# Velocity of an electric or magnetic field

1. Jun 10, 2008

### granpa

how does one calculate the velocity of an electric or magnetic field? obviously its not as simple as just taking the velocity of the particle that produces it since a given field may be produced by many particles with different velocities.

2. Jun 10, 2008

### cmos

Starting with Faraday's law and the Ampere-Maxwell law (in the absence of current), you realize that electric and magnetic fields induce each other. After some simple manipulations, you find that the electromagnetic field obeys the standard wave equation. The velocity is thus seen to be:

$$1/\sqrt{\epsilon_0\mu_0}$$,

which is exactly 299,792,458 meters per second.

3. Jun 11, 2008

### armis

cmos is right. You just need to get the wave equation.
Write down the Maxswell's equations in vacuum:

$$\nabla\times\vec{H}=\partial{\vec{D}}/\partial{t}$$ Ampere-Maxwell law in the absence of current

$$\nabla\times\vec{E}=-\partial{\vec{B}}/\partial{t}$$ Faraday's law

$$\nabla\cdot\vec{B}=0$$ This shows that the magnetic field has no sources and it's lines have no start nor end.

$$\nabla\cdot\vec{D}=0$$ That's the differential expression of Gause law which shows that charges are the sources of the electric field. We have a zero on the right side since there are no charges in vacuum.

Allright, now multiply the 2nd equation by $$\nabla$$ vectorically. You'll get

$$\nabla\times(\nabla\times\vec{E})=-\nabla\times(\partial{\vec{B}}/\partial{t})$$

Which is the same as

$$\nabla\times(\nabla\times\vec{E})=-\partial/\partial{t}\cdot\nabla\times\vec{B}$$

Since $$\vec{B}=\mu\mu_0\vec{H}$$ ; $$\vec{E}=\epsilon\epsilon_0\vec{D}$$ and $$\vec{a}\times(\vec{b}\times\vec{c})=\vec{b}\cdot(\vec{a}\cdot\vec{c})-\vec{c}\cdot(\vec{a}\cdot\vec{b})$$. Besides $$\mu$$ and $$\epsilon$$ are both =1 in vacuum

Thus you sould get

$$\nabla\cdot(\nabla\cdot\vec{E})-\vec{E}\cdot(\nabla\cdot\nabla)=-\epsilon_0\mu_0\cdot\partial^2\vec{E}/\partial{t^2}$$

$$\nabla\cdot(\nabla\cdot\vec{E})$$ according to the 4th Maxwell's equation in vacuum is zero. Thus:

$$\triangle{\vec{E}}-\epsilon_0\mu_0\cdot\partial^2{\vec{E}/\partial{t^2}=0$$

and it just so happens that the speed of an electromagnetic wave in vacuum according to the wave equation is the as the speed of light

$$v=1/\sqrt{\epsilon_0\mu_0}= 299,792,458 m/s$$ as pointed out by cmos.

In general the speed of the electromagnetic wave is

$$v=1/\sqrt{\epsilon\epsilon_0\mu\mu_0}$$

Another point worth atention is that in feromagnetics electromagnetic waves can't travel since most often they are good conductors.

I hope I didn't make any mistakes. Feel free to correct me

Last edited: Jun 11, 2008
4. Jun 11, 2008

### granpa

5. Jun 11, 2008

### Nerd

Magnetic and electric fields are waves.

6. Jun 11, 2008

### armis

granpa then the speed is c no matter what's the speed of the particle.
http://www.cco.caltech.edu/~phys1/java/phys1/MovingCharge/MovingCharge.html

Hmm, I wouldn't say magnetic and electric fields are waves. But an electromagnetic field could be propagated in a wavelike manner

Yet again, I might be wrong as I got a bit confused. Hopefully someone more advanced will help us out

7. Jun 11, 2008

### cmos

Loosely defined, a wave is a time-varying disturbance (I welcome a more rigorous definition). It turns out that light is a time-varying electromagnetic field, i.e. an electromagnetic wave.

Mathematically and conceptually, it is convenient to talk about the magnetic fields and electric fields as separate but related entities. In reality, however, they are one and the same - hence the electromagnetic field.

If you are familiar with the special theory of relativity, then you will recall the observers in different reference frames may observe slightly different occurrences. A consequence of this is that what appears to be an E-field in one reference frame may appear to be a B-field in another reference frame. In fact, when expressed in tensorial form (the language of relativity), Maxwell's equations are manifested as a set of two equations (in contrast to the set of four equations when expressed in vectorial form).

The applet posted by armis is awesome! One of the classical results is that a charged particle will release EM radiation when accelerated. Go into the "linear" mode and set v=0.0: a static electric field. Now grab the slider bar and immediately go to something like v=0.4 (a quick acceleration): you will see that a wavefront propagates. Now go into the "circular" mode and set v to something fast like v=0.5: you will see a continuous propagation of wavefronts due to the centripetal acceleration of the particle.

All the other modes are variants of the above - have fun!

8. Jun 11, 2008

### armis

Thanks cmos
I wish I would be familiar with the special theory of relativity. I think I'll have it next year though

9. Jun 13, 2008

### granpa

like i said, i am not referring to waves or to the propagation of changes in fields.

a wire carrying a current that is increasing (lets say at a rate proportional to time) produces a magnetic field that would produce an force upon a charge at point p. the velocity of the field at point p is the velocity that that charge would have to have at that same point to produce the same amount of force if the magnetic field wasnt increasing.

but i dont know how to calculate it.

10. Jun 13, 2008

### armis

Do you mean magnetic flux density?Cause the velocity of the field is c.

Sorry, If I don't get it but I am still confused with english termionology, although I am getting there!

Last edited: Jun 13, 2008
11. Jun 13, 2008

### cmos

Based on your previous posts, yes you are.

This is true only if the charged is moving; this is the old Lorentz force.

This is not true; see the above posts.

12. Jun 14, 2008

### Feldoh

That doesn't make much sense...

13. Jun 14, 2008

### HallsofIvy

First you will have to say what you mean by "velocity" or even "motion" of an electric or magnetic field! Fields in general do not "move". The strength of the field will vary at given points with time as the particles creating the field move. I would not call that "motion of the field" but if that is what you mean, then changes in an electric or magnetic field propogate with the speed of light.

(Nerd's statement "Magnetic and electric fields are waves" is incorrect. Light, radio waves, etc. are waves in the fields themselves. That may be what he was thinking.)

14. Jun 14, 2008

### Staff: Mentor

Hi granpa,

That is an odd and very non-standard concept, I would strongly recommend against the use of any standard term, like "velocity", to name it. Perhaps something like "force production velocity" or whatever similar phrase you feel encapsulates the essence of your concept.

In any case, I think the force on your charge is 0 in this set up. The Lorentz force is:
f = q (E + v x B)
Since E = 0 and v = 0 you get f = 0. Note that the Lorentz force does not depend on either temporal or spatial derivatives of the field.

So it appears that your "force production velocity" is also 0.

In general, I agree with HallsOfIvy that fields don't have velocity or "move". Changes in the field propagate at c, but you already expressly stated that is not what you are intersted in.

Last edited: Jun 14, 2008
15. Jun 14, 2008

### granpa

we can call it anything you like but it obviously isnt zero since this is how transformers work. its simple inductance.

16. Jun 14, 2008

### Staff: Mentor

Well, if it is simple inductance then why not call it "inductance". I really don't see the connection to velocity.

17. Jun 14, 2008

### granpa

i dont want to argue. i just need a formula.

18. Jun 14, 2008

### Staff: Mentor

It is your idea, I think you are going to have to derive it yourself. I already posted the only formula that I know which seems applicable, but it doesn't give an answer you want.

19. Jun 14, 2008

### cmos

What DaleSpam said is exactly correct. A transformer works as a results of Ampere-Faraday induction. The AC current in one solenoid induces an AC magnetic field (Ampere's law) which in turn induces an AC electric field in the second solenoid (Faraday's law). That AC electric field is what gives you current in the second solenoid (Ohm's law).

Granpa, from your completely convoluted posts, it seems that you may be asking what is the velocity of the electrons in that second solenoid. Well, assuming you have made it as far as solving Ohm's law (vectorial form, not scalar), then you should be able to relate the current to an average drift velocity.

20. Jun 14, 2008

### granpa

i need a formula for the force upon a stationary charge due to the magnetic field from a wire given the current as a function of time and the distance from the wire.