# How does Lenz law conserve energy in this case?

A magnet is falling into the coil as shown. The current in the coil creates a magnet as shown ( by the right-hand grip rule ). Now, as the magnet falls towards the coil, the magnetic field strength in the downward direction increases. The induced emf is such that it opposes this change, and so the induced emf decreases the current in the circuit, thus causing the magnetic field strength to decrease to counteract the increase from the magnet. Now, if the power in the circuit decreases, something else must be gaining that energy. But the magnetic attraction between the approaching pole and the pole due to the current has also decreased, so where has this energy gone?

I think your argument is fine up to the sentence:"Now, if the power in the circuit decreases, something else must be gaining that energy." There will be a back emf induced in the circuit by the approaching magnet, and that will indeed reduce the current and the power, that is the rate at which the battery supplies energy. Just because energy isn't being transferred from the battery at as high a rate doesn't mean that energy is going somewhere else!

• I don't understand... If the energy in the circuit is less than it should be, the energy of something else should be higher than it should be by energy conservation
– John
May 1, 2017 at 14:50
• What do you can by "the energy of the circuit"? This is not just a nitpick. In a circuit, energy is being transferred, in this case from the battery to – in the steady state – random energy in the wire of the coil. When the current is reduced because of the opposing emf induced by the approach of the magnet, the battery supplies energy at a lower rate (with respect to time). It's not a case of energy going somewhere else. If I may say so, I think it would help if you read up about energy conservation, from a source which discusses several examples. May 1, 2017 at 15:53
• But if we had just a coil as shown ( no battery ), then the induced current would create a magnetic field which would repel the incoming magnet. Some work would have to be done against this magnetic force, and this work would manifest itself in the form of the energy of the circuit. Why can we not apply a similar argument to this case?
– John
May 1, 2017 at 16:12
• In the no-battery case, work is done by the agency pushing the magnet, and this work indirectly supplies random energy to the wires of the circuit, because of the induced current. In the second case work is done by the coil on the magnet, either giving it KE or doing work on the holder of the magnet! The emf induced in the coil by the magnet tends to make the South pole (and therefore also the North pole) of the coil weaker, by inducing a back emf, and therefore reducing the current. Therefore the rate of transfer of energy from the battery is reduced. Unless I'm wrong! May 1, 2017 at 17:37
• Okay, so the 'energy' of the circuit remains unchanged, but it is the rate which decreases. However, as you said, the emf makes the North and South poles weaker, so the magnetic force of attraction they exert will decrease, , so the work done by this force will also decrease. Doesn't this 'lost' energy manifest itself in some other place?
– John
May 2, 2017 at 19:56

The falling magnet has a solenoidal electric field (lines of force form loops) which is oriented so that it decreases existing electric current. The induced emf due to that very current decreasing will actually oppose this decrease, but it can't stop it or deny it; the opposing emf can only appear if the current keeps decreasing. This means when the magnet gets close enough, the current actually decreases.

As a result the battery will push less charges and put in less energy per unit time into the circuit (which before got dissipated into heat). The energy that was already in the magnetic field is not lost, it just redistributes in space and part of it may end up as increase in kinetic energy of the magnet.