The picture shows the direction of the magnetic field around a current carrying wire. I wonder what will be the direction of the electric field with respect to the direction of the magnetic field ? I am trying to visualize both electric and magnetic field at the same time. I am getting the picture of the direction of the magnetic field everywhere but could not find any picture that shows both of them togather.
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See also: https://physics.stackexchange.com/questions/61884/does-a-current-carrying-wire-produce-electric-field-outside/ – BowlOfRed Jun 12 '18 at 23:12
4 Answers
In a circuit involving potential drop (so, not purely a current wo/voltage,) the e-field around conductors is perpendicular and radial, and the e-field around resistors is radial with some tilt. Inside a resistor (as well as just outside,) the e-field is parallel to the resistor.
For ideal conductors with arbitrarily small resistance, the flux-lines of e-field appear in the space outside the conductor, and are connecting the surface-charge with charges found upon other, distant parts of the circuit (e.g. parallel wires having opposite charge, where the two wires behave as opposite capacitor-plates.) Here's the oversimplified visual version:
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- Simple circuit:
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. Notice that the fields are those of a 2-wire waveguide or transmission line? Exactly right. The same physics applies at DC, Zero Hz, and also applies at 60Hz AC, and also at radio frequencies. (DC is really just an AC square wave, like an hours-long, flat-topped AC pulse which zooms along at c velocity.)
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Thanks alot for your answer. It was really great. Although i am not aware of the poyenting vector but you made many things clear to me. Is this how the electromagnetic waves look like ? they say that in an electromagnetic waves both electric and magnetic fields are perpendicular to each other. and the picture you have shown magnetic and electric field also look perpendicular to eachother. – Alex Jun 13 '18 at 10:52
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as the electromagnetic waves are produced by the acceleration of the charges. – Alex Jun 13 '18 at 10:54
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Yes, DC circuits employ electromagnetic waves. But the "wave" is just one large hump, many hours wide. This is little different than 60Hz power lines, where each hump is 8.33mS wide. The fields during the middle of each hump, they look like the diagram above. Two-wire waveguides function independent of frequency, and can transmit energy at zero Hz or at 50 GHz, or anything between. PS in the above, if the wires are resistive, then the e-field isn't at right angles to the wires. It tilts slightly, as EM energy from space is moving radially inwards to the wires, becoming heat in metal. – wbeaty Jun 13 '18 at 21:29
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Please could someone point me to the theoretical basis and experimental support for the Electric Field Flux Line patterns shown in the 3rd diagram? Oh, its OK - I've just seen the linked answer at https://physics.stackexchange.com/questions/61884/does-a-current-carrying-wire-produce-electric-field-outside?noredirect=1&lq=1. – steveOw Jan 11 '24 at 19:42
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@steveOw note that the 3rd diagram is simply the first two, plotted together. (It assumes ideal zero-ohm circuit, with no voltage-drop or axial e-field.) It's ancient, a part of Maxwell, and fully analyzed before 1900 by both Lodge and Heaviside. Today find it under "transmission line theory," in every electrical engineering, EM, and antenna-theory textbook. So, look for those tutorial websites, plus high-speed PCB design theory, high-freq grounding theory, etc. I first saw it in our physics book in high school. (But it's never mentioned in technician textbooks, only in physics and EE. ) – wbeaty Jan 11 '24 at 21:07
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Many thanks, I have yet to learn about Transmission Line Theory. Actually I counted the "simple circuit" as the 1st diagram, and the one I was referring to is "E-field flux, radial lines connecting surface-charge ". I am now wondering what the Electric field lines would look like for a single, very long, straight conductor carrying a steady DC current between two distant battery terminals. I guess it would look the same as the electric field between the two battery terminals when there is no intervening conductor present? – steveOw Jan 12 '24 at 01:09
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@steveOw using a wire to short out a battery? That's a resistor, getting hot, with energy flowing into the wire. Large voltage across the ends, zero voltage near the center. The e-field lines are radial at the ends of the wire. At the center they're parallel to the wire. They gradually change angle, all along the wire. Textbooks instead may show a very short resistor, with two ideal wires leading to the battery. In that case, the e-field lines at the resistor are parallel to the resistor, so the energy flows radially inwards. PS, oops, I counted wrong, FOURTH diagram, not 3rd – wbeaty Jan 12 '24 at 15:34
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Thanks. Isn't that E-field you describe (for the long thin wire acting as a conductor/resistor) much the same as the dipole E-field that would exist between the two battery terminals in the absence of the wire (ignoring any E-field produced by the charge distribution inside the battery). – steveOw Jan 12 '24 at 17:04
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1@steveOw nope, because surface-charge migrates onto the resistive wire, distorting the e-field. Near the ends of the wire, the e-field is mostly pointing outwards from the wire. Instead if the wire wasn't there, then the e-field in the same location would be pointing parallel to the (missing) wire. So, as we follow along the wire, we see a gradual tilting of the field-lines, until at the center of the wire, they're parallel to the wire. – wbeaty Jan 15 '24 at 07:39
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Thanks very much - I am beginning to understand this in the light of the Sherwood-Chabay text which you have linked for me elsewhere. – steveOw Jan 17 '24 at 00:14
Electric field lines describe the force experienced by positive point charge at a point. In a current carrying wire,the force experienced by a positive point charge is in the direction of current,so electric field is in the direction of current.But outside a current carrying wire there won't be any field lines as there is no electric force .
Electric field was produced inside the wire due to the potential difference created by the battery.But outside the wire there is no such thing.
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Imagine the nature of the electric field if the wires are not next to each other. In fact imagine the wires are going in the opposite direction away from the source for miles and then at the right angle for miles and then again at a right angle back around to the load somewhere and then keeps going for miles and finally turns back at the right angle to the source. I do not know the distance between the wires in the above schematics so the field on one wire knows where to be pointing towards. How far can the field detect the other wire and how quickly it knows that it better point that way. And if it is miles and miles square and not a nice neat rectangle as in the diagram..... IT is a natural inquiry.
I say there is more involved here than we are admitting. There is an electric field along the wires for charges to move. There is an electric field also in all directions and not just between the wires. There are also other influences on the charges so to keep them from accelerating in the direction of the field which is, of course, the direction of force too. The charges also most likely have to deal with the collective magnetic field, after all, the collective field is not an individual electron's magnetic field. They are supposedly moving through the center of the collective magnetic field
The current flows in the direction of the electric field. Since the current is the direction of positive charge flow and the electric field is defined by the direction a positive charge would move, the two are the same thing.
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