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Faradays law/Lens law: how does the induced emf 'act'

  1. Jun 10, 2012 #1
    Hello everyone.

    I'm trying to understand how an 'induced' emf behaves on a current carrying conductor.

    Basically the example that's giving me problems is in the case of a motor, if a current starts to pass through a coil it will get an induced 'emf' to oppose the motion. And hence it seems the current drops as the motor reaches a stable speed.

    My question lies in 'how' it acts, I'm imaging a current in one direction, being passed another current of opposite direction, but why doesn't it 'get stronger/reaches a emf to completely oppose the motion'?

    Thank you :smile:
  2. jcsd
  3. Jun 10, 2012 #2


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    hi thisischris! :smile:
    does this help? (from http://en.wikipedia.org/wiki/Hydraulic_analogy) …
    A heavy paddle wheel placed in the current. The mass of the wheel and the size of the blades restrict the water's ability to rapidly change its rate of flow (current) through the wheel due to the effects of inertia, but, given time, a constant flowing stream will pass mostly unimpeded through the wheel, as it turns at the same speed as the water flow. The mass and surface area of the wheel and its blades are analogous to inductance, and friction between its axle and the axle bearings corresponds to the resistance that accompanies any non-superconducting inductor.​
  4. Jun 10, 2012 #3
    So 'friction' acts as resistance which corresponds to a 'drop' in emf? Should I view this 'opposed' emf in the motor as a resistance to the larger current?

    I still don't quite understand why the induced emf does not keep building to completely oppose the motion however?

    Thank you.
  5. Jun 11, 2012 #4


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    Friction is another issue. The (inductive) reaction against a change in current doesn't involve any loss of energy.
    It is impossible to discuss, meaningfully, the totally 'ideal' situation, in which a voltage source of no resistance is connected to an inductor of no resistance because no current could pass. This is because the rate of current change would be such as to produce a back emf equal to the supply voltage.
    In reality, there is a limit to the conductance of a circuit so the current is limited by the resistance and the 'back' emf is never quite equal to the supply voltage so some current will always flow, eventually reaching a value of V/R, where R is the circuit resistance.
    So your confusion is quite justified, I think. :smile:
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