply more detail please.
  8. Sep 23, 2012 #7
    I am referring to this

    It Φ is in H/m X N X coulomb/sec X m^2. dΦdt is in (H/m X N X coulomb/sec X m^2)/sec

    On the left side, μ is in H/m, I is in A/sec, area is m^2. Then it is per second.
    On the right side, it is Web per second.

    I am still missing something.
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