alpha2cen Posted December 27, 2010 Share Posted December 27, 2010 When we flow electrons through the copper wire, the magnetic field is generated. But I' m curiously about the condition which electromagnetic waves leave the copper wire. The wave is called radio wave. How the electromagnetic wave to leave the copper wire? Link to comment Share on other sites More sharing options...
Nexium Tao Posted December 27, 2010 Share Posted December 27, 2010 (edited) Sorry Alpha, your question doesn't really make sense, at least not in the way you have asked it. All electrons emit a magnetic field perpendicular to it's movement, but usually electrons group up in domains(A group of electrons all moving in the same direction), all domains flow in different directions, counteracting other domains. This creates a net magnetic field of 0. When a current passes through a metal, all the electrons are more or less flowing in the same direction so all the domains of electrons in the metal become aligned, moving in the same direction instead of canceling one another out their magnetic fields join together. This is what I am assuming you meant by magnetic field through copper wire. But after that, what you are asking is how the electrons leave the wire and are calling it an electromagnetic wave(which it isn't), are you talking about when the net voltage becomes zero or are you talking about electrons that while traveling from point 1 to point B in the wire, break loose and leave the wire? Edited December 27, 2010 by Nexium Tao Link to comment Share on other sites More sharing options...
swansont Posted December 27, 2010 Share Posted December 27, 2010 When the electrons accelerate, which happens if the current changes value. That gives a changing magnetic field, which induces an electric field. Link to comment Share on other sites More sharing options...
Nexium Tao Posted December 27, 2010 Share Posted December 27, 2010 (edited) When the electrons accelerate, which happens if the current changes value. That gives a changing magnetic field, which induces an electric field. Is that what he was asking, how an electromagnetic field is produced through a conductor? Edited December 27, 2010 by Nexium Tao Link to comment Share on other sites More sharing options...
swansont Posted December 27, 2010 Share Posted December 27, 2010 Sorry Alpha, your question doesn't really make sense, at least not in the way you have asked it. All electrons emit a magnetic field perpendicular to it's movement, but usually electrons group up in domains(A group of electrons all moving in the same direction), all domains flow in different directions, counteracting other domains. This creates a net magnetic field of 0. When a current passes through a metal, all the electrons are more or less flowing in the same direction so all the domains of electrons in the metal become aligned, moving in the same direction instead of canceling one another out their magnetic fields join together. This is what I am assuming you meant by magnetic field through copper wire. But after that, what you are asking is how the electrons leave the wire and are calling it an electromagnetic wave(which it isn't), are you talking about when the net voltage becomes zero or are you talking about electrons that while traveling from point 1 to point B in the wire, break loose and leave the wire? Domains refer to a ferromagnetic material, and the magnetic field is due to the electron spin. That's a different situation from current flow. Oscillating current flow radiates EM waves; it's not the electrons leaving the wire. 1 Link to comment Share on other sites More sharing options...
alpha2cen Posted December 28, 2010 Author Share Posted December 28, 2010 (edited) Oscillating current flow radiates EM waves; it's not the electrons leaving the wire. Then, what is different from visible light emission and heat emission? They are all electromagnetic waves. Edited December 28, 2010 by alpha2cen Link to comment Share on other sites More sharing options...
swansont Posted December 28, 2010 Share Posted December 28, 2010 Heat emission can be conduction or convection, too. Radiation from a thermal source can be in the visible part of the spectrum. But for radiation that's outside the visible part, the difference is the wavelength and frequency. IR, for example, is a longer wavelength and lower frequency. Microwaves are longer still. Link to comment Share on other sites More sharing options...
alpha2cen Posted December 29, 2010 Author Share Posted December 29, 2010 Temperature dependent radiation follows the Planck radiation theory. High temperature electric wire emits very short wave length electromagnetic wave, whereas low temperature electric wire emits lower one. But molecular behavior is some different, short visible or UV waves are related to the electron behavior, external orbit electrons falling into inside orbit. IR waves are related to molecular motion, vibration, rotation, ... So UV or visible waves are very atomic dependent. Link to comment Share on other sites More sharing options...
mississippichem Posted December 29, 2010 Share Posted December 29, 2010 Temperature dependent radiation follows the Planck radiation theory. High temperature electric wire emits very short wave length electromagnetic wave, whereas low temperature electric wire emits lower one. But molecular behavior is some different, short visible or UV waves are related to the electron behavior, external orbit electrons falling into inside orbit. IR waves are related to molecular motion, vibration, rotation, ... So UV or visible waves are very atomic dependent. That's just because the energy release associated with those events correlates to that frequency of light because [math] E=\frac{hc}{\lambda} [/math]; where lambda is frequency, E is energy, h is Planck's constant, and c is the speed of light. So for example, the energy of molecular vibrations is of the magnitude to give off radiation in the IR region of the spectrum. Electron orbital transitions are more energetic and usually correspond to the visible/UV part of the spectrum. It all has to do with how energetic the event was that created the radiation. There's really not much difference between a radio wave and a gamma ray. Gamma rays are just way more energetic and therefore have a much shorter wavelength which translates to higher frequency. Link to comment Share on other sites More sharing options...
alpha2cen Posted January 1, 2011 Author Share Posted January 1, 2011 (edited) That's just because the energy release associated with those events correlates to that frequency of light because [math] E=\frac{hc}{\lambda} [/math]; where lambda is frequency, E is energy, h is Planck's constant, and c is the speed of light. So for example, the energy of molecular vibrations is of the magnitude to give off radiation in the IR region of the spectrum. Electron orbital transitions are more energetic and usually correspond to the visible/UV part of the spectrum. It all has to do with how energetic the event was that created the radiation. There's really not much difference between a radio wave and a gamma ray. Gamma rays are just way more energetic and therefore have a much shorter wavelength which translates to higher frequency. At the electric wire radio wave emitting condition is different from visible wave emitting condition. We can make visible light by using tungsten filament. The temperature condition is very high. Visible light generating procedure is like this. The electron current through the filament makes a high temperature, and at the hight temperature light is naturally emitted by electron orbital movement in the atom. Then, How the radio wave is emitted on the surface of the copper wire? Edited January 1, 2011 by alpha2cen Link to comment Share on other sites More sharing options...
alpha2cen Posted January 1, 2011 Author Share Posted January 1, 2011 (edited) Sorry computer problem. Edited January 1, 2011 by alpha2cen Link to comment Share on other sites More sharing options...
alpha2cen Posted January 1, 2011 Author Share Posted January 1, 2011 (edited) Sorry, computer problem. Something wrong, during the saving step, strange thing happen. Edited January 1, 2011 by alpha2cen Link to comment Share on other sites More sharing options...
timetes Posted January 1, 2011 Share Posted January 1, 2011 At the electric wire radio wave emitting condition is different from visible wave emitting condition. We can make visible light by using tungsten filament. The temperature condition is very high. Visible light generating procedure is like this. The electron current through the filament makes a high temperature, and at the hight temperature light is naturally emitted by electron orbital movement in the atom. Then, How the radio wave is emitted on the surface of the copper wire? i understand what your asking....id like to know too.... Link to comment Share on other sites More sharing options...
swansont Posted January 1, 2011 Share Posted January 1, 2011 Blackbody radiation is not due to electron transitions; it's a continuum. It's emitted due to charged particle acceleration, just as in a radio wave. But a radio wave has a specific driving frequency, and all of the electrons have the same acceleration, so they all emit at the same frequency. Link to comment Share on other sites More sharing options...
IM Egdall Posted January 1, 2011 Share Posted January 1, 2011 Blackbody radiation is not due to electron transitions; it's a continuum. It's emitted due to charged particle acceleration, just as in a radio wave. But a radio wave has a specific driving frequency, and all of the electrons have the same acceleration, so they all emit at the same frequency. And for blackbody radiation, the charged particles which are undergoing acceleration are the molecules and atoms of the object doing the radiating. Even though each molecule and atom is electrically neutral overall (number of electrons matches number of protons). It is primarily the acceleration of the protons inside the nucleii which produces this radiation. And this acceleration is the vibration of these molecules and atoms. We measure the average overall vibration as the temperature of the object. And that is why the blackbody frequency spectrum we see is determined soley by the temperature of the object radiating. Do I have this right? Link to comment Share on other sites More sharing options...
swansont Posted January 1, 2011 Share Posted January 1, 2011 Pretty much; I don't know if you can trace the source to the protons for the atoms. In a conductor, the electrons are free to undergo collisions, and these are usually the best approximations of a blackbody. Link to comment Share on other sites More sharing options...
timetes Posted June 20, 2011 Share Posted June 20, 2011 since the radio wave can not be absorbed is this harmful to humans........could go thru walls and floors...... Link to comment Share on other sites More sharing options...
swansont Posted June 20, 2011 Share Posted June 20, 2011 since the radio wave can not be absorbed is this harmful to humans........could go thru walls and floors...... Radio waves do get absorbed; it's a matter of degree. It's harmful via heating (e.g. a microwave oven). Link to comment Share on other sites More sharing options...
timetes Posted June 20, 2011 Share Posted June 20, 2011 Radio waves do get absorbed; it's a matter of degree. It's harmful via heating (e.g. a microwave oven). which is what alpa is asking about.....the radio wave leaving a wire causing an electrical current.........am i correct? Link to comment Share on other sites More sharing options...
swansont Posted June 20, 2011 Share Posted June 20, 2011 which is what alpa is asking about.....the radio wave leaving a wire causing an electrical current.........am i correct? I read it as a question about the source of the EM waves, not what happens after. Link to comment Share on other sites More sharing options...
timetes Posted June 20, 2011 Share Posted June 20, 2011 I read it as a question about the source of the EM waves, not what happens after. sorry your right...the condition.....id like to know both LOL Link to comment Share on other sites More sharing options...
swansont Posted June 20, 2011 Share Posted June 20, 2011 As I said before, absorption in the low-frequency part of the spectrum results in heating. At higher energy it results in ionization. Link to comment Share on other sites More sharing options...
Wilmot McCutchen Posted August 4, 2011 Share Posted August 4, 2011 Maybe the question is about near field. Link to comment Share on other sites More sharing options...
Wilmot McCutchen Posted August 7, 2011 Share Posted August 7, 2011 The difference between the near field and the far field is explained in this page on Wikipedia: http://en.wikipedia.org/wiki/Near_and_far_field Link to comment Share on other sites More sharing options...
DrRocket Posted August 8, 2011 Share Posted August 8, 2011 Then, How the radio wave is emitted on the surface of the copper wire? In classical electrodynamics, there is one single electromagnetic field that exists throughout all of space. It can be viewed as a sum (superposition) of the fields associated with each of the charged particles, including their motion, that exist. When electrons move in a copper wire, the position and velocity of those charged electrons changes, and with that change comes a change in the electromagnetic field associated with them. That local change propagates according to Maxwell's equations, and that is what is meant by "the emission of a radio wave". Link to comment Share on other sites More sharing options...
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