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Mutual Inductance: Coil 1 Excited, Coil 2 Open Circuited

  1. Jun 9, 2010 #1
    I have a very basic question regarding mutual inductance.

    Consider a basic square loop with two coils wound on opposite sides from each other, with coil one excited while coil two si left open circuited. The total flux linking coil 1 is

    Φ11 = Φl121


    Φ11 is the total flux linking coil 1
    Φl1 is the leakage flux and
    Φ21 is the mutual flux linking both coils

    My question is why there is mutual flux if the second coil is open-circuited, in other words, why is Φ21 non-zero? I would expect flux through the iron core due to coil one but how/why does coil two contribute to this flux?
  2. jcsd
  3. Jun 9, 2010 #2
    Here [tex]\phi_{l1}[/tex] is the leakage flux i.e., the flux which is linking only with first winding and not with second winding and is due to imperfect coupling. The flux [tex]\phi_{l21}[/tex] is the one which interlinks with both windings. Since [tex]\phi_{l21}[/tex] has nothing to do with the second coil's open or short circuit conditions it wont be zero. You may refer the Wikipedia's page below
  4. Jun 9, 2010 #3
    Let me take step back and ask this question:

    What would happen to the flux in the magnetic core if the second coil is removed? Does only the leakage flux remain, or would some flux still flow through the whole core?
  5. Jun 9, 2010 #4
    In fact when the second coil is open circuited, the first coil behaves as if it is the only inductor present in the core. So even if the second coil is removed, the flux going through the core [tex]\phi_{l21}[/tex] will be the same.
  6. Jun 10, 2010 #5
    That helps a great deal. I am going to ponder that for a little while...
  7. Jun 27, 2010 #6
    Hey Karthick,
    This is different you have been discussing,Should I use N1+N2 for the no.of turns in the B=U0NI/L where both the coils are one over the other , but one is NOT Connected?
    Or should I take only N1 or N2 which ever coil is driven by DRIVE CIRCUIT.

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