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Energy transfer in electromagnetic induction |
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| Jan7-13, 03:04 PM | #69 |
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Energy transfer in electromagnetic induction
Me too. Think about the field generated by the induced current in the first turn. How does that affect the total field seen by the second turn.
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| Jan8-13, 02:56 AM | #70 |
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The kinetic energy of the electromagnet cannot contribute to the induced current, as it does not decrease initailly. Lets consider the time interval between x/c and when the coil begins to feel resistance due to mutaul induction. It will be less than 2x/c since the electromagnet is moving towards the coil. The energy due to the current in the coil during this time cannot be greater than the energy in the electromagnetic wave intially radiated. But if we increase the value of v, the energy in the coil increases because of a larger change in flux. But there is no noticeable increase in the radiation energy. (v<<c) If we consider the frame of the moving electromagnet there is no Doppler effect. All the electromagnet sees is the coil moving towards it. Here again we can see that the energy in the coil (between x/c and 2x/c) increases with increase in the relative velocity. |
| Jan8-13, 06:50 AM | #71 |
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What is important, is that in the frame of the coil, the electromagnet is moving towards it. When a source of isotropic radiation moves in some direction, the radiation is released preferentially to that direction. Check https://en.wikipedia.org/wiki/Synchrotron_radiation The bunches of charged particles circling in synchrotron move so fast that the radiation is needle-like, similar to laser, only much brighter and not monochromatic. With the electromagnet, it is similar; even if it moves slowly, there is more radiation going to the coil than in the other directions. As the velocity is increased, coil receives greater and greater power. However, there is a limit: when v approaches c, the coil receives almost all the radiated power 1 J/s and this is the maximum. Of course, as the processes in the source are slowed down (dilatation) , it will receive it for a long time and thus the net amount of energy received in the end can be much greater than 1 J. Where did the extra energy came from? From the total energy of the electromagnet; as the net energy of the electromagnet decreases by radiation, in the frame of the coil the electromagnet loses also momentum via loss of its mas (the velocity is unaffected). |
| Jan8-13, 07:08 AM | #72 |
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| Jan8-13, 08:05 AM | #73 |
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Can we not increase the energy transferred initially by increasing the number of turns in the coil. More turns mean more current flowing in the coil. |
| Jan8-13, 10:17 AM | #74 |
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This goes back to the point before. I think DaleSpam's answer was: "Think about the field generated by the induced current in the first turn. How does that affect the total field seen by the second turn. My hint is that the equation:
[tex]\displaystyle{\varepsilon}=-N \frac{d \Phi}{dt} [/tex] Uses a lot of assumptions, and if you break those assumptions, you cannot expect the equation to give correct results. Edit: actually, it doesn't use a lot of assumptions, but the simple case of increasing number of turns to increase the current through the coil does introduce assumptions. Another Edit: and generally, it is assumptions used along with this equation that have caused the problems in this thread. For example, the assumption "that the magnetic field at the magnet and at the coil is approximately the same" is often used with this equation, but this assumption becomes false when the magnet and coil are far away from each other. |
| Jan8-13, 10:54 AM | #75 |
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| Jan8-13, 11:09 AM | #76 |
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| Jan8-13, 01:05 PM | #77 |
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![]() If each dot represents the energy extracted by a coil with i turns then it is both true that "there is a limit to the energy that can be transferred" and "we can increase the energy transferred by increasing the number of turns". |
| Jan9-13, 11:56 AM | #78 |
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Is this because the successive turns in the coil are linked to lesser magnetic flux? |
| Jan9-13, 12:39 PM | #79 |
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Yes, it is because each turn reduces the flux seen by the other turns. This can be seen through the superposition principle.
Suppose that you have two turns, consider them to be two separate loops. There is a current in loop A which creates a field which opposes the change in the external field. By superposition the field seen by loop B is the sum of the field from loop A and the external field, which is less than the change in the external field. So the current induced in B is a function of the external field and the current induced in A where current in A reduces the induced current in B. Then, to consider the loops as separate turns in a single coil simply equate the current in A to the current in B. |
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