
Physics II course sequence · Interactive lesson
Lenz's Law & Induced Direction
Faraday's law gives the size of an induced emf; the minus sign tells you which way the current runs. Here you fix a loop's orientation, decide whether its flux is growing or shrinking, and let the induced current fight that change — with energy conservation as the reason it has no choice.
UPII-10.03 · Lesson 3 of 8 in this unit. Position is not a completion score.Build the model
The minus sign is energy conservation wearing a direction rule.
The key idea
Lenz's law: a current induced in a circuit flows in the sense whose own magnetic field opposes the change in magnetic flux through that circuit — the change, not the flux itself.
Example. If the upward flux through a flat ring grows from 0 to 4.0 mWb in 20 ms, then |ℰ| = 0.20 V and the current runs clockwise seen from above, so its own field points down inside the ring and fights the growth.
The complete physical model and assumptions
Lenz's law is the physical content of the minus sign in ℰ = −dΦB/dt: an induced current flows in whichever sense its own magnetic field opposes the change in flux that produced it. Read those words exactly — it opposes the change, not the field. While the flux through a loop is growing, the induced current's field inside the loop points against the applied field; while the flux is dying away, the same rule turns that induced field around so it points along the applied field, propping up what is being lost. The rule is not an extra postulate. If an induced effect reinforced its own cause, a nudge would grow without limit and dissipate energy that nothing had supplied, so the sign is forced by energy conservation: the induced current always loads the agent driving the change. Lenz's law is also strictly a direction rule. It never supplies a magnitude — that is Faraday's law together with the circuit's resistance — and it supplies nothing at all until you have chosen a normal for the loop and the circulation sense that goes with it. In a circuit already carrying current from other sources, the induced term adds algebraically and may only reduce the current rather than reverse it.
1Fix a normal before you argue about direction
A direction answer has no meaning until the loop has an orientation. Choose a unit normal n̂ for the surface the loop bounds — either face will do — and the sense of circulation counted positive is the one your right hand's fingers curl when the thumb points along n̂. Now ΦB = ∫ B⃗ ⋅ n̂ dA carries a sign, dΦB/dt carries a sign, and ℰ = −N dΦB/dt returns a signed number: positive means the induced current runs along the positive sense, negative means against it. Flip the choice of n̂ and everything flips with it — ΦB changes sign, so does dΦB/dt, so does ℰ, and so does what "positive current" means. Two sign changes cancel, and the current still runs the same way round the bench. That invariance is the check that you did the bookkeeping rather than guessed it.
2Opposition is to the change, not to the field
This is where most sign errors live. Take a loop of area 0.020 m² in a uniform field pointing into the page. If B grows at 2.0 T s⁻¹, the into-page flux grows, |ℰ| = (0.020 m²)(2.0 T s⁻¹) = 0.040 V, and the induced current runs counterclockwise so that its own field inside the loop points out of the page, against the applied field. Now let the same into-page field decay at 2.0 T s⁻¹ instead. The emf has exactly the same size, but the flux is shrinking, so the induced current runs clockwise and its field inside the loop points into the page — along the applied field, not against it. The induced field is not a mirror of B⃗. It mirrors dΦB/dt, and a loop will happily reinforce a field that is dying away.
3One rule, three ways for the flux to change
ΦB = B A cos θ can change through any of its three factors, and Lenz's law reads each the same way. A coil inside a switched electromagnet sees B change. A loop dragged out of a field region, or a flexible loop being squeezed, sees A change. A coil turned on its axis sees θ change. In every case the only question is the sign of dΦB/dt at that instant. Pull a loop out of a region where the field points out of the page: the out-of-page flux falls, so the induced current is counterclockwise, and its own field props up the flux being lost. Push the loop back in and the current reverses. When two factors change at once the derivative is a sum, so a shrinking loop inside a strengthening field can have dΦB/dt pass through zero, and the current reverses exactly there, not when either factor does.
4From current direction to the force on the agent
Once the sense of the current is known, the loop's magnetic moment m⃗ = N I A n̂ is known too, and that fixes the mechanical consequence. Bring a bar magnet's north pole down toward a horizontal ring: the downward flux through the ring grows, the induced current circulates so that its own moment points up toward the magnet, and the ring presents a north face to the approaching north pole. The ring repels the magnet, and the magnet pushes the ring away in return. Lift the magnet again and every sign reverses: the flux falls, the moment turns to point down, the ring shows a south face, and it is dragged after the retreating magnet. Both cases carry the same content — the induced current arranges itself to make the magnet's job harder — and neither needs a new rule.
5Energy conservation is why the sign cannot be the other one
Suppose the induced current ran the other way. The approaching magnet would be pulled in rather than pushed back, so it would speed up. A faster magnet means a faster flux change, a larger emf, a larger current and a stronger pull. The magnet accelerates, the ring dissipates I²R as heat, and nothing has supplied either the kinetic energy or the heat. That is the contradiction Lenz's law removes. With the sign as it stands, the dissipated power P = ℰ²/R is positive whichever way the flux is moving, and it is drawn from whoever forces the change — the hand holding the magnet, or the supply driving the electromagnet. Moving a magnet toward a closed ring costs more work than moving it toward a cut one, and the difference appears as heat in the ring.
6Lenz gives one term, not always the whole current
A loop with no other source carries only the induced current, so its direction is the answer outright. Put the loop in a circuit that already carries current and the induced emf becomes one term among several in the loop equation: it adds algebraically to the battery's emf, and it may merely reduce the current rather than reverse it. A motor coil spinning up is the standard case — the induced emf opposes the supply and the current falls, without ever changing sign. The same caution applies to a loop acting on itself, since a changing current makes a changing flux through its own area; the loop opposes changes in its own current, which is why current in a real circuit cannot jump discontinuously. Lenz's law fixes the sign of the induced contribution, never the sign of the total.
Key equations
Read the units, assumptions and limits before substituting.
Faraday's law, sign included
ℰ = −N dΦB/dt
The minus sign is Lenz's law, and the rest of this topic is learning to read it.
ℰ in V, ΦB in Wb, t in s, N turns; ℰ > 0 drives current along the loop's chosen positive sense.Choosing the positive sense
ΦB = ∫ B⃗ ⋅ n̂ dA, with I positive along n̂'s right-hand circulation
Signs mean nothing until one normal is fixed, yet the physical direction never depends on which face you picked.
n̂ is a dimensionless unit normal, A in m², B in T; either face may be chosen if the choice is kept throughout.Lenz's law in sign form
sign(Iinduced) = −sign(dΦB/dt)
Flux growing, the induced field fights it; flux falling, the induced field lines up with the applied one.
I in A, counted positive along the right-hand circulation of n̂; dΦB/dt in Wb s⁻¹.Field of the induced current
Bind = μ₀ I / (2a) at the centre of a loop of radius a
Turns "opposes the change" into a field you can draw and compare with the applied one.
μ₀ = 4π × 10⁻⁷ T m A⁻¹, I in A, a in m; its direction comes from the right-hand grip rule.Induced magnetic moment
m⃗ = N I A n̂
Says which pole the loop turns toward a magnet: it repels one approaching and attracts one leaving.
m in A m², N turns, A in m², I in A; m⃗ lies along the right-hand circulation of the induced current.Energy check on the sign
Pdis = I²R = (N dΦB/dt)²/R
Positive for either sign of dΦB/dt, so the agent always pays — reverse Lenz and it would be paid.
P in W, I in A, R in Ω; this power is supplied by whatever agent forces the flux to change.Interactive lab
Predict first. Explore the model. Compare the result.
Explore the model
Try this. Drag the rate slider down through zero. The flux line tips the other way and both current arrows reverse: a field dying away drives a counterclockwise current whose own field points out of the page, along the very field it is propping up.
Live results
- LOOP AREA
- 0.0113 m²
- FLUX RATE dΦ/dt
- 0.0226 Wb s⁻¹
- INDUCED CURRENT
- 0.1131 A
- SENSE (+1 = CCW)
- -1
Results follow this lesson’s model and assumptions. Displayed values are rounded.
Predict → change one variable → check
Plan a change before running it. Every other input stays at the saved baseline.
Saved baseline. Rate of change dB/dt: 2.0 T s⁻¹; Loop radius: 6.0 cm; Loop resistance: 0.20 Ω
This experiment is kept only while this page is open. It does not change saved course progress.
Catch the common trap
The error is compressing Lenz's law into "the induced field opposes the applied field". That is right only while the flux is growing. When the flux is falling, the induced current's field inside the loop points along the applied field — a loop in a dying field works to keep that field alive, and a student who always draws the induced field antiparallel gets every decaying case backwards. A second version of the same slip is treating the minus sign as a formality to be dropped once |ℰ| has been found. The minus sign is not attached to the number; it is attached to a circulation sense you chose, and it is the only part of Faraday's law that answers "which way".
Practice & worked examples
3 authored examples. Make an attempt, then reveal the reasoning one step at a time.
EasyExample 1
A single circular loop of radius 8.0 cm and resistance 0.25 Ω lies flat on a bench. A uniform magnetic field points vertically downward through it and grows steadily from 0.10 T to 0.34 T in 0.12 s. Find the induced current, and state which way it circulates as seen from above.
Sketch your answer or reasoning, then compare it with the authored solution. Your writing is not automatically graded or saved.
0 of 4 steps revealed.
- Take the normal downward, along the field, so ΦB is positive and growing. The loop area is A = π(0.080 m)² = 0.0201 m².
- The field is uniform over the loop, so dΦB/dt = A dB/dt = (0.0201 m²)(0.34 − 0.10) T ÷ 0.12 s = (0.0201 m²)(2.0 T s⁻¹) = 0.0402 Wb s⁻¹.
- |ℰ| = |dΦB/dt| = 0.0402 V, so I = |ℰ|/R = 0.0402 V ÷ 0.25 Ω = 0.161 A.
- Direction: the downward flux is growing, so the induced current's own field inside the loop must point upward. Curling the right hand with the thumb up gives a counterclockwise circulation seen from above.
Answer. I = 0.16 A, counterclockwise seen from above — its own field inside the loop points up, against the growing downward flux.
MediumExample 2
A 120-turn coil of cross-sectional area 45 cm² and total resistance 18 Ω sits on the axis of a magnet whose axial field decays as B(t) = 0.60 e(−t/0.40 s) T. Take the coil's normal n̂ along that field. Find the current at t = 0.50 s, and say whether the induced current's own field inside the coil points along n̂ or against it.
Sketch your answer or reasoning, then compare it with the authored solution. Your writing is not automatically graded or saved.
0 of 5 steps revealed.
- With n̂ along B⃗, the flux through one turn is ΦB = B A = A B₀ e(−t/τ), where A = 4.5 × 10⁻³ m², B₀ = 0.60 T and τ = 0.40 s.
- Differentiate: dΦB/dt = −(A B₀/τ) e(−t/τ). This is negative at every time, so the flux is falling throughout.
- ℰ = −N dΦB/dt = N A (B₀/τ) e(−t/τ). With N A = 120 × 4.5 × 10⁻³ m² = 0.54 m² and B₀/τ = 1.5 T s⁻¹, at t = 0.50 s the factor is e(−1.25) = 0.2865, giving ℰ = (0.54 m²)(1.5 T s⁻¹)(0.2865) = 0.232 V.
- I = ℰ/R = 0.232 V ÷ 18 Ω = 0.0129 A, that is 12.9 mA.
- Keep the sign: ℰ came out positive, so the current runs along the right-hand circulation of n̂ and its own field inside the coil points along n̂ — the same way as the applied field, which at this instant is still 0.60 e(−1.25) = 0.172 T and falling.
Answer. I = 12.9 mA, running so that its own field inside the coil points along n̂, reinforcing the decaying applied field rather than opposing it.
HardExample 3
A square loop of side 15.0 cm and resistance 0.12 Ω lies between the poles of an electromagnet in a uniform field of 0.80 T pointing out of the page. The supply is ramped down so that the field falls linearly to zero in 0.25 s. Find the direction of the induced current, its size, the total charge that circulates and the heat dissipated, then say what changes if the ramp is made twice as fast.
Sketch your answer or reasoning, then compare it with the authored solution. Your writing is not automatically graded or saved.
0 of 6 steps revealed.
- Area A = (0.150 m)² = 0.0225 m², so the initial out-of-page flux is ΦB = (0.80 T)(0.0225 m²) = 0.0180 Wb, falling to zero.
- The out-of-page flux is falling, so the induced current must add out-of-page flux inside the loop: it runs counterclockwise as drawn.
- The ramp is linear, so dΦB/dt = −0.0180 Wb ÷ 0.25 s = −0.0720 Wb s⁻¹, giving |ℰ| = 0.0720 V and I = 0.0720 V ÷ 0.12 Ω = 0.600 A.
- Charge: q = ∫ I dt = |ΔΦB|/R = 0.0180 Wb ÷ 0.12 Ω = 0.150 C. It depends on the total flux change alone, never on how quickly the change happened.
- Heat: P = I²R = (0.600 A)²(0.12 Ω) = 0.0432 W for 0.25 s, so W = 0.0108 J = 10.8 mJ. The supply pays it, because the loop's moment m = I A = (0.600 A)(0.0225 m²) = 0.0135 A m² points out of the page and resists the ramp-down.
- Halve the ramp time: |ℰ| and I double to 0.144 V and 1.20 A, the charge stays at 0.150 C because ΔΦB is unchanged, and W = (ΔΦB)²/(R Δt) doubles to 21.6 mJ.
Answer. 0.600 A counterclockwise, q = 0.150 C, W = 10.8 mJ. At half the ramp time: 1.20 A, q still 0.150 C, and W = 21.6 mJ.
Quick check
Select an answer and explain it before checking.
Review & continue
Lenz's law: a current induced in a circuit flows in the sense whose own magnetic field opposes the change in magnetic flux through that circuit — the change, not the flux itself.
- Explain the relationship without looking at the equation.
- Check a sign, a unit and a model assumption in your working.
- Return to the lab or worked example if a step is still unclear.
Optional supporting lessons
These chapter resources can help with earlier ideas; they are not additional required Physics II completion tasks.