I have read that in simple words, Lenz's law states that:

The direction of current induced in a conductor is in such a fashion, that it opposes its cause.

This validates law of conservation of mass-energy.

I arranged the following thought experiment:

Let there be a pendulum with its bob being a small bar magnet. The pendulum is oscillating in a direction parallel to the horizontal axis of the bar magnet on which the North and South poles lie. Also, the pendulum is in complete vacuum. (But gravity is there to make the pendulum oscillate.)

At one of the extreme positions of the pendulum, we keep a solenoid, ends of which are connected to a load resistance.

As the North pole of the bar magnet approaches the solenoid, current is induced in the solenoid in such a fashion that a North pole is formed at the end near to the bar magnet's North pole, and the bar magnet gets repelled towards the other side.

The bar magnet then goes to the other end and then comes back (as a pendulum does) and again the same process is repeated. This should go on forever, and current should keep appearing across the load resistance.

How does the law of conservation of energy hold here?

I've made a small illustration depicting the key idea. If this is in coherence with what you've asked, we could summarize some important points about the case.

Thought experiment http://imageshack.com/a/img69/807/u337.png

Total energy of the Earth and Bar Magnet system is given by the equation: $KE + PE = \frac{1}{2}mv^{2} + \frac{GMm}{R}$

While PE is there for both Earth and magnet system (combined), KE is available for the magnet to spend on oscillation. If, while oscillating, it spends this KE by transforming it in to thermal energy in the resistor-inductor circuit, KE decreases slowly and becomes zero. $KE + PE = 0 + \frac{GMm}{R}$ ie, the oscillation seize to exist.

An easy way to look at it would be to consider change in KE. We will start with the magnetic pendulum:

Component of angular acceleration = $\alpha =\frac{-Lmg\ sin(\theta)cos(\theta)}{I}$

This clearly has its peak values at 45 degree amplitudes. Now, component force due to this restoring torque is give by,

$F =\frac{-L^{2}m^{2}g\ sin(\theta)cos(\theta)}{I}$

From this pushing or pulling force we can derive the power and thereby energy:

$P=Fv$

$\frac{dE}{dt} = Fv = \frac{-L^{2}m^{2}gv}{I}sin(\theta)cos(\theta)$

On integrating,

$\int dE = \int \frac{-L^{2}m^{2}gv}{I}sin(\theta)cos(\theta)\ dt$

$\int_{0}^{\frac{T}{2}}dE = k\int_{0}^{\frac{T}{2}} vsin(\theta)cos(\theta)\ dt$

This gives the push energy in one single push (half the oscillation period).

Now the energy burned by the resistor is given by:

$dE = Pdt$

$\int_{0}^{\frac{T}{2}}dE = \int_{0}^{\frac{T}{2}}Pdt$

$\int_{0}^{\frac{T}{2}}dE = \int_{0}^{\frac{T}{2}}I^{2}Rdt$

Current in the LR circuit is

$I = \frac{V}{R}\left (1-e^{\frac{-Rt}{L}} \right )$

Substituting for current,

$\int_{0}^{\frac{T}{2}}dE = \int_{0}^{\frac{T}{2}} V\left (1-e^{\frac{-Rt}{L}} \right )^{2}dt$

Thus the change in energy for a half cycle is

$\Delta E_{\frac{T}{2}} = k\int_{0}^{\frac{T}{2}} vsin(\theta)cos(\theta)\ dt - \int_{0}^{\frac{T}{2}} V\left (1-e^{\frac{-Rt}{L}} \right )^{2}dt$

This is how the RHS dampens energy change. The resistance in the second integral dominates in each integration. If it wasn't there, The remaining emf/potential term will cancel itself in a full cycle integration. Thus, there would not be an energy change. This isn't surprising because inductor can give back the magnetic force in each half cycle with the same magnitude. I hope this will make it more clear since my previous attempt was somewhat dubious.

• Why does a push while advancing and a pull while retreating never cause damping. From an energy point of view I get it, but it seems contradictional when you examine the forces as in both cases the force from the coil on the magnet tends to decelerate the magnet – KvdLingen Mar 8 '14 at 23:34
• @KvdLingen Corrected it. The absence of resistance causes an acceleration in the retreating phase. Net current in the inductor falls back to zero. (I rushed through that conclusion paragraph). – Renae Lider Mar 9 '14 at 7:48
• @AwalGarg It's an open source software called inkscape – Renae Lider Mar 9 '14 at 7:49
• @KvdLingen Also, wouldn't the damping be so quick if the circuit system had an infinite mass? – Renae Lider Mar 9 '14 at 7:54
• The expression for the current isn't remotely correct. It would be correct if this were just an RL circuit driven by a constant $V$ with initial current $i(0) = 0$ but that isn't the case here. Qualitatively, we know the current must alternate. – Alfred Centauri Mar 9 '14 at 14:57

As the magnet approaches the solenoid, a current is induced. The current generates a magnetic field. The field repels the magnet, slowing it's approach. The amplitude of the oscillations diminish.

If there was no resistance, this would work in reverse as the magnet receded from the solenoid. The magnetic field would accelerate the magnet. The magnet would induce a current in the other direction, reducing the current to 0. This would reduce the field of the solenoid to 0. The amplitude would not diminish.

• Hmm, I was thinking this as well. Just wanna ask one thing. I have read that electromagnetic fields propagate at light's speed. So it takes some time for the field to act. If the speed of movement of the bar magnet was suitably high and the distance between the magnet and solenoid was suitably large, wouldn't the magnet recede away before the magnetic field actually acts? In-fact, this way, the magnetic field would favor the magnet's recede and increase the amplitude of oscillations. (And we can balance that by friction of the pulley of the pendulum to keep the amplitude constant.) – user3459110 Mar 8 '14 at 14:18
• @AwalGarg no. As you said electromagnetic fields propagate at light speed. You want to make the speed of movement of the bar suitably high but in this case, you are saying it has to be high to "outrun" the electromagnetic field, and it cannot be faster than light. – Guy Mar 8 '14 at 16:27
• And just like that, we have once again accidentally confirmed that any body cannot propagate at more than light speed, in this case, it would violate energy conservation. – Guy Mar 8 '14 at 16:27

This should go on forever, and current should keep appearing across the load resistance.

This is a contradiction. Since there is current through (not across) the load resistance, there is work being done on the load: $p = i^2R$.

Let's be clear on this: the coil-load system does no work on the pendulum, the pendulum does work on the coil-load system.

As the pendulum approaches the coil, the induced magnetic field repels the magnet; as the pendulum recedes from the coil, the induced magnetic field attracts the magnet.

Thus, with each cycle, energy is transferred from the pendulum to the load; mechanical power is converted to electrical power which is converted to heat by the resistor.

• Keeping to the newtonian system, work equals force times displacement. As the pendulum motion is decelerated, there is work done on the pendulum. Otherwise it's energy could never be transferred to the coil. – KvdLingen Mar 8 '14 at 14:45
• @KvdLingen, you're not thinking clearly about this. The force on the pendulum due to the induced magnetic field is directed opposite the displacement and thus, the associated work 'on' the pendulum is negative; the transfer of energy is from the pendulum. – Alfred Centauri Mar 8 '14 at 15:06
• A misunderstanding might be due to our definitions of terms. I do not discriminate between negative and positive work. And I understand that the transfer is from the pendulum to the coil. – KvdLingen Mar 8 '14 at 15:13
• @KvdLingen, but I didn't say there is no work done, I wrote: "the coil-load system does no work on the pendulum, the pendulum does work on the coil-load system." thus, how could I possibly interpret your first comment other than the way I did? – Alfred Centauri Mar 8 '14 at 15:34
• I agree that energy is transferred from the pendulum to the coil. I think it's the use of the word 'work' that's causing the confusion. I interpret 'work being done' it is done by the object which exerts a force on another object, causing a change in that second objects velocity. Your use of the term work seems to be broader. I do not disagree with that usage. – KvdLingen Mar 8 '14 at 16:03

The short answer is that there is a induced force on the magnet. This induced force will make the pendulum loose energy in the same proportions as there is electrical energy being generated.

A good experiment to show this effect is by having a small bar magnet and a copper pipe Or solenoid. When you let a small bar magnet drop from a certain height it will hit the ground with maximum kinetic energy proportional to that height. However if you take a copper pipe or solenoid and you let the magnet free fall inside it there will be a current generated like you said in the OP. But there will be a force on the magnet slowing it down. which means you will notice it reaching the bottom of the pipe / solenoid latter than it would have in complete free fall.

A different way to look at it is with they law of momentum. You could use the voltage across the load resistance to accelerate a different magnet. Which means you are giving that magnet momentum. Which in order to be conserved should have been lost by the first pendulum. Show there was a feedback force.

• Didn't get your point... I think what you are trying to say is already a part of the experiment, a known fact. – user3459110 Mar 8 '14 at 13:44

Actually pendulum wont oscillate forever.Its energy turns into heat in resistor.In other word it's domain would be something like this figure. • Of course it won't oscillate. Law of conservation of energy would hold. But the question is, how? The thought experiment concerned suggests that electricity is being constantly generated at every tick of the pendulum. – user3459110 Mar 8 '14 at 13:46
• The graph seems to be that of a dampened harmonic motion. As the force only acts on one side of the swing it will not look like this. – KvdLingen Mar 8 '14 at 14:31
• @AwalGarg: Pendulum's energy(kinetic energy and potential energy) will force electrons to move.(but how?)as pendulum moves makes electromagnetic waves(its speed=$1/\sqrt{\mu \epsilon}$).waves and its energy carried by photons.(waves energy $Energy=hf$h=Planck's constant & f=wave's frequency). – rza Mar 9 '14 at 19:52

The force on the pendulum only applies when the pendulum is in the vicinity of the coil. At that moment the harmonic motion of the pendulum is distorted. It 's amplitude is lessened and with it the upward motion. So the kinetic energy of the pendulum is converted into gravitational energy and electric energy. But the gravitational energy is less than without the coil present. The total mechanical energy (gravitational and kinetic) is reduced. The pendulum swings back but next time it comes around the velocity in the equilibrium is less, the change of the Flux per unit time in the coil is less, so less energy is converted into electrical energy, but the gravitation energy is still reduced. This continues until the amplitude of the swing is that much reduced, that no flux is induced anymore in the coil. Furthermore when the magnet pulls away from the coil the flux in the coil recedes, the current in the coil turns around to form a attracting force, reducing the velocity of the pendulum even more, while the kinetic energy is converted to electric energy and from there into heat in the resistor.

• Thanks for your answer. Do you mind seeing my recent comment @mmesser314's answer. I want to convey that same point to you... – user3459110 Mar 8 '14 at 14:20
• That would require the bat magnet to move at a velocity exceeding the speed of light? While at the same time the velocity of the pendulum needs to go down to zero as it approaches it's greatest distance from the equilibrium. Even disregarding quantum effects, these two conditions and to exclude each other. – KvdLingen Mar 8 '14 at 14:25
• No. Only a small fraction of the velocity of light would work great. Because, the magnet is first travelling to the solenoid, then, the changing magnetic field induces current, and then, the changing current induces magnetic field, and finally, that induced magnetic field starts to act. – user3459110 Mar 8 '14 at 14:32
• I do not think you can describe this sequentially. The moment Flux is changed in the coil, current changes and magnetic field starts to rise. And without the magnetic field you expect the gravity to produce enough acceleration to change the velocity from a few percent lightspeed to zero and back to those few percent? As this is impossible it follows that the magnetic field acts faster than the bar moves – KvdLingen Mar 8 '14 at 14:39
• I was thinking of something else. But I think, @user3058846's answer sorts up everything... – user3459110 Mar 9 '14 at 2:52

When the bar magnet approaches the solenoid, the magnetic field oppose the cause and repel it. But when it is departing, the induced magnetic field will oppose that too. Thus magnet will be attracted to solenoid which decelerate its motion. It will be exactly equal to the previous acceleration. Therefore the solenoid have no effect on its motion and it will die out due to frictional forces.