Hemal Pansuriya

No, we don't have to. Because, magnet is not going to feel resistance during delay period.

I know transformers don't Violate energy conservation law...power in primary and secondary will be the same.....but by using transformer you can draw more power by the following way : if voltage induced in coil is varying with time than use coil as voltage source and connect it to transformer's primary coil suppose voltage produced in coil is V, then step up this voltage by factor of 1000 ( or arbitrarily large ). Suppose resistance connected in secondary is R, so current in secondary winding will be I =1000V/R. And in primary winding current will be drawn 1000*I ( because IpVp=IsVs ). So now the power output of the coil will be V*(1000*I).So now the power output of the coil will be V*(1000*I) . We have drawn 1000 times more power than previously, which was V*I.

If you have looked at it. Please tell me why can't we draw large power from coil with help of transformer.

please just look at what i have written.

Coil can be very large and massive such that coil will move negligible.And if voltage induced in coil is varying with time than we can use transformer to draw more power by following way :use coil as voltage source and connect it to transformer's primary coil suppose voltage produced in coil is V, then step up this voltage by factor of 1000 ( or arbitrarily large ). Suppose resistance connected in secondary is R, so current in secondary winding will be I =1000V/R. And in primary winding current will be drawn 1000*I ( because IpVp=IsVs ). So now the power output of the coil will be V*(1000*I).So now the power output of the coil will be V*(1000*I) . We have drawn 1000 times more power than previously.

If the velocity of magnet is constant, than change in magnetic flux is also constant. To get from 0 current to some value is due to changing magnetic flux, after that if that change in magnetic flux is constant than the EMF will also constant . The coil is not moving in this problem . only magnet is moving.

The magnetic flux is integral of B.da . As magnet moves towards coil with constant velocity, the B increases at the coil and so does the magnetic flux. But due to constant velocity of magnet, d∅/dt in coil is constant .

If the magnet has constant velocity towards coil, than the magnetic flux change in coil is constant ( d∅/dt = constant ). Hence the EMF and current are constant.

If the change in magnetic field is constant , than EMF produced is constant. Than current is also constant ( dI/dt = 0). So, there is no self inductance in this case. The resistance value can be put as low such that we get output energy V*I > input energy .

You can increase the current in the coil other way also, like by decreasing resistance in the circuit assuming fixed EMF during delay period. Since, the power output is = V*I , you can increase power output by decreasing resistance value in circuit.

can u tell me the reason and what have i ignored ?

The coil will see the magnetic field change When signal reaches from magnet to coil. As soon as coil feels magnetic field change it produces current. This current will generate magnetic field that will resist the motion of magnet. But it is not instantaneous, as signal from coil to magnet will take time to travel. During this delay period we can produce arbitrarily large EMF in coil by keeping number of turns of coil arbitrarily large and hence power output of coil arbitrarily large. We will not have to give any extra input energy to magnet during this delay period. You can refer to my PDF , in which i have described the problem more clearly. magnet.pdf

We are not moving COIL at all. We are moving magnet towards coil. So the magnet won't feel any resistance during delay period.

Work is only done when magnet feels any resistance, my friend. Energy in the coil produces due to change in magnetic field but for that change, magnet is not feeling any resistance during delay period.

The initial energy that we are giving to magnet is fixed , it will not change during the process because during the delay period magnet will not going to experience any resistance. So during this delay period whatever amount of current flows in the coil we don't have to put any extra input energy in magnet.