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If I move a magnet towards the coil then the magnet will experience a restive force because of Lenz's law. This way mechanical energy will convert into electrical energy.

But what if the magnet and the coil are a sufficiently large distance apart? The change in the magnetic field travels at the speed of light so the magnet will not feel any restive force instantly. Doesn't this violate the energy conservation law? Because we fix the magnet to a particular point before the restive force arrives at the magnet. And we will able to produce arbitrarily large energy in the coil.

If the magnet starts to move at time t=t1 and lets assume that t0 is the time taken for the flux linking the coil to change due to motion of the magnet.

t0 = (distance between the magnet and the coil/velocity of light)

the magnet will begin to experience resistance at time (t1+2t0). so the delay will be 2t0. In this time period (2t0), magnet is free without experiencing any resisting force. During this time period we can draw current arbitrarily large value by making arbitrarily large numbers of turns of the coil. If energy is conserved, then what mechanism stops us to draw any large amount of power from the coil ?

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    $\begingroup$ I mean there is a really easy answer to this, which is that the Maxwell Equations admit of a least-action principle which is time-independent, therefore by Emmy Noether's theorem, energy is conserved: therefore nothing you can do with classical electromagnetism can possibly violate energy conservation. However that does not tell you a microscopic reason why you are wrong, just a macroscopic reason why you cannot be right. $\endgroup$
    – CR Drost
    Commented Jul 17, 2017 at 17:25
  • $\begingroup$ @CRDrost That unedited comment would be a good answer. $\endgroup$
    – Selene Routley
    Commented Jul 17, 2017 at 22:06
  • $\begingroup$ > During this time period we can draw current arbitrarily large value by making arbitrarily large numbers of turns of the coil. This is not true, increasing number of turns increases both emf and ohmic resistance, so the current stays roughly the same. Current density in wire is proportional to induced rotational electric field, which is finite and determined by how fast the magnetic field changes. $\endgroup$
    – Ján Lalinský
    Commented Oct 18, 2017 at 22:18
  • $\begingroup$ @Ján Lalinský....current won't be same....suppose first time you use iron coil and in second time you use gold coil with more numbers of turns, since gold coil will have lesser resistance, current won't be same for both cases. $\endgroup$
    – Hemal Pansuriya
    Commented Oct 19, 2017 at 16:16
  • $\begingroup$ If you increase the number of turns and decrease the resistance, it is true you will achieve higher current, eventually. However, current in coil does not simply jump from 0 to some high value, it takes time to build up such current. With lower resistance it will take longer time to achieve same fraction of the maximum because of the counteracting effect of self-induction. For very small resistances, the time needed to achieve some fixed value of current is roughly the same, it does not depend on the coil resistance. The longer the process goes on, the more energy is the coil able to extract. $\endgroup$
    – Ján Lalinský
    Commented Oct 19, 2017 at 22:18

3 Answers 3

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The full problem involves an electromagnetic wave calculation. As you accelerate the magnet, the changing magnetic field launches an electromagnetic field and the magnet will indeed feel a force - the radiation resistance, very similar to the force opposing an accelerated charge. Likewise, as the wave passes through the coil, any field acting on the magnet propagates back as an electromagnetic wave.

The magnetostatic calculation of this force is valid only for small separations, as you understand. In the small-separation limit, one can show that the electromagnetic waves travel back and forth very swiftly to set up the magnetostatic conditions that allow the approximate calculation.

Any "time delayed" work through the effects that you note is accounted for as energy of propagating electromagnetic waves in transit. Energy conservation is upheld once one takes full account of the energy of the electromagnetic field.

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  • $\begingroup$ We can create current in the coil arbitrary large value when the magnetic field change reaches to the coil. If the magnet is feeling resistance only due to that radiation only it won't be related to that resistance which magnet would have felt if the interaction were instantaneous. The radiation resistance will be a fix amount . But current in the coil can be produced in arbitrary large amount. $\endgroup$
    – Hemal Pansuriya
    Commented Jul 17, 2017 at 13:46
  • $\begingroup$ @HemalPansuriya No, the radiation resistance will depend on the motion - see the Larmor formula, for example. The energy of the field thus accounts for all the work in transit. $\endgroup$
    – Selene Routley
    Commented Jul 17, 2017 at 22:04
  • $\begingroup$ @HemalPansuriya ... and see CRDrost's comment for the reason why the energy of the radiation must account for all the work in transit. $\endgroup$
    – Selene Routley
    Commented Jul 17, 2017 at 22:16
  • $\begingroup$ yes, you are right that radiation resistance will depend on the motion. But for given amount of motion (change in magnetic field ) the radiation resistance will be a fix amount.But we can create current in the coil arbitrary large value when the magnetic field change reaches to the coil. $\endgroup$
    – Hemal Pansuriya
    Commented Jul 18, 2017 at 9:20
  • $\begingroup$ @HemalPansuriya No, it's simply not possible to create an arbitrary large energy in the coil for a fixed energy input to the magnet - as in Chris's comment, Maxwell's equations must fulfill energy conservation because they have a time-invariant Lagrangian formulation, which guarantees an invariant Noether charge - i.e. constant total energy in the case of time shift Lagrangian invariance. So this is the definitive answer. My radiation resistance argument simply gives you intuition for how energy conservation is enforced, as it must be through the Noetherian argument. $\endgroup$
    – Selene Routley
    Commented Jul 18, 2017 at 11:06
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If the magnet starts to move at time t=t1 and lets assume that t0 is the time taken for the flux linking the coil to change due to motion of the magnet.

t0 = (distance between the magnet and the coil/velocity of light)

the magnet will begin to experience resistance at time (t1+2t0). so the delay will be 2t0. In this time period (2t0), magnet is free without experiencing any resisting force. During this time period we can draw current arbitrarily large value by making arbitrarily large numbers of turns of the coil. If energy is conserved, then what mechanism stops us to draw any large amount of power from the coil ?

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During this time period we can draw current arbitrarily large value by making arbitrarily large numbers of turns of the coil.

This is not a correct statement for the simple reason that the inductance $L$ of the coil goes as the square of the number of turns $N$ and thus, the time constant $\tau \equiv \frac{L}{R}$ increases with $N^2$.

Let $\Phi_{ext}$ be the externally generated (by a moving magnet or whatever) magnetic flux through the coil. Assuming a resistive load of resistance $R$ is connected across the coil, the differential equation for the system is

$$L\frac{di}{dt} + Ri = N\frac{d\Phi_{ext}}{dt}$$

where $L = \mu_0\frac{A}{l}N^2$ and $A, l$ are the area and length of the coil. Rewrite the equation as

$$\frac{di}{dt} + \frac{R}{L}Ri = \frac{N}{L}\frac{d\Phi_{ext}}{dt}$$

For simplicity, assume that the external flux is a ramp from zero over time $T$ so that

$$\frac{d\Phi_{ext}}{dt} = V_0\left[u(t) - u(t - T) \right]$$

where $u(t)$ is the unit step function. It follows that

$$i(t) = \frac{NV_0}{R}\left[(1 - e^{-t/\tau})u(t) - (1 - e^{-(t-T)/\tau})u(t - T) \right]$$

where $\tau = \frac{L}{R} = \frac{\mu_0A}{l R}N^2$.

Now, For very large $N$, the current is approximately

$$i(t) = \frac{V_0}{N}\frac{l}{\mu_0A}\left[tu(t) - (t-T)u(t - T) \right]$$

and so

$$i_{max} = \frac{V_0}{N}\frac{l}{\mu_0A}T$$

which goes to zero as $N \rightarrow \infty$. Thus, it is not the case that "we can draw current arbitrarily large value by making arbitrarily large numbers of turns of the coil"

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  • $\begingroup$ Self inductance is like inertia i.e it will take time for current to go to higher value. And my point is, our input energy requirement is not increasing if increase output from coil. $\endgroup$
    – Hemal Pansuriya
    Commented Oct 21, 2017 at 5:39
  • $\begingroup$ Also please consider this also,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) . $\endgroup$
    – Hemal Pansuriya
    Commented Oct 21, 2017 at 5:42
  • $\begingroup$ ....no, please tell me where i am wrong in transformer calculation. $\endgroup$
    – Hemal Pansuriya
    Commented Oct 21, 2017 at 15:53
  • $\begingroup$ " please tell me where i am wrong in transformer calculation" - @HemalPansuriya, it is your naive stipulation to use the coil is a voltage source. At some point, you must come to grips with the elementary fact that the coil is not a voltage source (and not even approximately so). $\endgroup$
    – Alfred Centauri
    Commented Oct 22, 2017 at 1:00
  • $\begingroup$ .....EMF is produced in the coil. Why can't we connect it to a transformer ? $\endgroup$
    – Hemal Pansuriya
    Commented Oct 22, 2017 at 3:34

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