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LC circuit with constant EMF

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  1. Aug 6, 2015 #1
    Hi,

    A capacitor has both its terminals connected together using a wire in a rather lengthy circular fashion, hence acting as a significant inductance.
    If the area encircled by this 'circuit' is subjected to a linearly rising magnetic field (B=k x t) where k is a constant, the emf induced -due to this external magnetic source- is a constant value proportional to k.
    A current will run and consequently begin to charge the capacitor. My question; will this charging process continue for as long as the external magnetic field is linearly rising, or will it take the capacitor up to a voltage related to (k) and then reach an equilibrium? I'm aware that the current that runs in the circuit will affect the magnetic field crossing it and hence affect the process, but my question is still the same. Does the capacitor's voltage only rise to some certain value or would it carry on doing so until breakdown occurs in the dielectric?

    Thanks
     
  2. jcsd
  3. Aug 6, 2015 #2

    BiGyElLoWhAt

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    The EMF is a voltage, and since it's a constant rate of change of field, it will have a constant voltage across the loop. Once the capacitor reaches the same voltage across it as there is across the loop, it can't charge any more; there is no gradient to the Electric field to cause charges to move any more.
     
  4. Aug 6, 2015 #3
    Thanks @BiGyElLoWhAt . My final question is this: If the voltage across the capacitor now reaches the constant value you mentioned and remains so, this would imply that no current could be running in the loop (after that constant voltage was achieved). So as far as the inductor is concerned at that moment and on-wards, i=0, and also, di/dt=0. The emf for the inductor has two definitions. V=L x di/dt (which is zero), and emf=-N x dphi/dt (which is not zero; there is a source continuously providing a linearly increasing magnetic field). So how are these two expressions of emf reconciled in this case, or where is the preceding logic erroneous.

    Many thanks again.
     
  5. Aug 6, 2015 #4

    BiGyElLoWhAt

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    I don't think the current would be zero. There's still a potential difference across the capacitor and there's still a potential difference across the loop.
    I think what's going on between these 2 equations is this: ##L \frac{dI}{dt}## is one segment of the overall voltage (internal to the circuit) and ##N\frac{d\phi}{dt}## is another segment (external to the circuit) and both of them sum together to give you a total voltage across the inductor. If you look at the 2 equations, I think it's obvious that the source of the voltages between the 2 cases are completely independant of each other, well kind of. One is current dependant (internal) and the other is magnetic field dependant (external). Does this help?
     
  6. Aug 6, 2015 #5

    BiGyElLoWhAt

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    In other words, the inductor isn't generating any voltage independantly of the magnetic field, as the current becomes constant.
     
  7. Aug 6, 2015 #6
    If the current is NOT zero, charges are then moving along the loop. This will mean they have to be accumulating somewhere; that somewhere necessarily having to be the capacitor plates. The voltage would then keep on rising (which you don't believe to be the case).

    In short, if a loop is terminated with a capacitor and carries a constant current, the voltage across that capacitor has to be rising continuously. This derives directly from the concept of current continuity. I cant see how the capacitor can hold a constant voltage when you assume the current doesn't drop to zero in the loop.
     
  8. Aug 6, 2015 #7

    BiGyElLoWhAt

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  9. Aug 6, 2015 #8
    I'm aware of the amendment to Ampere's Law through the addition of displacement current. Yet this doesn't really resolve the issue of how current can be present in the loop -unidirectionally- without accounting to a continuous rise in capacitor voltage.

    In any case, thanks a lot for giving this your time. I will try to think more of the matter and see how it could be resolved.
     
  10. Aug 6, 2015 #9

    BiGyElLoWhAt

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    Yes, I see what you're saying. Perhaps it would go to zero current. With there being a potential difference across the loop, I'm not sure how it could, but I might be overlooking something.
     
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