AP Physics C: E&M Unit 6: Electromagnetic Induction Notes

Introduction

Electromagnetic induction connects changing magnetic flux to circulating electric fields, induced current, mechanical resistance, inductive energy, and LR or LC circuit dynamics.

Flux is an oriented surface integral, so its sign comes from a chosen area normal. Faraday’s law then assigns emf relative to the matching positive loop traversal. Lenz’s law is not a separate numerical rule: it is the physical meaning of the negative derivative in that signed equation.

Inductors store magnetic energy and oppose changes in current. A resistor turns that transient into exponential rise or decay, while an ideal capacitor–inductor pair exchanges electric and magnetic energy sinusoidally. Initial conditions and real-component limits decide which solution describes the circuit.

Magnetic field through an oriented surface

magnetic flux

Flux changing in time

induced emf and circulating electric field

Closed conducting path

induced current, force, and energy transfer

Flux linked to the circuit’s own current

inductance

Resistance or capacitance with inductance

LR transients or LC oscillation

Current exam guidance: Aligned to the current 2026–27 AP Physics C: Electricity and Magnetism framework, with calculus-based field, circuit, and induction models, explicit assumptions, and experimental reasoning.

13.1

Magnetic Flux

How does an oriented surface convert a magnetic field into magnetic flux?

Short answer

Magnetic flux is the signed surface integral $\Phi_B=\int_S\vec B\cdot d\vec A$ ; for a uniform field through a flat surface it is $BA\cos\theta$ , where $\theta$ is measured from the area normal.

Know firstMagnetic fields · Dot products · Surface orientation

The area vector carries the sign convention

Divide a surface into patches and assign each patch a normal vector. The local contribution is $d\Phi_B=\vec B\cdot d\vec A$ , so only the field component normal to the surface contributes. For a flat area in a uniform field, the integral reduces to $BA\cos\theta$ . The angle is between $\vec B$ and the normal, not between the field and the plane.

An open surface has two possible normals. Reversing the chosen normal reverses flux but does not change the physical field. For a loop, the right-hand rule links the chosen normal to positive boundary traversal. This orientation pairing becomes essential when Faraday’s law assigns the sign of induced emf.

Magnetic flux is measured in webers, with $1\ \mathrm{Wb}=1\ \mathrm{T\,m^2}$ . It is not the number of drawn field lines or an amount of magnetic material. Since magnetic field lines close, the net flux through every closed surface is zero even when the field is strong and nonuniform on that surface.

Physics model

Magnetic flux depends on surface orientation

Rotate a 0.040 m² loop in a uniform 0.20 T magnetic field. Signed flux follows the field component along the chosen area normal.

Field-to-normal angle60 °

Angle: 60°
Normal field: 0.1 T
Magnetic flux: 4 mWb

magnetic fieldoriented surfacearea normalangle θ

\Phi_B=\int_S\vec B\cdot d\vec A=BA\cos\theta

Read this figure without the graphic

Uniform magnetic-field arrows meet an oriented loop surface. The chosen area normal and its angle with the field determine the signed magnetic flux.

Current quantity	Value
Angle	60°
Normal field	0.1 T
Magnetic flux	4 mWb
Model relation	$\Phi_B=\int_S\vec B\cdot d\vec A=BA\cos\theta$

Evaluate flux from vector components

System: A flat oriented surface in a uniform magnetic field
Frame: Right-handed Cartesian coordinates
Assumptions: The field and area vector are constant across the surface
Givens: $\vec B=(0.2,0.3,0.4)$ T and $\vec A=(0.01,-0.02,0.03)$ m $^2$

Multiply matching components.
Add with signs: $0.2(0.01)+0.3(-0.02)+0.4(0.03)$ .
Check that reversing $\vec A$ reverses the result.

Result

\Phi_B=\vec B\cdot\vec A=0.008\ \mathrm{Wb}

Units:

\mathrm{T\,m^2}=\mathrm{Wb}

Boundary cases

A field tangent to the surface gives zero flux.
Zero net flux through a closed surface does not require zero field on every patch.

Positive and negative patch contributions add to eight millwebers for the selected normal.

Calculate an oriented flux

Declare the surface and its positive normal.
Resolve the field into normal and tangential components.
Use a dot product for a uniform flat surface or integrate over nonuniform patches.
Keep negative contributions instead of replacing them by magnitudes.
Test the result by reversing the normal and checking units.

Keep these three ideas

Flux is a signed dot-product surface integral.
The area normal—not the plane—defines the angle.
Every closed surface has zero net magnetic flux.

Electromagnetic Induction