Can Switching Magnetic Fields in a Coil Generate Energy?

  • Thread starter Thread starter linux kid
  • Start date Start date
  • Tags Tags
    Magnetic
AI Thread Summary
Switching magnetic fields around a coil of wire at a certain frequency can generate electricity, akin to how traditional generators operate. The current produced would depend on the rate of change of magnetic flux, leading to different waveforms; a square wave from rapid on-off switching and a sine wave from conventional generators. If the magnetic field is switched instantaneously, it theoretically results in infinite current spikes, modeled by delta functions, though practically this would yield large current spikes over brief time intervals. The discussion also notes that square waves can be approximated by a series of sine waves, with the sharpness of the square wave depending on the frequency components included. Overall, the principles of electromagnetic induction apply regardless of the waveform generated.
linux kid
Messages
101
Reaction score
0
If magnetic fields aroung a coil of wire were turned on and off at a certain frequency, would it generate some form of energy?
 
Physics news on Phys.org
linux kid said:
If magnetic fields aroung a coil of wire were turned on and off at a certain frequency, would it generate some form of energy?
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.
 
WhyIsItSo said:
I suppose if you turned the magnetic fields on and off, you'd generate a square wave, while a generator creates a sine wave,

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.
 
ZapperZ said:
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).
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?
 
WhyIsItSo said:
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?

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

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.
 
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:
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.
 
Back
Top