EM Field Generation and Energy: A Simple Explanation

[worwhite]

Hi,

I understand that EM fields have energy and they travel at c. Light is an EM field so it travels at c as well. So say you run a current through a wire - an EM field is generated, propagating outwards at c. Since an EM field is generated, and EM fields have energy, that energy has to come from somewhere. So does it mean that the current keeps losing energy due to this generation of EM field (assuming a perfect conducting material for the current)?

Thanks.

 

[granpa]

its called inductance. inductance resists changes in current. it takes energy to overcome inductance.

the energy in each part of the field is proportional to the square of the field strength. but the total amount of energy in the field in finite even though the field extends forever.

i think changes in fields move at c. the fields themselves move at whatever velocity the thing that's creating them moves.

 

[DavidWhitbeck]

Hi,

I understand that EM fields have energy and they travel at c. Light is an EM field so it travels at c as well. So say you run a current through a wire - an EM field is generated, propagating outwards at c. Since an EM field is generated, and EM fields have energy, that energy has to come from somewhere. So does it mean that the current keeps losing energy due to this generation of EM field (assuming a perfect conducting material for the current)?

Thanks.

The em field was generated by the power supply in the circuit which is where the energy comes from. No power supply, no field.

 

[lbrits]

EM fields can be broken into a radiative component, that goes like 1/r, and a non-radiative component, that vanishes like some higher power of r. The radiative component makes things "lose energy" to infinity, while the higher components just sort of "fill space" but don't have enough strength to go to infinity. Only changes in current/charge distributions produce radiative components, so if you have an AC circuit you will lose energy, like an antenna, whereas a DC circuit, while it gives off a field, does not lose energy in the same way. Energy is still stored in the field somewhat, but it isn't dissipative.

 

[granpa]

"EM fields can be broken into a radiative component, that goes like 1/r, "

not sure what you mean by this. the intensity of radio waves follows an inverse square law.

"and...makes things "lose energy" to infinity, while the higher components just sort of "fill space" but don't have enough strength to go to infinity."

the magnetic field of a dc current extends to infinity (theoretically).

 

[DavidWhitbeck]

"EM fields can be broken into a radiative component, that goes like 1/r, "

not sure what you mean by this. the intensity of radio waves follows an inverse square law.

That's because intensity is proportional to the square of the field. It's really the Poynting vector, but it becomes like the same thing (because E is proportional to B).

Spherical waves in general go as 1/r I believe.

 

[granpa]

of course. the energy in the radio waves follows an inverse square law. and the energy in the field is proportional to the square of the field strength. so the field strength follows an inverse first power law.

 

[worwhite]

Thanks for your answers.

Granpa, David, combining your answers, I assume that you mean that generating current involves overcoming inductance, which is due to the generation of the EM field. Therefore besides providing energy for the current itself, the power supply has to provide additional energy for the inductance/EM field. And it doesn't just provide this extra energy at the start of the current flow - it has to keep providing it as the field keeps on growing (though maybe less and less).

So basically if I have a closed loop of a perfect conductor, and start a current in it but I turn off the power supply after that, the current won't keep on going indefinitely but will keep losing energy due to the growing field.

If there are any mistakes in my interpretation of your answers, please feel free to correct me.

 

[DavidWhitbeck]

Well... um...

Ok the power supply creates the field. the field is what accelerates the electrons in the wire, thus causing a current to flow. Remember that current is just the flow of charge. Charged particles accelerate from rest only in the presence of a force. Force is what causes objects to accelerate. In this case it must be an electromagnetic force. So the em field is what creates the current.

You need the power because of resistance-- most of the energy is being lost as heat. Turn off the power supply, and the energy is dissipated as heat.

So it goes power --> em fields --> current

 

[worwhite]

David - "You need the power because of resistance-- most of the energy is being lost as heat. Turn off the power supply, and the energy is dissipated as heat"

I'm a little confused here. If you recall, I mentioned the assumption of a perfect conductor. So if it's a perfect conductor, there shouldn't be any electrical resistance or heat lost. Theoretically, with a perfect conductor, the moving charges that form the current wouldn't encounter any resistance and the current would keep on flowing (unless energy was lost via some other means such as through an EM field).

And I've always thought that the EM field was generated by the moving charges in a current. True, there needs to be a field to generate the required potential difference to activate the current in the first place, but that should be different from the EM field that I'm talking about. If I had a power source, a loop of wire and a very high resistance resistor in the loop, there would be a potential difference but no EM field (spreading out from the setup) since there is no current due to the resistance.

So I thought that the EM field doesn't drive the current, but the current generates the EM field?

Sorry if I seem a little slow. I hope someone can help me clarify matters on this.