Suppose we have a circuit with a voltage source, a switch open and an inductor all in series. If we close the switch, the potential difference of the voltage source is instantaneously applied to the inductor. As the current starts to build up, induced voltage from the inductance opposes it. If the induced voltage (back-emf) is equal and opposite to the applied voltage, and the net voltage is zero, what drives the current then? All I could find on the web was this:
"...it is difficult to realise that there can be a current without a 'resultant' emf. Faraday's law states $e = LdI/dt$ and if there is no resistance $e = E$. It is the electrical analogy of constant velocity with no need for a resultant force. If there is resistance then $e = LdI/dt = E - Ir$ ... 'resultant' $\text{emf} = e$"
Could you expand a little bit more on this idea?
Answer
Well... When the back emf is equal to the voltage supplied by the battery, it is not really hard or anything counter intuitive to realize how the current exists in such a case. See, all you need to realize is what is actually the back emf? When the charges in motion, tries to pass through an inductor - the inductor converts its kinetic energy into magnetic energy and slows down the moving charges. The actual force which acts on the charges to slow them down is the induced electric field due to changing magnetic field associated with the inductor. Now what is back emf? It is simply the energy taken by the inductor per unit charge. According to the Kirchoff's law (Energy Conservation), a charge particle in motion expenses the same energy in its motion outside the battery as it gains inside the battery. So all the energy that an electron gains in the battery will be transferred to the magnetic energy of the inductor. So, back emf = source voltage. But, obviously, current exists, because, first the charge gets accelerated inside the battery and then it slows down working against the back emf.
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