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Faraday's magnetic induction |
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| Jan24-10, 10:43 PM | #1 |
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Faraday's magnetic induction
Hi everyone...
While going through some texts.. At one place i found that in faraday's law of magnetic induction, >a time varying magnetic field induces an electric field.. And, >a spatially varying electric field induces a magnetic field.. Which one of these is correct? I know of former, but by any chance, second one can also be correct?? |
| Jan24-10, 10:58 PM | #2 |
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A time varying magnetic field results in a spatially varying electric field and visa versa.
Now, as long as the electric field varies spatially, then it must be nonzero somewhere, therefore we obtain your first statement. Thinking that one field causes the other field is incorrect. That is, there is no cause and effect pairing. The word 'induce' is basically a grammatical error with historical roots. There is one field, called the electromagnetic field tensor, that has various components. These components are the electric and magnetic field strengths. |
| Jan24-10, 11:14 PM | #3 |
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Ok..
Suppose i have an electric and a magnetic field (constant).. [faraday law also shows that a spatially non varying E leads to a temporally non varying B].. If i vary B in time.. This will lead to a spatially varying E.. So- will this E add to new originally existing E field... Or, the it is the originally existing field that will start varying... ? |
| Jan26-10, 03:22 AM | #4 |
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Faraday's magnetic induction |
| Jan26-10, 03:34 AM | #5 |
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Am i correct? |
| Jan26-10, 06:04 AM | #6 |
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That would depend but on the whole I would say no. First, the orientation of the magnetic field that is varying will dictate the orientation of the varying electric field. This electric field could be oriented in such a way that it does not impact the flow of current in your wire.
Second, since we now have time-varying fields, we can only induce time-varying currents. The resulting fields are electromagnetic waves, and they will induce currents on the surface of the wire (assuming a perfect conductor). These currents will simply add to whatever current is already impressed upon the wire. If you can somehow induce a wave that is of the same frequency and will induce currents that are perfectly out of phase with the applied currents, then you could cancel out the applied currents. However, this really isn't feasible, and so you can filter out whatever induced signals are on the wire and regain your original signal should you wish. |
| Jan26-10, 06:20 AM | #7 |
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ok.. i get sm idea..
thx.. this may be a little off the topic, but do you know if a plasma, in stable state, has any net electric field? |
| Jan26-10, 08:29 AM | #8 |
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| Jan26-10, 08:32 AM | #9 |
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ok.. thanx for the help..
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| Jan26-10, 10:12 AM | #10 |
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| Jan26-10, 10:47 AM | #11 |
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My point is that any statement like this should make sense to a very educated person who understands the tensor/relativity aspects of field theory. But, it should also make sense to an undergraduate student in a first semester of EM field theory. That statement will confuse the heck out of the uninitiated student. |
| Jan27-10, 08:51 AM | #12 |
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| Jan27-10, 10:13 PM | #13 |
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It turns out to be non-zero curl. Grad x E + dB/dt = 0, in natural units. This is the Maxwell-Faraday equation. So where ever you can factor-out a nonzero curl(E) from the field, there will be an associated changing B field. The corollary is that wherever one finds a time changing B field, there is a circuital electric field. The electric field doesn't terminate on charge--or, at least not all of it, but wraps back on itself. |
| Jan28-10, 09:32 AM | #14 |
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I think the earlier part of this is getting close to some questions I've had. In Feynman's lectures - vol. 2 either chap 19, 21 or 23 - I think 23 (I'm at work and don't have the lectures here), Feynman shows in a capacitor that a changing E-Field induces a B-field, then he shows that the B-field induces a new E-Field (he calculates the E-field) and then adds the 2 E-fields together. Well he says that "new" E-field creates a new B-field - he calculates that new B and adds the two B's together and so on. He does 3 or so to see the pattern. He ends up showing the E-field and B-field is this complicated expression involving a Bessel function. I had under-grad electromagnetics - junior and senior level and this was never mentioned. This Feynman lecture and results does not seem obvious to me - Is this thread getting close to this topic of Feynman's lecture? Can you elaborate further on this? Why was it not mentioned in electrodynamics (under-grad level at least)? It seems so interestesting but I never would have known to go there, I would have stopped at the first calculation. Thanks -Sparky |
| Jan30-10, 10:56 AM | #15 |
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| Jan30-10, 01:14 PM | #16 |
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| Jan30-10, 09:04 PM | #17 |
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But since, the magnetic field in an EM wave is not the same as that due to a current carrying wire, I can't say the same thing for this field.. Right?? |
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