# Induced voltage in a coil

1. Nov 20, 2012

### daudaudaudau

Hello.

Say I have a single turn of a coil in an external magnetic field. How do I calculate the voltage induced in this turn ? Initially I would use Faraday's law

∇×E = -j ω B

But what should I use for B ? Will B not be a combination of both the external field as well as the field created by the current induced in the coil:

∇×B = μ J = μ σ E

?

Suddenly the problem seems more complicated!

2. Nov 20, 2012

### Enthalpy

That's correct. If current can flow in the loop - that is, the loop is closed somehow - then it does create its own induction and flux. This is called self-induction.

The distribution of the induction differs when the coil creates it, for instance it's stronger near the conductor, so detailed spatial comparison helps little.

Reasoning with the electric circuit helps more. The external field induces a potential difference across the loop. The loop has its own impedance, which includes its self-inductance and possibly other circuit elements. The potential difference divided by the (complex) impedance defines a current. This current tells you what field the loop creates by reaction.

3. Nov 20, 2012

### Enthalpy

Your question has very concrete practical implications.

In a good transformer, which has no gap and a permeable iron core, very little primary current suffices to create the flux. The voltage induced at the secondary can make current flow at an external use, and the secondary current creates a flux just as the primary current does. As current is used at the secondary, the primary current increases to keep the flux constant, following N1I1=N2I2 plus the magnetizing current.

In a powerful generator (10kW to 1600MW) little current is needed to create the initial induction, but the output current's effect can be much stronger than this initial current, even at low copper losses. The output current would bring the field to zero far before reaching this maximum, if it were not compensated by increasing the excitation current accordingly and actively to keep the desired induction. This resembles the transformer.

This is more easily done if the output current is resistive than inductive and is one excellent reason why electricity companies want customers to show a good cos(phi).

Same in a motor of non-negligible power. DC motors have compensations windings, which are perpendicular to the field windings, only to compensate the flux created by the rotor's current; the compensation winding is more massive and consumes more power than the field winding, beginning around 100kW. This is less visible - but very present and important! - in AC motors; there, one set of stator windings are both the field and compensation windings because the 90° phase at a rotating field means only a different mechanical angle shift at the rotor. Electronic control of AC motors avoids the time lag resulting from the mechanical angle shift by shifting the AC phase proactively - it's called vector control.

Just for fun: when I designed the 13.56MHz RFID (long ago...) I observed that the induction created by the chip card was stronger than the one made by the reader. In other words, the reaction exceeded the action, which was puzzling as the reaction should have compensated the action before being as strong, stopping the game.

But at the card, I had a resonant LC circuit which let the current flow in phase with the induced voltage, instead of lagging by 90° in a self-inductance alone, so the reaction field was in quadrature with the reader's field, and the reaction could exceed the action without compensating it. This took me some time to understand...

Apologies for the lengthy answer, electromagnetics is just puzzling...

4. Nov 20, 2012

### Enthalpy

In the second equation, E must be complemented by dA/dt, possibly with some sign if you like.

This is all-important in induction. For instance in a generator, copper wires shall have low loss, meaning E~0, but you get a V at the terminals thanks to the induction dA/dt summed over the conductor path (or d phi / dt if you prefer). Or even, E=0 in a superconductor, which is considered practically for generators and motors, at orientable pods for boats for instance. Though there, it would supposedly be a type II superconductor, which has a resistance.

5. Nov 20, 2012

### daudaudaudau

Thanks for the answers. Where exactly should A be included ? In the first equation, ∇×E=-j ω B, I would put $E=-\nabla V-j \omega A$ to give
$$\oint\left(\nabla V+j\omega A\right)\cdot dl=j\omega \Phi$$

So does this mean that my voltage is
$$V=j\omega\Phi-j\omega\oint A\cdot dl$$
?

6. Nov 20, 2012