Electronic – If induced voltage (back-emf) is equal and opposite to applied voltage, what drives the current

electromagnetisminductor

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?

Best Answer

The seemingly paradoxical image of "a current that is flowing despite nothing is driving it" stems from the misconception that you always need a force to move charges. You don't!

Just think about those old-fashioned cathode ray tubes, where you accelerate electrons inside a highly charged capacitor, and the electrons exit the capacitor through a small hole at the end with a certain velocity. Even in the absence of any electric field they will keep this velocity (current!) right until they hit the opposite phosphorescent side of the screen where they get decelerated abruptly and get their energy converted into visible light (and some X-ray!). This lack of resistance is all because these electrons move in vacuum and they move almost in the same direction, so they don't collide with each other.

This is very different from a solid state conductor, where electrons get scattered at the atom bodies, ie. the crystal structure of the solid, and thus constantly lose energy. This is similar to the molecules in a gas that will eventually lose any initially coordinated movement direction, and showing their undirected internal energy as temperature. In order to refuel the charge carriers in a solid resistive conductor (ie. non-superconductor), a persisting electric field is needed.

In an ideal inductor there is no resistance. Think of a coil shaped cathode ray tube as an approximate mental picture. Then what governs electron movement are the law of induction and Ampere's law:

$$\nabla \times E = -\frac{\partial B}{\partial t}$$ $$\nabla \times B = j$$

the latter of which is the simplified version for slowly varying fields (no EM waves).

So the only thing the laws of nature require for the existence of a current is a magnetic field with non-vanishing curl (second equation). If there is no magnetic field (i.e. in the beginning right after pulling a huge Frankenstein switch), there is certainly no current. Moreover, the shape of the current determines the shape of the magnetic field and vice versa. If the shape of the currents is given for all times (because we squeeze the charge carriers through the windings of a coil), the only thing that can vary about the magnetic field is its magnitude with time.

On the other hand, from the law of induction (first equation) we learn, that if the shape of the magnetic field is constant and only its magnitude changes with time, it will be the source of a rotational (closed) electric field of a given shape. In a coil the induction voltages along the individual windings (multiple rotations around an axis) add up to the total inductor voltage. Again, vice versa, any electric field with non-vanishing curl will give rise to a magnetic field changing in time.

Finally we can combine both equations into one by forming the curl of the law of induction and substituting Ampere's law into it:

$$\frac{\partial j}{\partial t}=-\nabla\times\nabla\times E$$

What we see here is the mathematical way of saying that a change of current in time is caused by (the second curl derivative of) a circular electric field. The remaining steps towards U=-L*dI/dt is just geometry applied to the specific inductor under consideration, and the boundary conditions it is subject to.

Conclusion: since there is no resistance, we don't need any voltage driving the current according to Ohm's law. Rather the current change rate is driven by the rotational electric field represented by the constant external voltage that is guided through the coil windings (as soon as the switch has been actuated). Most importantly the "net voltage" is not at all zero, but rather it is the external voltage that is defined by the given boundary conditions.