# Another dumb magnetic firlds question

1. Aug 22, 2006

### linux kid

If magnetic fields aroung a coil of wire were turned on and off at a certain frequency, would it generate some form of energy?

2. Aug 22, 2006

### WhyIsItSo

We call these generators. Ok, that's a little sarcastic, but the principle of spinning windings past magnets has the effect you are talking about.

I suppose if you turned the magnetic fields on and off, you'd generate a square wave, while a generator creates a sine wave, but both create good old electricity.

3. Aug 22, 2006

### ZapperZ

Staff Emeritus
Er.. you do?

The current generated in the coils would be proportional to the rate of flux across the coils' area, i.e. $\frac{d\Phi}{dt}$. So if you have a sine wave, the time derivative of that would also be a sinusoidal wave with a phase shift (ignoring self induction).

Zz.

4. Aug 22, 2006

### WhyIsItSo

Which part are you talking about?

In the hypothetical scenario where a magnetic field was either on or off, and oscillated between these states at a high enough frequency, would that not generate a square wave?

If referring to my comment about a generator, do they not create a sine wave? Is not a sine wave sinusoidal?

5. Aug 22, 2006

### ZapperZ

Staff Emeritus
Oh, sorry. I misread your statement. I thought you said that if the generator of the external field has a square wave, the induced current can still be sinusoidal.

Oy.. that's what I get for dipping my nose into more than a few threads at a time.

Zz.

6. Aug 23, 2006

### phun

I think you'd get a train of delta functions (pulses) of electricity that way.

7. Aug 26, 2006

### leright

This is true. The emf induced in the coil, and thus the current in the coil, is proportional to the rate of change of the magnetic flux through the coil. Well, if you are able to "immediately" switch the B-field between on and off states, the rate of change is infinity and this occurs in a no time, which would mean an infinite current for a time period of zero. This is modeled by a train of dirac delta functions. Practically, however, there would be a very large rate of change in the flux over a very small finite time period, which would produce very large spikes of current that occur over very small time periods, similar to the ideality of the delta function, but with finite height and width.

BTW, recall that the derivative of the unit step function is the delta function, and the derivative of the square wave is a delta pulse train function, which is intuitively obvious.

8. Sep 3, 2006

### leright

Also, to add, the pulses will alternate in sign. When you switch the field from the off position to the on position you will get a pulse current in one direction (which we'll call positive), and then when you switch from on to off, you will get a pulse current in the opposite direction (which we'll call positive.)

keep in mind these pulses are not square waves, but approximations to the dirac delta function.

Last edited: Sep 3, 2006
9. Sep 4, 2006

### NoTime

I'm not familiar with the dirac delta function.

But, a square wave can be treated as the sum of a collection of sine waves.
How square the wave is depends on highest sine frequency that can be included in the collection.
The corners start to round as you truncate the collection.