Electromagnetic Induction

AP Physics 2· AP Physics 2 CED — Magnetism and Electromagnetic Induction· 14 min read

1. What Is Electromagnetic Induction?★★☆☆☆⏱ 2 min

Electromagnetic induction is the generation of an electromotive force (emf) and resulting induced current in a conductor exposed to a changing magnetic environment. Unlike batteries that produce emf from chemical energy, induction converts mechanical or changing magnetic energy into electrical energy, forming the physical basis for all modern generators, transformers, and wireless charging technology.

Per the AP Physics 2 CED, this topic makes up 10-15% of the total exam score within Unit 5, and appears regularly in both multiple-choice (MCQ) and free-response (FRQ) sections, often combined with prior concepts of electric circuits and magnetic fields.

📘 Definition

Electromagnetic Induction

Generation of an electric potential (emf) and induced current in a conductor exposed to a changing magnetic environment

Example:

A moving bar magnet near a conducting loop induces a measurable current

2. Magnetic Flux★★☆☆☆⏱ 3 min

Magnetic flux is the measure of the total magnetic field passing through a given area, and it is the change in flux that drives electromagnetic induction, per Faraday's Law. For a flat surface of area $A$ in a uniform magnetic field, the formula for magnetic flux is:

Phi_B = \vec{B} \cdot \vec{A} = B A \cos\theta

where $θ$ is explicitly defined as the angle between the magnetic field vector $B$ and the normal vector (perpendicular) to the surface. If the magnetic field is parallel to the surface, $θ = 9 0^{\circ}$ , so $cos θ = 0$ , and flux is zero even for very large $B$ . If $B$ is perpendicular to the surface, $θ = 0^{\circ}$ , $cos θ = 1$ , so $Φ_{B} = B A$ , the maximum possible flux for given $B$ and $A$ . Units of flux are webers (Wb), where $1 Wb = 1 T \cdot m^{2}$ . Three independent changes can alter magnetic flux to produce induction: a change in $B$ , a change in $A$ , or a change in $θ$ .

📐 Worked Example

A square conducting loop with side length 0.2 m is placed in a uniform 0.5 T magnetic field. The plane of the loop is tilted so that the angle between the plane of the loop and the magnetic field is 30°. What is the magnetic flux through the loop?

1
The flux formula uses the angle between $B$ and the normal to the plane, not the angle between $B$ and the plane itself. For the given 30° angle between B and the plane, the angle between B and the normal is
$90^\circ - 30^\circ = 60^\circ$
2
Calculate the area of the square:
$A = (0.2\ \text{m})^2 = 0.04\ \text{m}^2$
3
Substitute into the flux formula:
$Phi_B = BA\cos\theta = (0.5\ \text{T})(0.04\ \text{m}^2)\cos(60^\circ)$
4
Simplify with $cos (6 0^{\circ}) = 0.5$ :
$Phi_B = 0.5 \cdot 0.04 \cdot 0.5 = 0.01\ \text{Wb}$

Exam tip:

Always confirm whether the problem gives the angle relative to the plane or the normal of the loop — 70% of student flux calculation errors come from mixing these two angles up.

3. Faraday's Law of Induction★★★☆☆⏱ 3 min

Faraday's Law is the core quantitative rule that relates changing magnetic flux to the magnitude of the induced emf. For a single conducting loop, the magnitude of the induced emf equals the magnitude of the rate of change of magnetic flux over time:

|\varepsilon| = \left| \frac{\Delta \Phi_B}{\Delta t} \right|

For a coil with $N$ identical turns of wire, the total emf is $N$ times the emf of a single turn, since each turn adds its emf in series:

|\varepsilon| = N \left| \frac{\Delta \Phi_B}{\Delta t} \right|

Faraday's Law gives the magnitude of induced emf, but not the direction of the resulting induced current. If the conductor forms a closed loop with total resistance $R$ , the magnitude of the induced current is $I = ∣ ε ∣/ R$ , directly from Ohm's Law. AP Physics 2 problems most commonly test Faraday's Law for three scenarios: uniform $B$ changing at a constant rate, loop area changing at a constant rate, or a loop rotating at constant angular velocity.

📐 Worked Example

A 50-turn circular coil has radius 0.1 m. The magnetic field through the coil is perpendicular to the plane of the coil, and increases at a constant rate from 0.1 T to 0.6 T over 0.25 s. The total resistance of the coil is 2.0 Ω. What is the magnitude of the induced current in the coil?

1
Calculate the area of the coil, with $θ = 0^{\circ}$ so $cos θ = 1$ :
$A = \pi r^2 = \pi (0.1\ \text{m})^2 \approx 0.0314\ \text{m}^2$
2
Calculate the change in flux per turn:
$\Delta \Phi_B = A\Delta B = 0.0314\ \text{m}^2 (0.6\ \text{T} - 0.1\ \text{T}) = 0.0157\ \text{Wb}$
3
Apply Faraday's Law for $N$ turns to find induced emf:
$|\varepsilon| = N |\Delta \Phi_B / \Delta t| = 50 \cdot (0.0157\ \text{Wb} / 0.25\ \text{s}) \approx 3.14\ \text{V}$
4
Use Ohm's Law to find current (2 significant figures):
$I = \varepsilon / R = 3.14\ \text{V} / 2.0\ \Omega \approx 1.6\ \text{A}$

Exam tip:

If the problem describes a multi-turn coil, always scan for the number of turns $N$ and explicitly include it in your calculation — forgetting the $N$ multiplier is the second most common error on AP induction problems.

4. Lenz's Law for Direction of Induced Current★★★☆☆⏱ 3 min

📘 Definition

Lenz's Law

The induced current will flow in a direction such that the magnetic field produced by the induced current opposes the change in the original magnetic flux that created it.

Example:

An approaching north pole induces a current that creates a repelling north pole facing the magnet

A common student misstatement is that the induced field "opposes the original magnetic field" — this is incorrect; it opposes the change in flux, not the field itself. To apply Lenz's Law correctly, follow this three-step process: (1) Determine the direction of the original magnetic field through the loop, and whether flux is increasing or decreasing. (2) The induced magnetic field will point opposite to the change: if original flux is increasing, induced $B$ points opposite original $B$ ; if original flux is decreasing, induced $B$ points in the same direction as original $B$ . (3) Use the right-hand rule for current-carrying loops to find the direction of induced current that produces the required induced $B$ .

📐 Worked Example

A bar magnet is moving toward a circular conducting loop along the loop's axis, with the north pole facing the loop. What is the direction of the induced current in the loop, as viewed from the side of the approaching bar magnet?

1
The original magnetic field from the bar magnet points outward from the north pole, so through the loop, original $B$ points toward the viewer (on the side the magnet is approaching from). The magnet is moving toward the loop, so flux through the loop is increasing.
2
By Lenz's Law, the induced magnetic field must oppose the increase in flux, so induced $B$ points opposite to the original $B$ , which is away from the viewer (toward the bar magnet).
3
Use the right-hand rule for current loops: point your right thumb in the direction of the induced $B$ (away from the viewer), and your fingers curl in the direction of current. This gives a clockwise current direction when viewed from the side of the approaching magnet.

Exam tip:

Always start with "is flux increasing or decreasing?" before predicting direction — this one question eliminates 90% of direction errors.

5. Motional Emf★★★☆☆⏱ 3 min

Motional emf is the emf induced in a moving conductor in a magnetic field, a special case of Faraday's Law where flux changes because the area of the conducting loop changes. For a conducting rod of length $L$ moving with speed $v$ perpendicular to a uniform magnetic field $B$ , the motional emf across the ends of the rod is:

\varepsilon = BLv

This formula can be derived directly from Faraday's Law: if the rod slides along parallel conducting rails to form a closed loop, the area of the loop increases by $Δ A = Lv Δ t$ in time $Δ t$ , so $Δ Φ_{B} = B Δ A = B Lv Δ t$ , and $ε = Δ Φ_{B} /Δ t = B Lv$ . For a closed loop with total resistance $R$ , the induced current is $I = ε / R = B Lv / R$ .

📐 Worked Example

A 0.5 m long conducting bar slides at constant speed 4 m/s along two parallel conducting rails connected by a 10 Ω resistor at one end. The entire system is in a uniform 0.2 T magnetic field perpendicular to the plane of the loop. What is the magnitude and direction of the current through the resistor, if the bar slides away from the resistor?

1
Calculate motional emf:
$\varepsilon = BLv = (0.2\ \text{T})(0.5\ \text{m})(4\ \text{m/s}) = 0.4\ \text{V}$
2
Use Ohm's Law for current magnitude:
$I = \varepsilon / R = 0.4\ \text{V} / 10\ \Omega = 0.04\ \text{A}$
3
For direction: flux into the plane of the loop increases as the bar slides away from the resistor. By Lenz's Law, induced B points out of the plane, so current flows counterclockwise around the loop, which means it flows from top to bottom through the resistor.

Exam tip:

Motional emf only exists for velocity components perpendicular to both B and the length of the rod. If the rod moves parallel to its own length, the induced emf is zero.

6. Common Pitfalls

Wrong move:

Using the angle between the magnetic field and the plane of the loop directly as $θ$ in the flux formula

Why:

$θ$ is defined relative to the normal vector, not the plane. Students often use the given diagram angle directly without adjustment.

Correct move:

Always subtract the given angle from 90° if it is provided relative to the plane of the loop before plugging into the flux formula.

Wrong move:

Forgetting to multiply induced emf by the number of turns $N$ for a multi-turn coil

Why:

Students remember Faraday's Law for a single loop and stop, missing that emf adds in series across multiple turns.

Correct move:

Scan any problem mentioning a coil for the number of turns, write $N$ at the top of your working, and always multiply by $N$ when calculating total emf.

Wrong move:

Stating that Lenz's Law requires the induced magnetic field to oppose the original magnetic field

Why:

The common phrasing "opposes the change" is often misremembered as opposing the field, leading to wrong direction when flux is decreasing.

Correct move:

Always first confirm if flux is increasing or decreasing; only oppose the original field when flux is increasing.

Wrong move:

Calculating motional emf as $B Lv$ when the rod moves parallel to its length

Why:

Students memorize $B Lv$ and forget the requirement that $v$ , $B$ , and $L$ are all mutually perpendicular.

Correct move:

If velocity is parallel to the rod's length, no flux change occurs, so induced emf is zero.

Wrong move:

Assuming any moving conductor in a magnetic field has an induced current

Why:

Students confuse induced emf with induced current.

Correct move:

Check if the conductor forms a closed loop; an isolated moving rod has induced emf (potential difference) but no induced current.

7. Quick Reference Cheatsheet

Category	Formula / Rule	Key Notes
Magnetic Flux	$Φ_{B} = B A cos θ$	$θ$ = angle between $B$ and normal to surface; units: webers (Wb)
Faraday's Law (N-turn coil)	$∣ ε ∣ = N \frac{Δ Φ _{B}}{Δ t}$	Always multiply by number of turns N
Lenz's Law	Induced B opposes change in flux, not original B	Check if flux is increasing or decreasing first
Motional Emf (perpendicular case)	$ε = B Lv$	v, B, L must all be mutually perpendicular
Induced Current	$I = \frac{∣ ε ∣}{R}$	Only exists for closed conducting loops

When this came up on past exams

AI-estimated based on syllabus patterns — cross-check with official past papers for accuracy. Use only as revision-focus signals.

2023 · AP Physics 2
FRQ: motional emf terminal speed
2022 · AP Physics 2
MCQ: Lenz's Law direction

What's Next

Electromagnetic induction is the foundation of all modern electrical power generation and a high-weight topic on the AP Physics 2 exam, often combined with concepts from electric circuits, forces, and energy conservation. Mastering the core rules of flux, Faraday's Law, and Lenz's Law will prepare you for both multiple-choice and multi-part free-response questions that combine multiple concepts in this unit. To build mastery, next explore applications of induction, review foundational concepts from earlier in the unit, or practise mixed problem types to identify knowledge gaps ahead of exam day.