Lenz's Law & Conservation of Energy

The Law of Opposition

Faraday's law gives the magnitude of the induced emf. Its polarity comes from a beautifully concise rule deduced in 1834 by the German physicist Heinrich Friedrich Lenz:

The polarity of the induced emf is such that it tends to produce a current which opposes the change in magnetic flux that produced it.

This is exactly what the negative sign in $\varepsilon = -\frac{d\Phi_B}{dt}$ encodes.

The bar-magnet picture (the exam classic):

• N-pole approaching the coil: flux through the coil increases → the induced current opposes the increase → it flows counter-clockwise as seen from the magnet's side → the coil's near face becomes a north pole → it repels the approaching magnet.
• N-pole receding: flux decreases → induced current flows clockwise (seen from the magnet) → near face becomes a south pole → it attracts the receding magnet, opposing its departure.

Either way, the coil fights the change — never the flux itself. And the rule works even for an open circuit: an emf of the corresponding polarity appears across the open ends, ready to drive that opposing current.

Lenz's Law IS Conservation of Energy

Suppose Lenz's law were reversed: the approaching N-pole induced a south face on the coil. The magnet would be attracted, accelerate, increase the flux faster, induce more current, attract harder… A gentle push would make its velocity and kinetic energy grow without any energy being spent. You could build a perpetual-motion machine.

That violates the law of conservation of energy — so it cannot happen.

In the correct (Lenz) case:

• The magnet is repelled, so the person pushing it must do work against the opposition.
• Where does that work go? It is dissipated as Joule heat by the induced current in the coil.

Key Point: Energy bookkeeping is exact: mechanical work in = electrical energy induced = heat out. Lenz's law is not a separate axiom; it is energy conservation wearing electromagnetic clothes.

[NEET Important] A magnet dropped through a conducting ring/coil falls with acceleration less than g while interacting (opposed approaching and leaving); the same magnet dropped through a cut (open) ring falls at exactly g because no current can flow through the gap (though an emf still appears across the cut).

The Direction Toolkit & Reasoning Set

Standard direction calls (field into the page, loop in the plane of the page):

Situation	Flux (into page)	Induced current
Loop entering the field region	increasing	anticlockwise (makes out-of-page flux)
Loop fully inside, moving	constant	zero
Loop leaving the field region	decreasing	clockwise (makes into-page flux)

This is the leaving-the-field case in compact form — and note its punchline: no current flows while the loop is completely inside or completely outside the field region, however fast it moves.

Three reasoning gems:

1. Stationary loop between very strong fixed magnets: no current, however strong the magnets. No change of flux, no induction.
2. Loop moving in the uniform electric field of a capacitor: no magnetic flux anywhere in the problem, so no induced current — whether wholly inside or partially outside the plates.
3. Rectangular vs circular loop leaving a field region at constant v: the rectangular loop gives a constant emf during exit (the in-field length, and so $d\Phi/dt = Blv$, is constant), while the circular loop's in-field width keeps changing — its emf varies during the passage.

[JEE Tip] For any 'find the direction' question, run the three-step drill: (1) which way is $\vec{B}$ through the loop? (2) is $\Phi_B$ increasing or decreasing? (3) the induced current makes flux opposing that change — fix its sense by the right-hand rule. Three seconds, full marks.

Eddy Currents

When a solid conductor (not a thin wire) sits in a changing magnetic flux, the induced currents are not confined to a single loop — they swirl through the bulk of the metal in closed paths, like little whirlpools. These are eddy currents (also called Foucault currents). Their direction always obeys Lenz's law: they oppose the very change of flux that creates them.

Because the metal has resistance, eddy currents dissipate energy as heat ($I^2R$). Depending on the situation that is a nuisance, a brake, or the whole point:

As a brake (electromagnetic damping): in magnetic braking of trains, electromagnets over the rails induce eddy currents whose Lenz-law drag brings the train to a smooth, frictionless halt. The same damping makes a galvanometer dead-beat — a metal frame moving in the field is braked, so the pointer settles at once instead of oscillating.

As a heater: an induction furnace uses high-frequency eddy currents to melt a metal charge for preparing alloys; an induction cooktop heats the pan the same way.

As an unwanted loss: in transformer and motor cores eddy currents waste energy. The cure is lamination — the core is built from thin, insulated sheets, which chops the eddy loops into many small high-resistance paths and sharply cuts the dissipation.

[NEET Important] An eddy current is nothing exotic: it is an ordinary induced current in a bulk conductor. Faraday's law still fixes its size and Lenz's law still fixes its direction.

Solved Examples

Example 1: Direction AND magnitude at entry [JEE Numerical]

A rectangular loop of width 0.2 m and circuit resistance 0.4 $\Omega$ enters a 0.5 T field region (into the page) at 2 m/s. Find the induced current's magnitude and sense.

Solution:

1. Magnitude: only the leading side cuts lines: $\varepsilon = Blv = 0.5 \times 0.2 \times 2 = 0.2$ V, so $I = \varepsilon/R = 0.5$ A.
2. Direction (Lenz): into-page flux is increasing → oppose with out-of-page flux → current anticlockwise.
3. Answer: 0.5 A, anticlockwise — direction calls earn full marks only with the magnitude attached.

Example 2: Inside, then leaving [NEET Numerical]

The same loop now (a) moves wholly inside the region, (b) exits at the same 2 m/s. Find the current in each phase.

Solution:

1. (a) Fully inside, the enclosed flux is constant: $\varepsilon = 0$, so $I = 0$ — however fast it moves.
2. (b) Exiting, the into-page flux decreases → the loop supports it → current clockwise, magnitude again $Blv/R = 0.5$ A.
3. Pattern: entry pulse (anticlockwise), silence, exit pulse (clockwise) — equal magnitudes, opposite senses.

Example 3: The retarding force on the loop [JEE Numerical]

For the entering loop of Example 1 (B = 0.5 T, l = 0.2 m, v = 2 m/s, R = 0.4 $\Omega$), find the magnetic force on it and the external force needed for constant velocity.

Solution:

1. The induced current $I = 0.5$ A flows through the in-field side: $F = BIl = 0.5 \times 0.5 \times 0.2 = 0.05$ N.
2. By Lenz's law this force opposes the motion (it retards entry; it would also retard exit).
3. Answer: an equal external force of 0.05 N ($= B^2l^2v/R$) must be applied to maintain v.

Example 4: Strong magnets, stationary loop

A closed loop is held stationary between the poles of two fixed, very strong permanent magnets. Can we generate a current by using stronger magnets?

Solution:

1. Induction requires a time-varying flux, not a large flux.
2. Everything here is stationary and steady: $d\Phi_B/dt = 0$.
3. No current — no matter how strong the magnets. Strength is not change.

Example 5: A loop in a capacitor's electric field

A closed loop moves normal to the constant electric field between capacitor plates. Is a current induced (i) wholly inside, (ii) partially outside the plates?

Solution:

1. Electromagnetic induction responds to changing magnetic flux. Here there is an electric field but no magnetic flux at all.
2. Moving the loop changes nothing magnetic — in both cases the induced current is zero.
3. A constant electric flux through a loop also does not produce electromagnetic induction; the relevant quantity here is $d\Phi_B/dt$.

Example 6: Rectangle vs circle leaving the field

A rectangular and a circular loop move out of a uniform field region with the same constant velocity. In which is the induced emf constant during the passage out?

Solution:

1. The emf is $\varepsilon = Blv$ where $l$ is the length of the side still cutting field lines.
2. Rectangle: the in-field side length is fixed, so $\varepsilon$ is constant during exit.
3. Circle: the chord inside the field keeps changing as it exits, so the emf varies. Answer: the rectangular loop.

Example 7: Lenz's energy bill [JEE Numerical]

The loop of Example 1 is dragged out of the field region through 0.1 m at the constant 2 m/s. Compute (a) the heat generated, (b) the work done by the puller, and compare.

Solution:

1. (a) $I = 0.5$ A, so $P = I^2R = 0.25 \times 0.4 = 0.1$ W. Exit time $t = 0.1/2 = 0.05$ s. Heat $= Pt = 5 \times 10^{-3}$ J.
2. (b) Work $= Fd = (B^2l^2v/R) \times d = 0.05 \times 0.1 = 5 \times 10^{-3}$ J.
3. They match exactly — the puller's work against the Lenz force is precisely the Joule heat. Conservation of energy, audited.

Example 8: Magnet falling towards a ring [NEET Numerical]

A magnet dropped towards a horizontal ring (R = 0.1 $\Omega$) raises the downward flux through it from 0 to 8 mWb in 0.2 s. Find the average emf and current, and the current's sense seen from above.

Solution:

1. $\varepsilon = \Delta\Phi/\Delta t = 8 \times 10^{-3}/0.2 = 4 \times 10^{-2}$ V.
2. $I = \varepsilon/R = 0.04/0.1 = 0.4$ A.
3. Sense: downward flux increasing → upper face turns N to repel → anticlockwise seen from above. (And the magnet falls with a < g.)

Example 9: Reversing a solenoid's current [NEET Numerical]

A 20-turn coil is wound around a solenoid; each turn links $5 \times 10^{-4}$ Wb. The solenoid's current is reversed in 0.05 s. Find the average emf induced in the coil.

Solution:

1. Reversal doubles the change: $\Delta\Phi$ per turn $= 2 \times 5 \times 10^{-4} = 10^{-3}$ Wb.
2. $\varepsilon = N\frac{\Delta\Phi}{\Delta t} = 20 \times \frac{10^{-3}}{0.05}$.
3. Answer: $\varepsilon = 0.4$ V, its polarity set by Lenz's law (opposing the collapse, then the rebuild, of the flux).

Example 10: The forbidden alternative

Show that reversing Lenz's law would permit a perpetual-motion machine.

Solution:

1. Reversed law: the approaching N-pole induces a south near face — the magnet is attracted.
2. It accelerates, the flux changes faster, the induced current and attraction grow — velocity and kinetic energy increase without any energy input after a gentle push.
3. Free, ever-growing kinetic energy is a perpetual-motion machine — forbidden by conservation of energy. Hence the induced effects must oppose the change.