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Polarity of back EMF in an inductor....

  1. Nov 9, 2015 #1
    Hello Forum,
    the law of electromagnetic induction states that a time-changing current causes a back (counter) emf in an inductor. The back emf opposes the change in current by generating an induced current I_ind.

    If dI/dt>0, the induced current is opposite in direction to the main changing current. If dI/dt<0, the induced current is in the same direction as the changing current trying to keep it going.

    That said, I am confused about the polarity of the emf (please see figure below): the emf points to the left and that makes sense since the induced current must oppose the increasing current. A current always moves from high potential to lower potential.
    But why is V_ab positive and directed opposite to the emf? Point a has a higher potential than point b. Current usually goes from high potential to lower potential (unless it is traversing the inside of a battery)...


  2. jcsd
  3. Nov 9, 2015 #2

    Just found an old book with the following explanation: if the increasing current is going from left to right, the induced emf generates an induced current that carries positive charges to the right and leaves negative the right side of the inductor. That is equivalent to having a virtual battery with V_a > V_b. This battery is is parallel with the actual battery in the circuit and opposes it (the batteries fight each other).

    If the current decreased, the induced current would make V_a< V_b, which is equivalent to having a battery in series with the actual battery in the circuit. These two series batteries have their voltages added together. When we unplug a device, we see a spark because of this large voltage that is generated at the opening point....

    So the emf and the virtual battery with voltage V_ab =V_a-V_b are opposite in sign....

    Did I answer myself correctly?
  4. Nov 10, 2015 #3


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    Staff: Mentor

    Is there a mistake here? Otherwise it reads okay. The back emf always opposes the applied external voltage.
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