# Understanding Electric Generators

1. Apr 9, 2013

### Number2Pencil

I think I am starting to understand various aspects of electrical generators, and I wanted to run this by the forum and have people correct my explanations, if needed. I understand there are many types of motors and this thought-experiment is just a permenant magnet near a coil. If anyone can verify this explanation, I'd be most grateful:

According to Faraday's Law, Induced voltage is the negative of the change in magnetic flux linkage. Flux linkage is only related to the magnetic B-field linked within a conductor (such as a coil loop). The direction of the voltage/current induced (eddy currents) will be induced to create a magnetic field which opposes the change in the external field.

So basically, if I have a magnet moving through a coil, as long as I move it at a specific speed through the loop, it will induce a specific voltage from the loop no matter what the load is. The difference that the load makes, however, is that it dictates how much mechanical energy you need to input into the generator in order to overcome the counteracting magnetic field created by the coil.

This can be seen by looking at Ampere's Law, which shows that a magnetic field created by a loop is proportional to amount of current in the conductor.

Ideally, mechanical power in = electrical power out.

Let's say I have a generator connected to a 5Ohm load. I am able to move the permenant magnet at a certain speed to produce 40V.

P = V^2 / R
P = 320 Watts
Electrical current = 8Amps

In an ideal situation, I should be able to calculate how much force/torque I need to apply to the permenant magnet based on the power equivalency, the mass of the PM, the field strength of the PM, the geometry of the coil relative to the PM, and the velocity needed to get that specific rate-of-change of flux linkage.

If I increase the resistance, I lower the required power:

Load = 50 Ohms
P = V^2/R
P = 32 Watts
Electrical current = 0.8 Amps

It should be mechanically easier to move the PM around the coil since it is unable to create a large counteracting field due to the smaller coil current.

What if I shunt the generator output?

Load = 0 Ohms
P = infinity

I would have to apply an infinite amount of mechanical force/torque to produce this electrical output. I either wouldn't be able to move the magnet at all, or in even trying would produce such a large current through the non-ideal resistances of the wire that it would create a huge current and likely burn up the shunt.

What if I connect a capacitor to the output?

From a discharged state, the capacitor would look like a 0-ohm shunt (probably want to add some limiting series resistance due to the prior thought-experiment). As the capacitor charges the voltage accumulated on the capacitor causes the overall current to exponentially decay, requiring less mechanical power over time.

Once the capacitor is charged, I can let go of the PM, and now that there is energy stored in the capacitor, it will want to discharge through the same coil, and everything works in reverse (the cap is now acting as a power source for an electric motor). If I wanted to keep the energy in the cap, I should probably build a full-wave rectifier circuit on the output of the generator and a diode to prevent the reverse currenting.

2. Apr 10, 2013

### jim hardy

3. Apr 13, 2013

### Number2Pencil

Onto a question: How does Faraday's law work with a magnet moving inside a long single layered solenoid? I guess I have a problem defining a 2D linkage contour when the magnetic flux is being spread out in a 3D volume. Here is my theory:

1) Magnet outside the solenoid completely: No change in flux at all, no induced voltage at all.

2) North side of magnet starts moving through solenoid entrance: The flux lines from the magnet will spread out to a certain distance (distance "d"), which will effect a certain number of coils since those are experiencing changing magnetic flux. To calculate the induced voltage, I must integrate Faradays' law from the entrance to distance d. Basically, the induced voltage from one coil in the solenoid adds to many other coils in the solenoids (only the ones affected by changing magnetic flux), and the overall induced voltage and current is a product of them all.

3) The magnet is moving completely inside the solenoid: Now the coils near the North side of the magnet are producing a current in direction +I to creating an opposing magnetic field (trying to keep the magnet from entering), and the coils near the South side of the magnet are producing a current in direction -I to create an attracting magnetic field (trying to keep the magnet from leaving). The net voltage and current is equal to zero since they cancel each other out. The net forces on the magnet cancel each other out as well, and the magnet will require no mechanical power to push while it is completely inside the solenoid.

Is this correct?

4. Apr 13, 2013

### Number2Pencil

Also, how does Lorentz Force come into play?

F = q(E+vXB)

When the voltage and current is induced in the coil, we have all of these terms at play (maybe the electric field one can be ignored?) So is the lorentz force is acting on the coil, causing it to try and move (acting like a motor), but if we firmly clamp our coils down, does the force work against the magnet trying to move?

Are Lorentz Forces the force of which the mechanical power source overcomes?

5. Apr 13, 2013

### jim hardy

you seem to have it.

Open circuit there's only changing flux coupling turns so there's voltage induced as magnet enters and leaves...

http://www.a-levelphysicstutor.com/field-electro-mag-ind.php

short circuit there's a plenty of simple demonstrations on youtube using copper pipe which might represent your solenoid with the ends connected together.

seeing is believing - try it yourself. junk hard disk drives have an ultra strong magnet and coil assembly for head positioner....

Last edited by a moderator: Sep 25, 2014
6. Apr 14, 2013

### Number2Pencil

Just to clarify, you are saying that the answer to:

"Are Lorentz Forces the force of which the mechanical power source overcomes? "

is "yes?"

7. Apr 14, 2013

### jim hardy

Yes. The mechanical force experienced by the conductors is from the Lorentz force acting on the moving charge(current) inside them.
http://hyperphysics.phy-astr.gsu.edu/hbase/magnetic/magfor.html

as I recall forces always come in pairs - so there's an equal opposite force experienced by the moving magnet.

That's where the actual conversion from mechanical to electrical energy takes place.

In large electric machines the conductors are firmly blocked in place so they won't vibrate.

old jim

Last edited: Apr 14, 2013