Back EMF in Solenoid: Explaining Current Flow

[QwertyXP]

Consider an ideal solenoid (no resistance, no leakage reactance etc.) connected across an AC supply. The back EMF induced in it will be exactly equal and in opposite direction to the source voltage (which means that when a certain terminal of the AC supply is positive, the side of solenoid connected with it would also be positive, and vice versa).

My question is, how will current flow at all when the EMFs of AC source and solenoid are cancelling each other out? It's like having having a circuit with only two batteries and terminals of similar polarities shorted with each other.

I've read quite a few explanations on the internet but have yet to fully understand what's happening here.

 

[NascentOxygen]

The back emf is proportional to di/dt, so if no current flowed there would be no back emf. Current will rise to the value where di/dt is such that back emf exactly opposes applied voltage.

This is not much different from current in a resistance. The voltage across a resistance increases as the current increases (obeying Ohms Law) until the voltage across the resistance exactly opposes the applied voltage.

 

[QwertyXP]

i didn't really get this. I was comparing ''back EMF'' of a solenoid with the emf induced across a DC motor's armature. When the motor's back EMF = source voltage, no current flows. So it's like connecting a battery with opposite polarity.

Considering a resistor directly connected across a source, once the voltage across it is equal to source, the current through it does not increase because for that there would have to be a difference in potential. However, if you connect a solenoid across a DC source, the current through it continues to increase for a certain time even though source voltage = back EMF. With no difference in potential, how is this possible?

Also, the solenoid circuits's equation doesn't appear to be balanced:
V(source)=Back EMF (which is equal to source) + CurrentxReactance
when back EMF = source, the CurrentxReactance part should be zero!?

 

[NascentOxygen]

i didn't really get this. I was comparing ''back EMF'' of a solenoid with the emf induced across a DC motor's armature. When the motor's back EMF = source voltage, no current flows. So it's like connecting a battery with opposite polarity.

In an "ideal lossless" motor, yes. In a practical motor, there is armature resistance, so that drop has to be accounted for.

Considering a resistor directly connected across a source, once the voltage across it is equal to source, the current through it does not increase because for that there would have to be a difference in potential. However, if you connect a solenoid across a DC source, the current through it continues to increase for a certain time even though source voltage = back EMF. With no difference in potential, how is this possible?

v=L.di/dt so if the applied voltage is maintained constant, di/dt will be constant, meaning current continues to ramp up forever (in a lossless inductor).

Also, the solenoid circuits's equation doesn't appear to be balanced:
V(source)=Back EMF (which is equal to source) + CurrentxReactance

when back EMF = source, the CurrentxReactance part should be zero!?

For sinusoidal AC, in the steady state, the back emf is the current x reactance term.

 

[alphy]

I can assure you that current will still flow in an ideal solenoid connected to an AC supply, despite the back EMF being equal and opposite to the source voltage. This is due to the fact that the back EMF is induced in the solenoid coil as a result of the changing magnetic field created by the alternating current in the solenoid. This changing magnetic field induces a current in the solenoid coil, which will flow in the opposite direction to the applied voltage.

In an ideal solenoid, there is no resistance or leakage reactance, meaning that the induced current will flow freely without any hindrance. This is similar to a superconductor, where current can flow without any resistance. Therefore, even though the back EMF may be equal and opposite to the source voltage, the induced current will still flow in the solenoid.

To better understand this concept, it may be helpful to think of the solenoid as a simple circuit with a resistor and an inductor in series. The back EMF can be thought of as the voltage drop across the inductor, while the source voltage is the voltage applied to the entire circuit. In this analogy, the resistor represents the resistance of the solenoid, which in an ideal solenoid is negligible. Therefore, the induced current will still flow through the solenoid, despite the back EMF and source voltage being equal and opposite.

In summary, the back EMF in an ideal solenoid does not cancel out the current flow, but rather it is a result of the changing magnetic field created by the alternating current in the solenoid. This induced current will flow in the opposite direction to the applied voltage, but will not be hindered by any resistance or leakage reactance in an ideal solenoid. I hope this explanation helps to clarify the concept for you.