# Faraday's Law of Induction

> AP Physics C: Electricity and Magnetism · AP CED E&M Unit 5
> Source: https://www.owlsprep.com/study/ap-physics-c-em-u5-faraday-s-law-of-induction/

This module covers core concepts of electromagnetic induction for AP Physics C E&M, including magnetic flux calculation, Faraday's Law, Lenz's Law, motional emf, and induced electric fields, with worked examples aligned to AP exam expectations.

**Prerequisites:** Magnetic field calculation for common geometries; Vector dot product operations; Ohm's law for DC circuits

## Learning objectives

- Calculate magnetic flux through closed loops
- Apply Faraday's Law to find magnitude of induced emf
- Use Lenz's law to determine direction of induced current
- Solve motional emf problems for moving conductors
- Calculate induced electric fields from changing magnetic fields

## What Is Faraday's Law of Induction?

Faraday’s Law of Induction describes electromagnetic induction, the process by which a changing magnetic field generates an electromotive force (emf) in a nearby conductor. This topic makes up 10–15% of the total AP Physics C E&M exam score, appearing in both multiple-choice and free-response questions, often integrated with circuit analysis, forces, or kinematics.

Faraday’s key experimental observation 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, confirming the reciprocal effect of Ampère’s law: changing magnetic fields produce electric effects.

## Magnetic Flux and Faraday's Core Formula

**Magnetic Flux** — A measure of the total magnetic field passing through a given closed area, equal to the integral of the dot product of the magnetic field vector and the area vector over the entire area. SI unit is the weber (Wb = 1 T·m²).

*Notation:* $\Phi_B$

*Example:* For uniform $\vec{B}$, simplifies to $\Phi_B = BA\cos\theta$, where $\theta$ is the angle between $\vec{B}$ and the normal to the loop area.

For any arbitrary area, the differential flux through a small area element $d\vec{A}$ is:

$$d\Phi_B = \vec{B} \cdot d\vec{A}$$

Integrating over the entire loop area gives total magnetic flux:

$$\Phi_B = \int_A \vec{B} \cdot d\vec{A}$$

Faraday’s law of induction states that the induced emf $\varepsilon$ 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:

$$\varepsilon = -N \frac{d\Phi_B}{dt}$$

Any change that alters $\Phi_B$ will produce an induced emf: 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:

   $$A = \pi r^2 = \pi (0.10)^2 = 0.01\pi \text{ m}^2$$
2. Since $\vec{B}$ is parallel to the normal of the coil, $\cos\theta = 1$, 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:

   $$|\varepsilon| = N \left|\frac{d\Phi_B}{dt}\right| = 50 \times 0.01\pi \times \frac{d}{dt}\left(0.5t^2 + 0.2t\right) = 0.5\pi (t + 0.2)$$
4. Substitute $t=3.0$ s to get the final magnitude:

   $$|\varepsilon| = 0.5\pi (3.0 + 0.2) = 1.6\pi \approx 5.0 \text{ V}$$

> **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.

## 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*.

> **note**
>
> A critical point: 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.

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, always use the 4-step Lenz process above—avoid guessing based on intuition that leads to mistakes.

## 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:

$$\varepsilon = BLv$$

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 $\frac{d\Phi_B}{dt} = BLv$, matching the force-derived result. This confirms motional emf is just a special case of Faraday’s law, not a separate rule. If velocity is not perpendicular to B, the general expression is $\varepsilon = BLv\sin\theta$, 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:

   $$\varepsilon = BLv = (0.40\ \text{T})(0.50\ \text{m})(2.0\ \text{m/s}) = 0.40\ \text{V}$$
3. Apply Ohm’s law to the closed loop (assuming negligible resistance for the rod and rails):

   $$I = \frac{\varepsilon}{R} = \frac{0.40\ \text{V}}{10.0\ \Omega} = 0.040\ \text{A}$$
4. Confirm via Faraday’s law: area of the loop is $A = L x(t)$, so $\frac{dA}{dt} = Lv$, giving $\frac{d\Phi_B}{dt} = BLv$, the same emf result.

> **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.

## 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:

   $$\oint \vec{E} \cdot d\vec{l} = E \times (2\pi r)$$
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) and solve for E:

   $$E (2\pi r) = \frac{dB}{dt} \pi r^2 \implies E = \frac{r}{2} \frac{dB}{dt} = \frac{0.10\ \text{m}}{2} \times 0.10\ \text{T/s} = 0.0050\ \text{N/C}$$

> **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.

## Common pitfalls

- **Wrong:** Stating that the induced magnetic field always points opposite the direction of the original magnetic field through the loop.
  - Why it fails: Students memorize 'oppose' from Lenz's law and forget it opposes the change in flux, not the original field.
  - Correct: 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:** Forgetting to multiply by N (number of turns) when calculating emf for a multi-turn coil.
  - Why it fails: Students remember Faraday's law as $\varepsilon = -d\Phi/dt$ and leave out the N factor for multi-turn coils, which is common in AP problems.
  - Correct: 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:** Calculating flux as $BA\sin\theta$ instead of $BA\cos\theta$, when $\theta$ is measured between B and the plane of the loop.
  - Why it fails: Students mix up the angle definition: the formula uses the angle between B and the normal to the loop, not the plane.
  - Correct: 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:** Using $\varepsilon = BLv$ when v is parallel to B or parallel to the rod length.
  - Why it fails: Students use the simplified formula for all motional emf problems without checking angles, leading to wrong magnitudes.
  - Correct: Always confirm v, B, and L are mutually perpendicular; if not, include $\sin\theta$ or use general Faraday's law to calculate flux change.
- **Wrong:** Claiming induced electric fields are conservative, like electrostatic fields.
  - Why it fails: Students generalize properties of electrostatic fields to all electric fields, which is incorrect for induced fields.
  - Correct: 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.

## Cheatsheet

| Concept | Formula | Key Notes |
| --- | --- | --- |
| Magnetic Flux (uniform B) | $\Phi_B = BA\cos\theta$ | $\theta$ = angle between B and normal to area |
| Faraday's Law (multi-turn) | $\varepsilon = -N \frac{d\Phi_B}{dt}$ | Negative sign encodes direction (Lenz's) |
| Motional Emf (mutually perpendicular) | $\varepsilon = BLv$ | Special case of Faraday's Law |
| Induced E (r < R, cylindrical B) | $E = \frac{r}{2} \frac{dB}{dt}$ | R = radius of the cylinder |
| Induced E (r > R, cylindrical B) | $E = \frac{R^2}{2r} \frac{dB}{dt}$ | Flux only through area $\pi R^2$ |
| Lenz's Law Core Rule | Opposes change in flux, not flux itself | Increase: opposite B; Decrease: same B as original |

## What's next

Faraday's Law is the foundation for all electromagnetic induction concepts in AP Physics C E&M, and it connects directly to next topics including Lenz's law applications, advanced motional emf problems, inductance, inductors, RL circuits, and Maxwell's equations, which complete the description of electromagnetic phenomena. Mastery of Faraday's Law and Lenz's Law is critical for solving all subsequent induction problems, from energy stored in inductors to electromagnetic wave propagation, and makes up a significant portion of the exam score.

- [Lenz's Law](https://www.owlsprep.com/study/ap-physics-c-em-u5-lenz-s-law/)
- [Inductance](https://www.owlsprep.com/study/ap-physics-c-em-u5-inductance/)
- [Maxwell's Equations](https://www.owlsprep.com/study/ap-physics-c-em-u5-maxwell-s-equations/)

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