Faraday's Law vs Lenz's Law: The Difference

Faraday's law gives the size of an induced emf: the rate at which magnetic flux through a loop changes. Lenz's law gives its direction: the induced current opposes the change that caused it. They are not two laws. The minus sign printed in Faraday's law on the Physics C sheet is Lenz's law.

AP Physics: Unit 13 (topics 13.2 Electromagnetic Induction, 12.4 Electromagnetic Induction and Faraday's Law). In AP Physics C: Electricity and Magnetism both statements sit in Topic 13.2, Electromagnetic Induction, in Unit 13, weighted at 10 to 20 percent of the multiple-choice section over approximately 10 to 20 class periods. Faraday's law is essential knowledge 13.2.A.1 with the relevant equation printed as emf equals minus d phi B by dt equals minus d of B dot A by dt; 13.2.A.1.i and 13.2.A.1.ii split it into the constant-area and constant-field cases; 13.2.A.1.iii gives the solenoid form with N loops in magnitude only. Lenz's law is essential knowledge 13.2.A.2 under the same learning objective 13.2.A, with 13.2.A.2.i stating that an induced emf generates a current that creates a magnetic field that opposes the change in magnetic flux, and 13.2.A.2.ii naming the right-hand rule. Essential knowledge 13.2.A.3 records that Maxwell's third equation is Faraday's law of induction. AP Physics 2 carries the same pair as 12.4.A.3 and 12.4.A.4 under learning objective 12.4.A, but its equation sheet prints Faraday's law with absolute value bars and no minus sign. Lenz's law has no learning objective of its own in either course.

One law and its sign

Almost every treatment of induction presents these as two separate laws that happen to be taught together. They are not. Look at what the AP Physics C: Electricity and Magnetism sheet actually prints:

\mathcal{E} = \oint \vec{E} \cdot d\vec{\ell} = -\frac{d\Phi_B}{dt}

There is one equation there. The $\frac{d\Phi_B}{dt}$ part is Faraday's law: the emf is the rate of change of the magnetic flux. The minus sign in front of it is Lenz's law: the induced effect opposes the change that produced it. Take the minus sign away and you have an equation that gets every magnitude right and every direction wrong.

So the honest description is a law with two halves that do different jobs:

Faraday's half answers "how big". Volts. It cares only about how fast the flux is changing, not how large the flux is, not how strong the field is.
Lenz's half answers "which way". No number attached. It tells you the sense of the induced current around the loop.

They are usually first met as separate rules, because Lenz's law is easier to apply as a sentence than as a sign, and that is a reasonable teaching order. It becomes a problem when a question needs both at once, or when a student writes an emf of $-0.060 \ \mathrm{V}$ and also argues the direction from Lenz's law, applying the same physics twice and reversing the answer.

Side by side

	Faraday's law	Lenz's law
What it delivers	The magnitude of the induced emf	The direction of the induced current and emf
Written as	$\mathcal{E} = -\dfrac{d\Phi_B}{dt}$	The minus sign in that equation
Is it a number?	Yes, in volts	No, it is a sense of circulation
Depends on	The rate of change of flux	Which way the flux is changing
Independent of	The size of $\Phi_B$ itself	Everything numerical
CED status	Essential knowledge 13.2.A.1	Essential knowledge 13.2.A.2, under the same learning objective
Tool you apply	Differentiate $\Phi_B = \vec{B} \cdot \vec{A}$	The right-hand rule, per essential knowledge 13.2.A.2.ii
Named in Maxwell's equations?	Yes, the third	Not separately, it is the sign in the third
On the C: E&M sheet	Yes, with the minus sign	Yes, as that minus sign
On the AP Physics 2 sheet	Yes, as $\lvert \mathcal{E} \rvert = \lvert \frac{\Delta\Phi_B}{\Delta t} \rvert$	No, the printed form has no sign at all

The last row is the sharpest evidence for the claim this page makes, and the next section is about it.

The minus sign is Lenz's law, and the two courses print it differently

Compare the two equation sheets line for line.

AP Physics C: Electricity and Magnetism prints $\mathcal{E} = \oint \vec{E} \cdot d\vec{\ell} = -\frac{d\Phi_B}{dt}$ . The minus sign is there. A C student who substitutes correctly gets a signed answer whose sign already encodes the direction, relative to whatever positive sense the right-hand rule assigned to the loop.

AP Physics 2 prints $\lvert \mathcal{E} \rvert = \left\lvert \frac{\Delta\Phi_B}{\Delta t} \right\rvert$ . Absolute value bars on both sides and no minus sign anywhere. A Physics 2 student can never get a direction out of the printed equation, because the sign has been deliberately removed from it.

What replaces it in that course is Lenz's law stated as prose. AP Physics 2 essential knowledge 12.4.A.4 reads that Lenz's law is used to determine the direction of an induced emf resulting from a changing magnetic flux, and 12.4.A.4.i adds that an induced emf generates a current that creates a magnetic field that opposes the change in magnetic flux. The C: E&M CED carries the identical statements as 13.2.A.2 and 13.2.A.2.i.

So the same physics is packaged two ways. The algebra-based course strips the sign out of the equation and hands you the direction rule as a separate sentence. The calculus-based course keeps the sign in and expects you to see that the sentence and the sign say the same thing. Neither course treats Lenz's law as an independent law about a different phenomenon, and the sheets are the proof.

One wrinkle worth knowing: even the C sheet prints a magnitude-only version for coils. $\lvert \mathcal{E}_{\mathrm{sol}} \rvert = N\left\lvert \frac{d\Phi_B}{dt} \right\rvert$ has bars and an $N$ for the number of turns, and no sign. When a problem gives you a solenoid or a coil of $N$ turns, that is the line you use, and you supply the direction from Lenz's law in words.

What the CED does with them

Both statements live under one learning objective, and in both courses that objective is worded the same way: describe the induced electric potential difference resulting from a change in magnetic flux. It is 13.2.A in AP Physics C: Electricity and Magnetism and 12.4.A in AP Physics 2.

Underneath it, the C: E&M CED lays them out in order:

13.2.A.1 Faraday's law describes the relationship between changing magnetic flux and the resulting induced emf in a system, with the relevant equation $\mathcal{E} = -\frac{d\Phi_B}{dt} = -\frac{d(\vec{B} \cdot \vec{A})}{dt}$ .
13.2.A.1.i When the area is constant, the induced emf is the area multiplied by the rate of change in the component of the magnetic field perpendicular to the surface.
13.2.A.1.ii When the magnetic field is constant, the induced emf is the field multiplied by the rate of change in area perpendicular to the field.
13.2.A.1.iii For a long solenoid, the total induced emf is the emf in a single loop times the number of loops, with the relevant equation $\lvert \mathcal{E}_{\mathrm{sol}} \rvert = N\left\lvert \frac{d\Phi_B}{dt} \right\rvert$ .
13.2.A.2 Lenz's law is used to determine the direction of an induced emf resulting from a changing magnetic flux.
13.2.A.3 Maxwell's third equation is Faraday's law of induction.

Lenz's law has no learning objective of its own in either course. It is one essential knowledge statement inside Faraday's, sitting between the magnitude cases and the Maxwell framing.

The second form printed in 13.2.A.1 is worth extracting on its own: $\mathcal{E} = -\frac{d(\vec{B} \cdot \vec{A})}{dt}$ . Written that way, the two subpoints 13.2.A.1.i and 13.2.A.1.ii are just the product rule applied to $\vec{B} \cdot \vec{A}$ , taking one factor as constant at a time. That is the whole taxonomy of induction problems: the field changes, the area changes, or the angle between them changes.

The case that separates them: one loop, one changing field

Take a single square loop lying flat on the page, in a magnetic field pointing out of the page whose strength is increasing steadily.

Faraday's half notices only $\frac{d\Phi_B}{dt}$ . Increase $B$ twice as fast and the emf doubles. Make the loop twice the area and the emf doubles. Make $B$ enormous but hold it steady and the emf is zero. None of that says anything about which way current flows.

Lenz's half notices only that the flux out of the page is increasing. The induced current must oppose that, so it must create flux into the page inside the loop, and by the right-hand rule that current runs clockwise as you look at the page. None of that says how many amperes.

The two halves are genuinely independent pieces of information, and a question can ask for either alone. That is why they feel like separate laws. But run the experiment again with the field decreasing instead of increasing: Faraday's half gives the same magnitude, and Lenz's half flips to counterclockwise. The magnitude never knew about the direction and the direction never knew about the magnitude, and both came out of one equation.

Worked example one runs the numbers on exactly this setup. Worked example two does the other case in the CED's pair, where the field holds still and the area changes instead.

The four sign cases, in one table

Every direction question in the course reduces to one of these four rows. Read "out of the page" as toward you and "into the page" as away from you.

Flux through the loop	What it is doing	Induced current must make flux	Induced current direction
Out of the page	Increasing	Into the page	Clockwise
Out of the page	Decreasing	Out of the page	Counterclockwise
Into the page	Increasing	Out of the page	Counterclockwise
Into the page	Decreasing	Into the page	Clockwise

Two things to notice. First, the direction of $\vec{B}$ alone never decides the answer; you need to know whether the flux is growing or shrinking. Second, rows one and four give the same answer, and so do rows two and three, so there are only two distinct outcomes and the rule that selects between them is: the induced current tries to keep the flux the way it was.

There is a fifth case that the table cannot show and that questions love: the flux is not changing at all. A steady field through a stationary loop, or a loop moving entirely within a uniform field, or a loop at the instant its plane is at a maximum of flux while rotating. In every one of those, the emf is zero regardless of how strong the field is. Worked example three is the moving-loop version.

The angle case deserves its own line. If the loop rotates in a steady field, $\Phi_B = BA\cos\theta$ and the emf comes entirely from $\theta$ changing. The emf is largest when the loop's plane contains the field, which is when the flux itself is momentarily zero, and the emf is zero when the loop faces the field squarely, which is when the flux is largest. That inversion between flux and emf follows directly from the law being about a rate, and it catches people who expect the two to peak together.

Why the sign has to be a minus

Suppose Lenz's law ran the other way, so that an induced current reinforced the change in flux that created it. Push a magnet toward a loop and the loop would generate a field that pulls the magnet in harder, which speeds the magnet up, which increases the rate of change of flux, which increases the current, and so on. Energy would appear from nowhere.

The minus sign is what forbids that. It means the induced current always sets up a field that resists the change, so the loop pushes back on the magnet you are moving, and the work you do against that push is exactly the electrical energy that shows up in the circuit. Induction is a conversion, not a source, and the sign is the bookkeeping that keeps it one.

This is also why an induced current in a resistive loop is a braking effect. Drop a magnet down a copper pipe and it falls slowly; the changing flux drives currents in the pipe wall, those currents oppose the change, and the magnet loses gravitational potential energy to resistive heating instead of to kinetic energy. Nothing in that description needs a new law: it is $\mathcal{E} = -\frac{d\Phi_B}{dt}$ with the minus sign read out loud.

The CED touches the same accounting on the inductor side, where essential knowledge 13.4.A.2.ii states that the transfer of energy generated in an inductor to other forms of energy obeys conservation laws. An inductor is a device built to exploit this sign: essential knowledge 13.4.A.3 relates the induced emf to the rate of change of current as $\mathcal{E} = -L\frac{dI}{dt}$ , the same minus sign again, now opposing a change in current rather than a change in external flux. That relationship is what makes an inductor the mirror image of a capacitor, which is its own comparison.

When it costs a mark

Applying the minus sign and Lenz's law both. The commonest way to invert a correct answer. If you substitute into $\mathcal{E} = -\frac{d\Phi_B}{dt}$ and get a signed result, that sign already is the direction, relative to the positive sense you chose for the loop. Then arguing separately from Lenz's law and flipping it again gets you back to wrong. Pick one route per problem and declare which.

Reporting a negative emf as if it were a physical quantity. An emf of $-0.060 \ \mathrm{V}$ means nothing until you have said what positive means for that loop. On a free-response answer, name the direction in words (clockwise, or from a to b through the resistor) rather than leaning on a sign the grader cannot interpret.

Confusing large flux with large emf. A loop sitting in the strongest field you can build has zero emf if nothing is changing. The equation is a derivative. Questions are built around this and the correct answer is often simply zero.

Forgetting the $N$ for a coil. A single loop and a 200-turn coil in the same changing field have emfs differing by a factor of 200. The C: E&M sheet prints the coil version separately as $\lvert \mathcal{E}_{\mathrm{sol}} \rvert = N \lvert \frac{d\Phi_B}{dt} \rvert$ ; the single-loop version has no $N$ in it.

Using only the perpendicular component of the field, then also using $\cos\theta$ . $\Phi_B = \vec{B} \cdot \vec{A}$ already contains the angle. Resolving the field first and then multiplying by $\cos\theta$ again applies it twice.

Assuming a moving loop always has an emf. Motion in a magnetic field induces nothing if the flux does not change. A loop translating entirely inside a uniform field has a constant flux and zero emf, however fast it moves.

When the distinction stays hidden

Whole stretches of a course go by without needing Lenz's law, and that is what makes its absence hard to notice.

Any question that asks only "find the magnitude of the induced emf" is answerable from Faraday's half alone, and the AP Physics 2 sheet's absolute-value form is built for exactly those questions. A student can work through an entire unit on induction, get every numerical answer right, and never once have to decide which way a current goes.

The distinction also stays hidden whenever the problem hands you the direction. "The induced current in the loop is clockwise. Find its magnitude." There, Lenz's law is the given, not the task.

Three situations force it open, and those three are what the exam uses.

A direction is asked for explicitly. Which way does the current flow, or which end of the rod is at higher potential.
A force or a motion follows from the direction. An induced current in a magnetic field feels a force, and whether a loop is pulled in or pushed out depends entirely on the sign. This is Topic 13.3, which exists because the sign has consequences.
The change reverses partway through. A loop entering a field region and then leaving it produces two emfs of opposite sense. A magnitude-only treatment gives two identical answers and misses that the current reversed.

If you can handle those three, the pair is finished, because there is nothing else Lenz's law is for.

Where these sit in the course

Both statements live in Topic 13.2, Electromagnetic Induction, in Unit 13, which the CED weights at 10 to 20 percent of the AP Physics C: Electricity and Magnetism multiple-choice section over roughly 10 to 20 class periods. Topic 13.1 sets up magnetic flux first, because you cannot differentiate a quantity you have not defined.

In AP Physics 2 the same pair sits in Topic 12.4, Electromagnetic Induction and Faraday's Law, the closing topic of Unit 12. That course stops there: it has no inductance, no inductors and no LR circuits anywhere in its framework, so induction is the end of its electromagnetism rather than the start of a new unit.

For the C course, Topic 13.2 is the hinge of the whole unit. Topic 13.3 turns the induced current into a force, which is where the sign starts to matter physically. Topic 13.4 applies the same law to a coil opposing its own current change. And Topic 13.5 puts that coil in a circuit, which is where the minus sign stops being a convention and starts controlling how fast a current can turn on.

A changing field: magnitude from Faraday, direction from Lenz

A square loop of side $0.20 \ \mathrm{m}$ lies flat in the plane of the page. A uniform magnetic field points out of the page and grows steadily from $0.30 \ \mathrm{T}$ to $0.90 \ \mathrm{T}$ in $0.40 \ \mathrm{s}$ . (a) Find the magnitude of the induced emf. (b) The loop has a total resistance of $0.50 \ \Omega$ ; find the induced current. (c) State the direction of that current and justify it.

(a) The area is fixed and the field is perpendicular to the loop, so this is the CED's case 13.2.A.1.i: the emf is the area times the rate of change of the perpendicular field component.
Area: $A = (0.20 \ \mathrm{m})^2 = 0.040 \ \mathrm{m^2}$ .
Rate of change of field: $\dfrac{dB}{dt} = \dfrac{0.90 - 0.30}{0.40} = \dfrac{0.60}{0.40} = 1.5 \ \mathrm{T/s}$ .
$\lvert \mathcal{E} \rvert = A\dfrac{dB}{dt} = (0.040)(1.5) = 0.060 \ \mathrm{V}$ . Check the units: $\mathrm{m^2 \cdot T/s} = \mathrm{Wb/s} = \mathrm{V}$ .
(b) $I = \dfrac{\lvert \mathcal{E} \rvert}{R} = \dfrac{0.060}{0.50} = 0.12 \ \mathrm{A}$ .
(c) Direction, from Lenz's law. The flux out of the page is increasing, so the induced current must produce flux into the page inside the loop, to oppose that increase.
Right-hand rule: to make a field into the page inside the loop, the current runs clockwise as seen from the reader's side.
Sanity check on the whole answer: nothing here depended on the value of $B$ itself, only on how fast it changed. If the field had held steady at $0.90 \ \mathrm{T}$ , the emf would be zero.

$\lvert \mathcal{E} \rvert = 0.060 \ \mathrm{V}$ and $I = 0.12 \ \mathrm{A}$ , flowing clockwise. Faraday's half gave both numbers; Lenz's half gave the word clockwise and nothing else.

A changing area: a rod on rails, derived rather than quoted

Two long horizontal rails are $\ell = 0.25 \ \mathrm{m}$ apart, joined at the left end by a resistor, in a uniform magnetic field of $0.80 \ \mathrm{T}$ directed into the page. A conducting rod lies across the rails and is pulled to the right at a constant $4.0 \ \mathrm{m/s}$ . Find the induced emf and the direction of the induced current, working from Faraday's law rather than from a memorised shortcut.

The field is constant here and the area of the circuit is what changes, so this is the CED's case 13.2.A.1.ii.
Let $x$ be the distance from the resistor to the rod. The enclosed area is $A = \ell x$ , so the flux is $\Phi_B = B\ell x$ .
Differentiate: $\dfrac{d\Phi_B}{dt} = B\ell\dfrac{dx}{dt} = B\ell v$ .
So $\lvert \mathcal{E} \rvert = B\ell v = (0.80)(0.25)(4.0) = 0.80 \ \mathrm{V}$ .
Worth noting where that expression lives. The AP Physics 2 sheet prints $\mathcal{E} = B\ell v$ directly. The AP Physics C: Electricity and Magnetism sheet does not print it at all, so on that exam you get it from $-\frac{d(\vec{B} \cdot \vec{A})}{dt}$ in the three lines above, which is why the derivation is shown rather than the formula quoted.
Direction, from Lenz's law. The rod moving right enlarges the circuit, so the flux into the page is increasing.
The induced current must oppose that, producing flux out of the page inside the loop, which by the right-hand rule means the current runs counterclockwise around the circuit.
Consequence check: that counterclockwise current in the rod, sitting in the field into the page, feels a magnetic force. Work out its direction and it points left, against the pull. The rod resists being dragged, which is the minus sign showing up as a force.

$\lvert \mathcal{E} \rvert = B\ell v = 0.80 \ \mathrm{V}$ , with the induced current running counterclockwise. On the C exam this comes from differentiating $B \cdot A$ with the area changing, because the sheet for that course does not print $B\ell v$ .

Zero emf, then not: a loop crossing the edge of a field region

A square loop of side $0.20 \ \mathrm{m}$ moves to the right at a constant $3.0 \ \mathrm{m/s}$ through a region where a uniform $0.50 \ \mathrm{T}$ field points into the page. The field region ends at a sharp vertical boundary. (a) Find the emf while the loop is entirely inside the field region. (b) Find the emf while the loop is crossing the boundary and leaving. (c) Give the direction of the current in part (b).

(a) While the whole loop is inside, the flux is $\Phi_B = BA = (0.50)(0.20)^2 = (0.50)(0.040) = 0.020 \ \mathrm{Wb}$ , and it stays at that value as the loop slides along.
The flux is constant, so $\dfrac{d\Phi_B}{dt} = 0$ and the emf is zero. The loop is moving fast, in a strong field, and nothing is induced.
(b) Once the leading edge crosses the boundary, only part of the loop is still in the field, and that part shrinks. In a time $dt$ the loop moves $v\,dt$ and loses an area $\ell v\,dt$ from the field region.
$\left\lvert \dfrac{dA}{dt} \right\rvert = \ell v = (0.20)(3.0) = 0.60 \ \mathrm{m^2/s}$ .
$\lvert \mathcal{E} \rvert = B\left\lvert \dfrac{dA}{dt} \right\rvert = (0.50)(0.60) = 0.30 \ \mathrm{V}$ .
(c) The flux into the page is now decreasing, so the induced current opposes the decrease by producing flux into the page inside the loop. By the right-hand rule that current is clockwise.
Compare with the entering phase, when the flux into the page was increasing and the current would run counterclockwise. Same speed, same field, same magnitude of emf, opposite direction of current.

Zero while the loop is fully inside, because the flux is constant. $0.30 \ \mathrm{V}$ while it crosses the boundary, with the current running clockwise. Motion through a magnetic field induces nothing on its own; a change in flux does.

Frequently asked questions

What is the difference between Faraday's law and Lenz's law?

Faraday's law gives the magnitude of an induced emf: it equals the rate at which the magnetic flux through a loop changes. Lenz's law gives the direction: the induced current flows so that the magnetic field it creates opposes the change in flux that produced it. They are not two independent laws. The minus sign in the printed equation, emf equals minus d phi B by dt, is Lenz's law, and the rest of the equation is Faraday's. The AP Physics C CED places both under the same learning objective, 13.2.A.

Is Lenz's law just the minus sign in Faraday's law?

Yes. On the AP Physics C: Electricity and Magnetism equation sheet, Faraday's law is printed as emf equals the closed line integral of E dot dl equals minus d phi B by dt. That minus sign is the whole content of Lenz's law: the induced effect opposes the change that caused it. The AP Physics 2 sheet prints the same law with absolute value bars and no sign, and then states Lenz's law separately in words as essential knowledge 12.4.A.4, which shows that the sign and the sentence are the same statement packaged two ways.

Why is there a negative sign in Faraday's law?

Because an induced current has to oppose the change that created it, or energy would not be conserved. If the induced current reinforced the change instead, moving a magnet toward a loop would produce a field that pulled the magnet in harder, which would increase the current, which would pull harder still, generating energy from nothing. The minus sign means the loop pushes back, so the work you do moving the magnet is exactly the electrical energy that appears in the circuit.

How do I use Lenz's law to find the direction of an induced current?

Three steps. First, decide which way the magnetic flux through the loop points and whether it is increasing or decreasing. Second, the induced current must create a field inside the loop that opposes that change, so it points opposite to the flux if the flux is growing and along the flux if the flux is shrinking. Third, use the right-hand rule: curl your fingers in the direction of current flow and your thumb gives the field it produces inside the loop, so pick the circulation that makes the field you need. The AP CED names the right-hand rule for this in essential knowledge 13.2.A.2.ii.

Can there be a magnetic field through a loop with no induced emf?

Yes, and this is a standard exam trap. The emf depends on the rate of change of flux, not on the flux itself. A loop sitting in the strongest steady field you can produce has zero induced emf. So does a loop translating entirely within a uniform field, since its flux never changes. So does a rotating loop at the instant when its plane faces the field squarely, which is when the flux is at a maximum and its rate of change is momentarily zero.

Is Lenz's law on the AP Physics 2 exam?

Yes. It appears as essential knowledge 12.4.A.4 in AP Physics 2, in Topic 12.4, Electromagnetic Induction and Faraday's Law, worded identically to the Physics C statement: Lenz's law is used to determine the direction of an induced emf resulting from a changing magnetic flux. What differs is the equation sheet. The Physics 2 sheet prints Faraday's law only in magnitude form with absolute value bars, so a Physics 2 student must supply the direction from Lenz's law in words rather than from a sign.

Should I write a negative emf as my answer?

Only if you have said what positive means for that loop. A signed emf is meaningless without a declared positive sense of circulation, and a grader cannot interpret a lone minus sign. The safer habit on free response is to compute the magnitude and then state the direction in words: clockwise as viewed from above, or from point a to point b through the resistor. What you must not do is carry the minus sign through the algebra and then also flip the answer using Lenz's law, since that applies the same physics twice and reverses a correct result.