I'll try and explain my confusion as best I can:Lenz's law says that a changing current through a solenoid produces a magnetic field that generates an emf that opposes the current, right?Current through an inductor (when it is constant) produces a magnetic flux that 'favors' the constant current, right?So my question is; when you connect a inductor to a voltage source (DC), what exactly happens while the inductor is resisting the current (or 'charging up')? And what makes it stop?Also, how do you physically explain the phenomenon of an inductor producing a voltage HIGHER than the source it was connected to, when the circuit is rapidly open (please try not to use the calculus of inductors to explain this)?Any help would be hugely appreciated.
When first connected, the coil produces a 'back EMF' opposing the voltage applied. This reverse voltage is proportional to the rate of change of the field strength. The reverse voltage appears to the source as a high resistance, limiting the current initially. When the field is fully established, the rate is zero, and the current is simply E/R. When disconnected, the field tries to collapse instantaneously, thus causing a voltage spike in the forward direction. Depending on the values of wire resistance, inductance, etc., this voltage can be extremely high even though only a small voltage was used initially to establish the field. The energy expended charging the field is essentially released over a very short time, thus driving the voltage up.
Assuming zero resistance, when you connect the battery, the current will start increasing, but only at a rate the generates a back emf equal to the voltage. Theoretically, with no resistance, it will increase for ever. However in actuality the battery will start generating an internal resistance at large currents and there will be a voltage drop across that resistance.
