Calculating EMF in a Solenoid: Understanding the Relationship of Turns and EMF

[Adesh]

If I have a solenoid with N number of turns in total. And if I say that in each

turn the EMF is equal to e then can I conclude that the total EMF in the solenoid i.e. from A to B is N \times e .
I’m asking this because whenever a current I flows in each turn of the solenoid then we always say that total current flowing through the solenoid is N \times I . So can we say this even for EMF .

Thank you.

 

[berkeman]

the EMF is equal to e then can I conclude that the total EMF in the solenoid i.e. from A to B is N×e N \times e .

Yes.

I’m asking this because whenever a current I flows in each turn of the solenoid then we always say that total current flowing through the solenoid is N×I N \times I .

No. The current is I. The number of "Amp-Turns" is NI.

Does that make sense?

 

[Adesh]

Yes.

No. The current is I. The number of "Amp-Turns" is NI.

Does that make sense?

“Amp-turns” means ?

 

[berkeman]

“Amp-turns” means ?

The current in the coil multiplied by the number of turns. That is directly related to the magnetic flux in the coil.

https://en.wikipedia.org/wiki/Ampere-turn

 

[Adesh]

Yes.

No. The current is I. The number of "Amp-Turns" is NI.

Does that make sense?

Can you please explain why the EMF got added up?
Is it due to the total work done on a 1~C of charge will be N \times e as in each turn the work done is e ?

 

[berkeman]

EMF is like a voltage. Since each turn has its own EMF or voltage generated by a changing flux through the coil, the little voltage sources are in series so they add up.

The same current flows through the whole coil, so it's all the same current. The flux is generated by the current multiplied by the number of turns, hence the concept of Amp-Turns.

 

[Adesh]

EMF is like a voltage. Since each turn has its own EMF or voltage generated by a changing flux through the coil, the little voltage sources are in series so they add up.

The same current flows through the whole coil, so it's all the same current. The flux is generated by the current multiplied by the number of turns, hence the concept of Amp-Turns.

Thank you. I had a misconception that the current was N\times I. Thank you once again

 

[sophiecentaur]

EMF is like a voltage. Since each turn has its own EMF or voltage generated by a changing flux through the coil, the little voltage sources are in series so they add up.

We could also point out that, once the Field has built up to its final value, there is no Induced ('back') emf and the Resistance of the Coil is the only thing defines the current through it. The Potential drop across each turn is Total PD / N and the Current = V/R. Solenoids tend to have as low resistance as possible (thickest wire for the space available for your N turns) so the final current can be very high.
'Amp Turns' is a term that may not make sense at first but many turns with low current can produce the same flux as few turns with high current. The choice of N depends on available power supply and how hot the coil can be allowed to get (resistive heating).

 

[vanhees71]

Thank you. I had a misconception that the current was N\times I. Thank you once again

The current is of course constant along the wire (in DC). The reason is simply charge conservation. In the stationary case at any cross section of the wire the charge per unit of time flowing through must be the same. However, using Ampere's Law you have to count how many times this current runs through the area in the integral, and that's $N$ times, i.e., you have (in SI units)
$$\int_{\partial A} \mathrm{d} \vec{r} \cdot \vec{B}=\mu_{0} \int_A \mathrm{d}^2 \vec{f} \cdot \vec{j}=\mu_0 N I.$$

 

[rude man]

Each turn is like a battery of so many V emf.
What happens when you hook up N batteries in series with equidirectional polarity?