Faraday's Law of Induction — AP Physics C: E&M Study Guide

For: AP Physics C: E&M candidates sitting AP Physics C: E&M.

Covers: magnetic flux calculation, Faraday’s Law of Induction formula and interpretation, Lenz’s law for induced current direction, motional emf, induced electric fields, and core problem-solving techniques for exam questions.

You should already know: Magnetic field calculation for common geometries, dot product for vector operations, Ohm’s law for DC circuits.

A note on the practice questions: All worked questions in the "Practice Questions" section below are original problems written by us in the AP Physics C: E&M style for educational use. They are not reproductions of past College Board / Cambridge / IB papers and may differ in wording, numerical values, or context. Use them to practise the technique; cross-check with official mark schemes for grading conventions.

1. What Is Faraday's Law of Induction?

Faraday’s Law of Induction is the fundamental relationship describing electromagnetic induction, the process by which a changing magnetic field generates an electromotive force (emf) in a nearby conductor. Per the AP Physics C: E&M Course and Exam Description (CED), this topic makes up 10–15% of the total exam score, and it appears in both multiple-choice (MCQ) and free-response (FRQ) sections, often integrated with circuit analysis, force, or kinematics problems.

The core observation from Faraday’s experiments is that only a changing magnetic flux through a closed loop produces an induced emf; a constant magnetic flux (even a very strong one) produces no emf. This principle completed the unification of electricity and magnetism: prior to Faraday’s work, it was known that electric currents produce magnetic fields (via Ampère’s law), and Faraday’s law confirmed the reciprocal effect: changing magnetic fields produce electric effects. On the AP exam, this topic tests both conceptual understanding (direction of induced current) and quantitative calculation (magnitude of induced emf, current, or force).

2. Magnetic Flux and the Faraday's Law Core Formula

Magnetic flux is the quantity that captures how much magnetic field passes through a closed loop, and it is the core input to Faraday’s law. For any arbitrary area $A$, the differential magnetic flux through a small area element $d\vec{A}$ (where the direction of $d\vec{A}$ is normal to the area element by convention) is: $$d{\Phi }_B=\vec{B}\cdot d\vec{A}$$ Integrating over the entire area of the loop gives the total magnetic flux: $${\Phi }_B={\int }_A\vec{B}\cdot d\vec{A}$$ If the magnetic field is uniform over the entire area, this simplifies to ${\Phi }_B=BA\cos \theta$, where $\theta$ is the angle between the $\vec{B}$ vector and the normal to the loop area. The SI unit of magnetic flux is the weber (Wb), equal to 1 T·m².

Faraday’s law of induction states that the induced emf $\epsilon$ in a closed coil of $N$ identical loops is equal to the negative of the total rate of change of magnetic flux through the coil: $$\epsilon =-N\frac{d{\Phi }_B}{dt}$$ Any change that alters ${\Phi }_B$ will produce an induced emf: this includes changing the magnitude of $\vec{B}$, changing the area of the loop, rotating the loop (changing $\theta$), or moving the loop into or out of a magnetic field. The negative sign in the formula encodes the direction of the induced emf, which is governed by Lenz’s law.

Worked Example

A circular coil with 50 turns has a radius of 0.10 m, and lies flat in the $xy$-plane. A uniform magnetic field aligned along the $+z$-axis through the coil changes with time as $B(t)=0.5t^2+0.2t$ (in tesla, for $t$ in seconds). What is the magnitude of the induced emf in the coil at $t=3.0$ s?

1. Calculate the cross-sectional area of the coil: $A=\pi r^2=\pi (0.10{)}^2=0.01\pi$ m².
2. Since $\vec{B}$ is parallel to the normal of the coil, $\cos \theta =\cos 0=1$, so the flux through one turn is ${\Phi }_B=B(t)A=(0.5t^2+0.2t)(0.01\pi )$.
3. Apply Faraday’s law for magnitude: $$|\epsilon |=N\left|\frac{d{\Phi }_B}{dt}\right|=50\times 0.01\pi \times \frac{d}{dt}\left(0.5t^2+0.2t\right)=0.5\pi (t+0.2)$$
4. Substitute $t=3.0$ s: $|\epsilon |=0.5\pi (3.0+0.2)=1.6\pi \approx 5.0$ V.

Exam tip: When the AP exam asks only for the magnitude of induced emf, you do not need to include the negative sign from Faraday’s law—graders only award points for the correct numerical magnitude in these cases.

3. Lenz's Law for Induced Current Direction

The negative sign in Faraday’s law is interpreted by Lenz’s law, which gives the direction of the induced emf and induced current in a conducting loop. Lenz’s law states: The induced current flows in a direction that creates an induced magnetic field that opposes the change in magnetic flux that produced the induced current. A critical point here is that Lenz’s law opposes the change in flux, not the flux itself: an increasing flux gets an opposing induced field, while a decreasing flux gets an induced field that reinforces the original field to oppose the decrease.

A reliable 4-step process to apply Lenz’s law on exam problems is:

1. Identify the direction of the original external magnetic field through the loop.
2. Determine if the total magnetic flux through the loop is increasing or decreasing over time.
3. Find the direction of the required induced magnetic field: if flux increases, induced B is opposite original B; if flux decreases, induced B is in the same direction as original B.
4. Use the right-hand rule for current-carrying loops to get the direction of the induced current from the induced B direction.

Worked Example

A rectangular conducting loop is pulled at constant speed to the right out of a region of uniform magnetic field that points into the page. Only the left portion of the loop is still in the field region when it is being pulled. What is the direction of the induced current in the loop?

1. Original magnetic field through the loop is directed into the page.
2. As the loop is pulled to the right out of the field, the area of the loop inside the field decreases, so the total magnetic flux into the page through the loop is decreasing.
3. To oppose the decrease in flux into the page, the induced magnetic field must point in the same direction as the original field: into the page.
4. Right-hand rule: point your right thumb in the direction of the induced B (into the page), and your fingers curl clockwise around the loop. So the induced current is clockwise.

Exam tip: If a problem asks for the direction of induced current around a loop, you can always use the 4-step Lenz process above—avoid guessing based on intuition, as this leads to mistakes for decreasing flux problems.

4. Motional Emf

Motional emf is the emf induced in a conductor moving through a constant magnetic field, and it is one of the most common special cases of Faraday’s law tested on the AP exam. Motional emf arises because the magnetic force acts on free charges in the moving conductor: ${\vec{F}}_B=q\vec{v}\times \vec{B}$, which separates positive and negative charges to opposite ends of the conductor, creating a potential difference (emf) across the conductor.

For a straight conducting rod of length $L$ moving with constant speed $v$ through a uniform magnetic field $B$, with $\vec{v}$, $\vec{L}$, and $\vec{B}$ all mutually perpendicular, the magnitude of the motional emf is: $$\epsilon =BLv$$ This result can be derived directly from Faraday’s law for the case of a rod sliding on a fixed U-shaped conducting rail (forming a closed loop): the area of the loop changes at a rate $\frac{dA}{dt}=Lv$, so the rate of change of flux is $\frac{d{\Phi }_B}{dt}=B\frac{dA}{dt}=BLv$, giving $\epsilon =BLv$, matching the force-derived result. This confirms motional emf is just a special case of Faraday’s law, not a separate rule. If the velocity is not perpendicular to B, the general expression is $\epsilon =BLv\sin \theta$, where $\theta$ is the angle between $\vec{v}$ and $\vec{B}$.

Worked Example

A conducting rod of length 0.50 m slides at constant speed $v=2.0$ m/s without friction on two parallel horizontal conducting rails. The rails are connected at one end by a $10.0$ Ω resistor, forming a closed rectangular loop. A uniform 0.40 T magnetic field points perpendicular to the plane of the loop. What is the magnitude of the induced current in the resistor?

1. All three quantities $B$, $L$, and $v$ are mutually perpendicular, so we can use the motional emf formula directly.
2. Calculate the induced emf: $\epsilon =BLv=(0.40\text{T})(0.50\text{m})(2.0\text{m/s})=0.40$ V.
3. Apply Ohm’s law to the closed loop (assuming the rod and rails have negligible resistance): $I=\frac{\epsilon }{R}=\frac{0.40\text{V}}{10.0\Omega }=0.040$ A.
4. Confirm via Faraday’s law: the area of the loop is $A=Lx(t)$, where $x(t)$ is the position of the rod along the rails. $\frac{dA}{dt}=L\frac{dx}{dt}=Lv$, so $\frac{d{\Phi }_B}{dt}=BLv$, giving the same emf as above.

Exam tip: For problems where a conducting rod moves at an angle to the magnetic field, remember that only the component of velocity perpendicular to both B and the rod length contributes to the motional emf—always check the angle before plugging into $\epsilon =BLv$.

5. Induced Electric Fields

Faraday’s law tells us that a changing magnetic field creates an electric field, even in empty space where there is no conductor and no current. This induced electric field is fundamentally different from the electrostatic field produced by stationary charges: it is non-conservative, meaning the work done to move a charge around a closed path is non-zero.

The general form of Faraday’s law, written in terms of the induced electric field, is: $$\oint \vec{E}\cdot d\vec{l}=-\frac{d{\Phi }_B}{dt}$$ The left-hand side is the line integral of the induced electric field around a closed loop, which equals the induced emf around the loop. Symmetry is almost always used to solve for induced electric fields, since the magnitude of E is constant along concentric loops for symmetric changing magnetic fields (like a uniform B changing inside a cylinder).

Worked Example

A uniform magnetic field is confined to the volume of a long cylinder of radius $R=0.20$ m. The magnitude of the field increases at a constant rate $\frac{dB}{dt}=0.10$ T/s, aligned along the cylinder axis. What is the magnitude of the induced electric field at a point $r=0.10$ m from the cylinder axis?

1. By symmetry, the induced electric field is tangential to any circular loop of radius $r$ centered on the cylinder axis, and has constant magnitude E along the loop.
2. Evaluate the line integral: $\oint \vec{E}\cdot d\vec{l}=E\times (2\pi r)$, since $\vec{E}$ is parallel to $d\vec{l}$ everywhere on the loop.
3. Calculate the rate of change of flux through the loop: $\frac{d{\Phi }_B}{dt}=\frac{dB}{dt}\times A=\frac{dB}{dt}(\pi r^2)$.
4. Equate the two sides from Faraday’s law (taking magnitude): $E(2\pi r)=\frac{dB}{dt}\pi r^2$. Solve for E: $$E=\frac{r}{2}\frac{dB}{dt}=\frac{0.10\text{m}}{2}\times 0.10\text{T/s}=0.0050\text{N/C}$$

Exam tip: For points outside the cylinder (r > R) where B is zero, the flux through the loop is only $B\pi R^2$, so $E=\frac{R^2}{2r}\frac{dB}{dt}$—don't use the r < R formula for outside points.

6. Common Pitfalls (and how to avoid them)

• Wrong move: Stating that the induced magnetic field always points opposite the direction of the original magnetic field through the loop. Why: Students memorize "oppose" from Lenz's law and forget it opposes the change, not the original field. Correct move: Always check if flux is increasing or decreasing: if flux is decreasing, induced B points in the same direction as original B to oppose the decrease.
• Wrong move: Forgetting to multiply by N (number of turns) when calculating emf for a coil with multiple loops. Why: Students remember Faraday's law as $\epsilon =-d\Phi /dt$ and leave out the N factor for multi-turn coils, which is standard in many exam problems. Correct move: Always check the problem statement for number of turns, and multiply the rate of change of single-turn flux by N before calculating emf.
• Wrong move: Calculating flux as $BA\sin \theta$ instead of $BA\cos \theta$, because $\theta$ is measured between B and the plane of the loop. Why: Students mix up the angle definition: the formula uses the angle between B and the normal to the loop, not the plane. Correct move: Always draw the normal vector to the loop, measure the angle between B and the normal, then plug $\cos \theta$ into the flux formula.
• Wrong move: Using $\epsilon =BLv$ when v is parallel to B or parallel to the rod length. Why: Students use the simplified formula for all motional emf problems without checking angles, leading to wrong magnitudes. Correct move: Always confirm v, B, and L are mutually perpendicular; if not, include $\sin \theta$ where $\theta$ is the angle between v and B, or use general Faraday's law to calculate flux change.
• Wrong move: Claiming induced electric fields are conservative, like electrostatic fields. Why: Students generalize properties of electrostatic fields to all electric fields, which is incorrect for induced fields. Correct move: Remember that induced electric fields are non-conservative: work done around a closed loop is non-zero, so they cannot be described by a scalar potential like electrostatic fields.

7. Practice Questions (AP Physics C: E&M Style)

Question 1 (Multiple Choice)

A square loop of side length $a$ is moved at constant speed $v$ from left to right through a region of uniform magnetic field $B$ perpendicular to the loop, pointing into the page. The region of the field is wider than $2a$, so the entire loop is fully inside the field before any part of the loop exits. Which of the following correctly describes the induced emf $\epsilon$ as a function of the position $x$ of the right edge of the loop, where $x=0$ when the right edge enters the field? A) $\epsilon$ is constant non-zero for $0<x<a$, zero for $a<x<2a$ B) $\epsilon$ is constant non-zero for $0<x<2a$, zero for $x>2a$ C) $\epsilon$ is zero for $0<x<a$, constant non-zero for $a<x<2a$ D) $\epsilon$ is linearly increasing for $0<x<a$, constant for $a<x<2a$

Worked Solution: We analyze flux change in each interval: For $0<x<a$, only part of the loop is inside the field, and the area of the loop inside the field increases at a constant rate $\frac{dA}{dt}=av$, so the rate of change of flux $\frac{d{\Phi }_B}{dt}=Bav$ is constant, giving a constant non-zero induced emf. For $a<x<2a$, the entire loop is inside the uniform field, so total flux through the loop is constant, hence $\frac{d{\Phi }_B}{dt}=0$ and induced emf is zero. This matches option A. Correct answer: A

Question 2 (Free Response)

A rectangular loop of mass $m$, resistance $R$, height $h$, and width $w$ is dropped from rest from a height $y_0$ above the top edge of a region of uniform magnetic field $B$ pointing into the page, perpendicular to the plane of the loop. The magnetic field region is very deep, so the entire loop eventually falls into the field. Answer the following: (a) Find the speed of the bottom edge of the loop just as it enters the magnetic field region, after falling $y_0$ from rest. (b) Find the magnitude of the induced emf and induced current in the loop at this instant (before the entire loop is inside the field). (c) Find the magnitude of the magnetic force on the loop at this instant, and state its direction.

Worked Solution: (a) Use kinematics for free fall from rest: $v^2=v_0^2+2g{y}_0$. With initial speed $v_0=0$, solve for $v$: $$v=\sqrt{2g{y}_0}$$

(b) The bottom edge of the loop is a moving conductor of length $w$, with speed $v$ perpendicular to $B$, so motional emf is: $$\epsilon =Bwv=Bw\sqrt{2g{y}_0}$$ By Ohm's law, induced current is: $$I=\frac{\epsilon }{R}=\frac{Bw\sqrt{2g{y}_0}}{R}$$

(c) Magnetic force on a current-carrying conductor is $F=IwB$. Substitute $I$ from part (b): $$F=\left(\frac{Bw\sqrt{2g{y}_0}}{R}\right)wB=\frac{B^2{w}^2\sqrt{2g{y}_0}}{R}$$ By Lenz's law, the force opposes the change in flux (the motion of the loop falling into the field), so the direction of the force is upward.

Question 3 (Application / Real-World Style)

A small portable generator rotates a 100-turn circular coil of radius 0.1 m at 60 Hz (60 rotations per second) in a uniform 0.1 T magnetic field. Estimate the peak (maximum) induced emf produced by this generator, which is designed to power small household devices.

Worked Solution: When a coil rotates at angular frequency $\omega =2\pi f$, the flux through the coil is ${\Phi }_B=NBA\cos (\omega t)$, where $A=\pi r^2$ is the area of the coil. Calculate values: $A=\pi (0.1{)}^2=0.01\pi$ m², $\omega =2\pi (60)=120\pi$ rad/s. By Faraday's law, $\epsilon =-\frac{d{\Phi }_B}{dt}=NBA\omega \sin (\omega t)$, so peak emf is ${\epsilon }_{\text{max}}=NBA\omega$. Substitute values: ${\epsilon }_{\text{max}}=(100)(0.1)(0.01\pi )(120\pi )\approx 12\times 9.87\approx 120$ V. This small generator produces a peak emf of ~120 V, which matches the standard mains voltage required for small household devices in North America.

8. Quick Reference Cheatsheet

9. What's Next

Faraday’s Law of Induction is the foundational prerequisite for all remaining topics in the AP Physics C: E&M electromagnetism unit. Immediately after this, you will study self-induction, inductors, and RL circuits, where the induced emf from changing current in an inductor is directly derived from Faraday’s law. Without mastering flux change calculation and Lenz’s law direction from this chapter, you will not be able to correctly set up and solve the differential equations for RL circuits, which are a common multi-point FRQ topic. Faraday’s law is also one of the four core Maxwell’s equations that unify all of electromagnetism, so it is central to the entire conceptual framework of the course.

Follow-up topics to study next: Inductance and RL Circuits Maxwell's Equations Electromagnetic Induction Applications