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granpa

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- #1

granpa

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- #2

cmos

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[tex]1/\sqrt{\epsilon_0\mu_0}[/tex],

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

- #3

armis

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cmos is right. You just need to get the wave equation.

Write down the Maxswell's equations in vacuum:

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

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

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

[tex]\nabla\cdot\vec{D}=0[/tex] 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 [tex]\nabla[/tex] vectorically. You'll get

[tex]\nabla\times(\nabla\times\vec{E})=-\nabla\times(\partial{\vec{B}}/\partial{t})[/tex]

Which is the same as

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

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

Thus you sould get

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

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

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

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

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

In general the speed of the electromagnetic wave is

[tex]v=1/\sqrt{\epsilon\epsilon_0\mu\mu_0}[/tex]

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

Write down the Maxswell's equations in vacuum:

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

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

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

[tex]\nabla\cdot\vec{D}=0[/tex] 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 [tex]\nabla[/tex] vectorically. You'll get

[tex]\nabla\times(\nabla\times\vec{E})=-\nabla\times(\partial{\vec{B}}/\partial{t})[/tex]

Which is the same as

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

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

Thus you sould get

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

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

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

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

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

In general the speed of the electromagnetic wave is

[tex]v=1/\sqrt{\epsilon\epsilon_0\mu\mu_0}[/tex]

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

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- #4

granpa

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I`m.not.talking.about.waves

- #5

Nerd

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Magnetic and electric fields are waves.

- #6

armis

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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

cmos

- 367

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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

armis

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

- #9

granpa

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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 don't know how to calculate it.

- #10

armis

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the velocity of the field at point p is the velocity that that charge would have to have

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!

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- #11

cmos

- 367

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

Based on your previous posts, yes you are.

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.

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

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.

This is not true; see the above posts.

- #12

Feldoh

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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.

That doesn't make much sense...

- #13

HallsofIvy

Science Advisor

Homework Helper

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I`m.not.talking.about.waves

First you will have to say what you

(Nerd's statement "Magnetic and electric fields are waves" is incorrect. Light, radio waves, etc. are waves

- #14

Dale

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Hi granpa,

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 don't know how to calculate it.

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.

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- #15

granpa

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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.

we can call it anything you like but it obviously isn't zero since this is how transformers work. its simple inductance.

- #16

Dale

Mentor

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- #17

granpa

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i don't want to argue. i just need a formula.

- #18

Dale

Mentor

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- #19

cmos

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we can call it anything you like but it obviously isn't zero since this is how transformers work. its simple inductance.

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

granpa

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- #21

Dale

Mentor

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- #22

cmos

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You would need some pretty large currents though...

- #23

granpa

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the time varying E-field is.what.i.need

- #24

Dale

Mentor

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See "[URL [Broken] equations[/URL].

Last edited by a moderator:

- #25

- #26

Dale

Mentor

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f = q (E + v x B)

The reason is that the cross product has no unique inverse. In other words, given a fixed force, charge, E, and B, there are many v which will satisfy the above equation.

- #27

Dale

Mentor

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This is an error. The E field is not equal to zero, so the force is non-zero.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.

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