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4)Let me xplain my plan of constructing a perpetual motion machine according to my plan although there are some problems..Take 4 magnets and a loop of coil(what ever it's shape may be).2 of the magnets are to be arranged on one side of the loop and other 2 on the other side of the loop.Let us consider one side of the loop(Say left).On the left side the magnets are to be arranged such that they both are horizontal and repel each other.Let us call the magnet nearer to the loop as magnet1 and the other magnet be magnet2.Let us suppose that south pole of the magnet1 is facing towards the coil.Now arrange the other 2 magnets on the right in such a way that the new magnet that is adjacent to the coil has it's south pole facing the coil and let it be magnet3.The other magnet that is magnet4 is placed such that it repels magnet3.Now after arranging in such a way slightly pull any one of the magnets that are away from the coil(say magnet2) and leave it.It repels the magnet adjacent to the coil and the magnet1 goes into the coil and there is an induced current in that coil.Now the magnet3 repels this magnet1 and magnet1 now goes back and there is again an induced current but in opposite direction.Now this repeats on the other side of the loop and the same repeats.In this way an AC is setup.In this way the perpetual motion machine can be prepared..Although the magnets slightly stop in the middle due to the gravity effect and air resistance etwe can construct a machine like this by just spending a little energy..Isn't it.
- Prudhvi Raj Borra (age 17)
Instead of me telling you it won't work because of the first and second laws of thermodynamics, why don't you think about it yourself and try to figure out the possible sources of energy loss. You should consider friction, radiation losses, ohmic losses in wires and induced currents, possible heating mechanisms, etc, etc. Also think about how you could possibly extract energy from this system without slowing it down. Let us know the result of your thoughts.
(published on 05/06/2010)
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