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New answers tagged electromagnetic-induction

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Your proposal is a type of perpetual motion device which is popularly referred to as a "magnetic motor" (NOT an "electric motor", which is a real thing). Every one of these magnetic motors is impossible because they all presuppose that it is in some way possible to have one permanent magnet "sneak up" on another and then suddenly experience repulsion. The ...

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If an alternating voltage is applied at one end of your capacitor, then an alternating current will flow in (and out) of the central conductor at that end, with an opposite current in the outer conductor. The current will drop off to zero at the far end. An alternating magnetic field will be wrapped around the current inside the conductors and in the the ...

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If you had the hypothetical situation of a constant current travelling around a loop of wire, then any changing magnetic flux through the loop is calculated as $-\oint \vec{E}\cdot d\vec{l}$. For an ideal conductor then $\vec{J} = \sigma \vec{E}$ and the current and electric field are of constant magnitude and in the same direct, which is also the direction ...

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I may be missing your point, but current carrying loops produce varying magnetic fields in all kinds of situations: AC motors, transformers, inductors, demagnifiers, and many others. In an electromagnetic wave, the interaction of varying electric and magnetic fields determines the rate of propagation of the wave.

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Your questions seems to ask whether there is an electromagnetic force between the Sun and the Earth which has an appreciable impact on our orbit. The Earth's magnetic field is very weak locally, barely enough to align a compass point.The Sun's is about twice as strong, on average, but has intense variations associated with sunspots etc. The affect of the ...

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It should be noted that in a real transformer working in frequencies it is designed for, the second description is correct. I have played with transformers for a few years. I have winded my own one power transformer and 4 audio transformers. From what I have learned and from my experiments, the second description is correct. There is a fact that may or may ...

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Your problem statement is not entirely clear, but if the rod is perpendicular to the field, and is moving sideways across the field, then each free electron in the rod will be subject to a force (evB) pushing it toward one end of the rod. They will move until the separation of charge produces an electrostatic force opposing the magnetic force (eE = evB). E ...

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If during self-induction the inductor produces a current in opposite direction of that of battery in order to resist the change of current through it then how could the current after some reasonable amount of time gets to its peek point? If you apply a DC voltage to an ideal inductor, it will never reach its peak current. The current will continue to ...

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The whole point is that for a normal conductor Lenz can never win. It is true that at the start there is no current but there is a finite rate of change of current with time and so with the passage of time the current will change from its initial zero value. Another way of looking at the situation is that if Lenz did stop the change of current, and ...

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Here is some guidance. You can't change the current in an ideal inductor instantaneously (in zero time). Therefore, whatever the current is in the inductor just prior to closing the key in part (a) it will be the same the instant after the key is closed. It will then change over time. Likewise, whatever the current in the inductor is just prior to opening ...

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I will only answer the first part of the question since the rest is very much homework-like. When you open the switch, there is an EMF from the inductor because of the changing current, but you simultaneously lose the EMF from the voltage source E. Therefore the total EMF after opening the switch is not greater than before, hence the current does not ...

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I found the answer to my question here In summary, the field winding does not have to provide all the magnetic flux due to the fact that the current-carrying armature, in turn, creates a magnetic field. So the net magnetic flux becomes $B_{net} = B_{rotor} + B_{stator}$ In an imaginary totally resistive stator, the power balances out without any rotor ...

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I'm not sure I understand the nature of your question. Generator efficiency depends on shaft support bearing losses, ohmic losses in all the windings, and flux leakage in the gaps between the pole pieces in the stator and the armature. Efficiency is maximized when these are all minimized.

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But, shouldn't we include this in the flux calculation? You can think about this problem like coupled inductors where the 'voltage across' the 'secondary' (inner loop of wire) is given by $$v_s = k\sqrt{L_1L_2}\frac{di_1}{dt} + L_2\frac{di_2}{dt}$$ where $L_1$ and $L_2$ are the self-inductance of the solenoid and inner loop respectively, and $0\lt k \lt 1$...

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Why is it the flux change which induces the ac current of primary coil equal to the flux change that primary coil induces on the secondary coil? Because you're only studying or modeling an ideal transformer. In a real transformer, flux through the secondary is somewhat less than the flux through the primary. Some of the flux produced by the current through ...

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I've done some research and spent more time thinking about this. Provide feedback if any of this is inaccurate. Basically, magnets get their magnetism from the magnetic fields of all of the atoms lined up together. The magnetic field is generated by the motion of the electrons in the atom itself and with a sufficiently amount of atoms (e.g. quadrillions) ...

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