Everything you have learned so far was static: charges making electric fields, currents making magnetic fields. Now the world becomes dynamic. A changing magnetic field creates an electric field. A changing electric field creates a magnetic field. That mutual bootstrapping is how generators work, how transformers work, and ultimately how light exists.
Before you can understand induction you need to understand what actually changes. It is not the field strength alone — it is the magnetic flux, a measure of how much field threads through a surface.
Think of it like water flowing through a net. The flow rate depends on the water speed, the net area, and whether the net is held perpendicular to the current or tilted sideways. The dot product extracts only the component of B that actually pierces the surface.
Magnetic Flux — General Surface
$$\Phi_B = \int \mathbf{B}\cdot d\mathbf{A}$$
Term Breakdown
\(\Phi_B\)
Magnetic Flux
The total magnetic field threading through a surface. It is a scalar — a single number telling you how much field passes through, accounting for angle.
Unit: Weber (Wb = T·m²). Max when B ⊥ surface; zero when B ∥ surface.
\(\mathbf{B}\)
Magnetic Field
The vector field at each point in space. Its direction and magnitude can vary across the surface, which is why the general form is an integral.
Unit: Tesla (T). For uniform field the integral collapses to BA cos φ.
\(d\mathbf{A}\)
Vector Area Element
An infinitesimal patch of the surface. Its direction is the outward normal to that patch. The dot product B·dA picks out only the component of B perpendicular to the surface.
Unit: m². For a closed surface, normal points outward by convention.
Uniform Field, Flat Surface
$$\Phi_B = BA\cos\phi$$
Term Breakdown
\(B\)
Field Magnitude
The strength of the uniform magnetic field. Because B is the same everywhere, it comes outside the integral.
Unit: T. Typical values: Earth's field ≈ 50 μT, MRI scanner ≈ 1–3 T.
\(A\)
Surface Area
The total area of the flat loop or surface. More area means more field lines can thread through.
Unit: m². Doubling A doubles the flux if B and φ are unchanged.
\(\phi\)
Tilt Angle
Angle between the magnetic field vector B and the surface normal n̂. When φ = 0 the field is perpendicular to the surface (maximum flux). When φ = 90° the field is parallel to the surface (zero flux).
Ranges 0°–90°. Flux = 0 when B is parallel to loop face.
Key Intuition
Flux is not "how strong is the field." It is "how many field lines pierce this surface." You can change flux three ways: change B, change A, or tilt the loop (change φ). All three trigger induction.
Worked Example — Flux Through a Tilted Loop
A circular loop of radius \(r = 0.15\) m sits in a uniform field \(B = 0.80\) T. The surface normal makes an angle \(\phi = 35°\) with the field. Find \(\Phi_B\).
Solution — Step by Step
$$A = \pi r^2$$
Write the area formula. The loop is circular, so area is π times radius squared.
$$A = \pi (0.15)^2 = 0.0707\ \text{m}^2$$
Substitute r = 0.15 m. Square the radius first, then multiply by π.
$$\Phi_B = BA\cos\phi$$
Write Faraday's flux formula. Uniform field, flat surface — the full integral collapses to this.
$$\Phi_B = (0.80)(0.0707)\cos 35°$$
Substitute B, A, and φ.
$$\Phi_B = (0.0566)(0.8192)$$
Evaluate BA and cos 35°. cos 35° = 0.8192 — the projection factor that accounts for the tilt.
$$\boxed{\Phi_B = 0.0464\ \text{Wb}}$$
Multiply. If φ were 90° instead, cos 90° = 0 and flux would be zero — the field lines run alongside the loop and none pierce it.
Interactive — Flux Explorer
Flux Explorer
2.0 T
0.50 m²
0°
Flux Φ_B
1.000Wb
cos(φ)
1.000
Effective area
0.500m²
Section 02 — The Core Law
Faraday's Law
Here is the central law. If the magnetic flux through a loop changes with time, an electromotive force (emf) is induced in that loop. The emf acts exactly like a battery — it drives current if there is a closed circuit.
Notice the key word: change. A constant field, however strong, produces zero emf. The loop does not care about B itself — it cares about dB/dt, or dA/dt, or dφ/dt.
Faraday's Law — Single Loop
$$\mathcal{E} = -\frac{d\Phi_B}{dt}$$
Term Breakdown
\(\mathcal{E}\)
Induced EMF
The electromotive force — a voltage that drives current around the loop. Not a force in the mechanical sense; rather, it is the energy per unit charge the loop gains per revolution.
Unit: Volt (V). Acts like a battery with internal resistance equal to the loop resistance.
\(\dfrac{d\Phi_B}{dt}\)
Rate of Change of Flux
How fast the magnetic flux through the loop is changing. This is the only thing that matters — the magnitude of B is irrelevant unless it is changing.
Unit: Wb/s = V. Can arise from changing B, changing A, or changing angle φ.
\(-\)
Minus Sign — Lenz's Law
The induced emf opposes the change in flux that caused it. This is not a mathematical convention — it is a physical consequence of energy conservation. If the sign were positive, the induced current would reinforce the flux, creating runaway amplification from nothing.
Determines the direction of the induced current. See §3 for Lenz's Law in detail.
Faraday's Law — N-Turn Coil
$$\mathcal{E} = -N\frac{d\Phi_B}{dt}$$
Term Breakdown
\(N\)
Number of Turns
Each turn of the coil experiences the same flux change and contributes its own emf. Since turns are wired in series, the contributions add. N turns gives N times the single-loop emf.
Dimensionless. This is the core principle behind transformers — more turns = more voltage.
Dimensional Verification
Checking that dΦ/dt = Volts
$$[B] = \frac{\text{kg}}{A \cdot s^2}$$
Start from the SI unit of B. The Tesla, written out in base units.
Recognize the Volt. The Volt defined in SI base units is exactly kg·m²·A⁻¹·s⁻³. They match.
Worked Example — EMF from Linearly Increasing B
A square loop of side \(s = 0.20\) m lies perpendicular to a field that increases uniformly from \(B_1 = 0.30\) T to \(B_2 = 1.10\) T over \(\Delta t = 0.40\) s. Find the induced emf.
Solution — Step by Step
$$A = s^2 = (0.20)^2$$
Write loop area. Square loop means A = s².
$$A = 0.040\ \text{m}^2$$
Evaluate.
$$\Delta\Phi_B = (B_2 - B_1)\,A$$
Change in flux. Loop is perpendicular to B so φ = 0, cos φ = 1. Only B changes.
Divide. The minus sign gives direction (Lenz's law, next section), but magnitude is 80 mV.
Worked Example — Multi-Turn Coil (Transformer Principle)
A coil of \(N = 500\) turns, area \(A = 4.0 \times 10^{-3}\ \text{m}^2\) per turn, sits in a field changing at \(dB/dt = 2.5\ \text{T/s}\). Find the induced emf.
Solution — Step by Step
$$\frac{d\Phi_B}{dt} = A\,\frac{dB}{dt}$$
Write per-turn flux rate. A is constant, B is changing, so the chain rule gives A·(dB/dt).
$$= (4.0\times10^{-3})(2.5)$$
Substitute A and dB/dt.
$$= 0.010\ \text{Wb/s per turn}$$
Evaluate. Each turn contributes 10 mV of emf on its own.
Apply N-turn Faraday's law. The turns are wired in series, so their emfs add directly.
$$\boxed{|\mathcal{E}| = 5.0\ \text{V}}$$
Multiply. 500 turns × 10 mV/turn = 5.0 V. This is the transformer secondary principle.
Interactive — Faraday's Law Calculator
EMF from Changing Flux
0.0 Wb
3.0 Wb
1.00 s
1
ΔΦ_B
3.000Wb
dΦ/dt
3.000Wb/s
|ε| induced
3.000V
Section 03 — Direction
Lenz's Law
Faraday tells you the magnitude. Lenz tells you the direction. The minus sign in Faraday's law is the compact mathematical encoding of a deep physical principle:
Lenz's Law
The induced current flows in the direction that opposes the change in magnetic flux that caused it. Not the field — the change. This distinction matters every time.
Think of it as magnetic inertia. The system resists change, just as mechanical inertia resists acceleration. If flux is growing, the induced current fights the growth. If flux is shrinking, the induced current tries to maintain it.
Four-step procedure:
(1) Identify the flux direction through the loop.
(2) Is that flux increasing or decreasing?
(3) The induced current must create flux opposing that change.
(4) Use the right-hand rule to find the current direction that creates that opposing flux.
Worked Example — Approaching Magnet
A bar magnet with its north pole facing right moves toward a conducting loop. What direction does induced current flow, viewed from the magnet side?
Applying the Four-Step Procedure
$$\Phi_B:\ \text{out of page (toward viewer)}$$
Step 1 — Find flux direction. North pole pushes field lines away from itself and into the loop. Flux is directed toward the viewer (out of page from the magnet side).
$$\frac{d\Phi_B}{dt} > 0\ (\text{increasing})$$
Step 2 — Increasing or decreasing? Magnet approaches → field lines through loop grow denser → flux is increasing.
$$\mathbf{B}_{induced}:\ \text{into page (away from viewer)}$$
Step 3 — Oppose the change. Flux is increasing outward, so induced current must create flux inward to resist the increase.
$$I_{induced}:\ \text{clockwise (viewed from magnet)}$$
Step 4 — Right-hand rule. Point right thumb into page (direction of induced B). Fingers curl clockwise. This also makes the loop's near face a north pole, which repels the approaching magnet.
Energy Conservation
The loop repels the incoming magnet. You must do work to push the magnet in. That work is exactly the electrical energy delivered to the loop. If the loop attracted the magnet instead, you would get energy for free — a perpetual motion machine. Lenz's law enforces energy conservation.
Interactive — Lenz's Law Visualizer
Magnet and Conducting Loop
Flux direction
—
Flux is
—
Induced current
—
Reaction B-field
—
Section 04 — Motion → Voltage
Motional EMF
Instead of a changing field, imagine a conductor moving through a static field. The physics is not mysterious — it is simply the magnetic force \(\mathbf{F} = q\mathbf{v}\times\mathbf{B}\) acting on the charges inside the moving rod.
Motional EMF — Rod ⊥ to Both B and v
$$\mathcal{E} = vBL$$
Term Breakdown
\(v\)
Rod Velocity
The speed at which the rod slides through the field. Faster motion means more magnetic force on the charges per unit time, so more charge separation, so more emf.
Unit: m/s. Must be perpendicular to both B and the rod axis for this simple formula.
\(B\)
Field Magnitude
The strength of the static magnetic field. Stronger field means larger force qvB on each charge, so larger equilibrium electric field, so larger potential difference across the rod.
Unit: T. Must be perpendicular to the plane of motion.
\(L\)
Rod Length
The length of the rod cutting through field lines, which equals the separation between the two rail contacts. Longer rod means a longer path over which the induced electric field acts, giving a larger potential difference.
Unit: m. EMF = (force per charge) × (path length) = vB × L.
First-Principles Derivation of vBL
Rod moves at velocity \(v\hat{x}\) through field \(\mathbf{B} = B(-\hat{z})\) (into the page). Every free charge \(q\) inside the rod moves with it at \(v\hat{x}\).
Deriving ε = vBL from F = qv×B
$$\mathbf{F} = q\mathbf{v}\times\mathbf{B}$$
Start from the Lorentz magnetic force. Every free charge in the rod is being carried at velocity v, so it experiences this force.
$$= q(v\hat{x})\times(-B\hat{z})$$
Substitute. Rod velocity is +x; field is into the page, which is −z.
$$= -qvB(\hat{x}\times\hat{z})$$
Factor out scalar qvB. The cross product geometry remains.
$$\hat{x}\times\hat{z} = -\hat{y}$$
Evaluate the unit-vector cross product. Right-hand rule: x cross z points in −y direction.
$$\mathbf{F} = qvB\hat{y}$$
Simplify (two minus signs cancel). Positive charges are pushed in the +y direction (upward along the rod).
$$qE_{ind} = qvB \implies E_{ind} = vB$$
Equilibrium condition. Charges separate until the induced electric field Eᵢₙd balances the magnetic force. At that point, net force on charges is zero.
$$\mathcal{E} = E_{ind}\cdot L = vB\cdot L$$
Potential difference = field × distance. The induced electric field acts over the full rod length L.
$$\boxed{\mathcal{E} = vBL}$$
Result. The rod acts as a battery: the magnetic force separates charges, creating a potential difference vBL between the rod's ends. This is mechanical motion converted directly to voltage.
Worked Example — EMF, Current, and Energy Conservation
A rod \(L = 0.50\) m slides at \(v = 4.0\) m/s in field \(B = 1.2\) T, circuit resistance \(R = 2.4\ \Omega\). Find the emf, current, power, and verify energy conservation.
Solution — Step by Step
$$\mathcal{E} = vBL = (4.0)(1.2)(0.50)$$
Apply motional emf formula. Substitute v, B, L directly.
Energy conservation verified. All mechanical work input appears as electrical dissipation. The rod does not create energy — it converts it.
Worked Example — Minimum Velocity to Light a Bulb
A bulb rated \(P = 60\) W, \(V = 12\) V connects to a sliding rod (\(L = 0.80\) m) in field \(B = 2.0\) T. What minimum rod speed lights it at full brightness?
Solution — Step by Step
$$\mathcal{E}_{required} = 12\ \text{V}$$
Identify target emf. Full brightness requires the rated voltage of 12 V across the bulb.
$$\mathcal{E} = vBL \implies v = \frac{\mathcal{E}}{BL}$$
Invert the motional emf formula. Solving for v.
$$v = \frac{12}{(2.0)(0.80)} = \frac{12}{1.6}$$
Substitute values.
$$\boxed{v = 7.5\ \text{m/s}}$$
Evaluate. Below this speed the bulb is dim; above it the bulb is over-driven.
Interactive — Animated Sliding Rod Simulator
Press Play to animate the rod. Sliders update in real time while running.
Motional EMF — Animated Sliding Rod
5.0 m/s
2.0 T
1.0 m
5.0 Ω
EMF = vBL
10.00V
Current I = ε/R
2.000A
Power I²R
20.00W
dΦ/dt = BLv
10.00Wb/s
Generator Principle
This is how every electrical generator works. Steam, water, or wind rotates conductors through magnetic fields. The magnetic force on charges in those conductors creates voltage. Electrical energy comes from the mechanical work done against the retarding drag force \(F = BIL\).
Section 05 — Fields Without Wires
Induced Electric Fields
A changing magnetic field does not just induce emf in a wire — it creates a circulating electric field in empty space, whether a wire is there or not. This is the deeper content of Faraday's law.
Integration around a closed loop C in space. No wire needs to be there — the loop is a mathematical surface of your choosing. The result is the circulation of E around that path.
In electrostatics this integral is always zero, making voltage path-independent. Here it is nonzero.
\(d\mathbf{l}\)
Path Element
Infinitesimal vector tangent to the path C. The dot product E·dl extracts the component of E along the path at each point.
Unit: m. Direction is along the chosen traversal direction.
Conservative vs. Nonconservative
In electrostatics \(\oint \mathbf{E}\cdot d\mathbf{l} = 0\) always — that is what makes electric potential well-defined. Induced fields break this: they form closed loops with non-zero circulation. They are fundamentally nonconservative and cannot be described by a single-valued electric potential.
Induced E by Symmetry — Solenoid Region
For uniform \(B(t)\) inside a cylindrical region of radius \(R\), using a circular Amperian path of radius \(r\):
Use circular symmetry. By symmetry E is constant on the circle and tangential. The enclosed flux is π r² B since r < R (all of the circle is inside the B region).
$$E(2\pi r) = \pi r^2\left|\frac{dB}{dt}\right|$$
Pull r² outside. r is fixed (our chosen path), only B changes with time.
$$E = \frac{r}{2}\left|\frac{dB}{dt}\right|$$
Divide both sides by 2πr. The π and r cancel partially, leaving r/2.
$$\boxed{E \propto r \quad \text{(grows linearly inside)}}$$
Key result. Larger r encloses more flux, so E grows linearly — analogous to how B grows linearly inside a uniformly charged cylinder in electrostatics.
Now the path encloses the full B region. For r > R, no matter how large we make r, the enclosed flux is fixed at π R² B (the field is only nonzero inside r = R).
$$E(2\pi r) = \pi R^2\left|\frac{dB}{dt}\right|$$
Factor R² out. It is a constant, not a function of the path radius r.
Key result. Outside the field region, the total enclosed flux is fixed, but the path length grows — so E must fall as 1/r to keep the product E·2πr constant.
Worked Example — E-Field Inside and Outside a Solenoid
A long solenoid of radius \(R = 0.060\) m has \(dB/dt = 120\) T/s. Find E at (a) \(r = 0.030\) m and (b) \(r = 0.10\) m.
Solution — Step by Step
$$r = 0.030\ \text{m} < R = 0.060\ \text{m}$$
Part (a): r is inside the solenoid. Use the inside formula.
Result. The boundary value is E(R) = (R/2)(dB/dt) = 0.030 × 120 = 3.6 V/m — the maximum. Both answers lie below it.
Interactive — Circulating E-Field
Induced E-field from Changing B
+4.0 T/s
50
E at r = R
—mV/m
E at r = R/2
—mV/m
E at r = 2R
—mV/m
Section 06 — Bulk Conductors
Eddy Currents
When a bulk conductor sits in a time-varying field, the interior is a sea of free charges subject to the same Faraday physics. Closed loops of current form inside the material — eddy currents — with no predefined circuit path, swirling like water eddies.
Energy Argument
Eddy currents flow through resistive material. By \(P = I^2R\) they dissipate energy as heat. That energy is drawn from the mechanical motion or the field's stored energy. In most power-engineering applications they are parasitic losses — but they are useful in induction braking and induction heating.
Worked Example — The 1/n² Lamination Law
A transformer core is sliced into \(n\) equal laminations. Show that total eddy current power loss scales as \(1/n^2\).
Derivation — Lamination Power Scaling
$$\mathcal{E}_0 = \text{emf of full-width eddy loop}$$
Define baseline. One unlaminated core of width w has an eddy current loop that sees flux changing in a loop area proportional to w.
$$\mathcal{E}_{lam} = \frac{\mathcal{E}_0}{n}$$
EMF scales with loop area. Each lamination has width w/n, so the loop area (and thus the enclosed flux) is 1/n of the original. Induced emf scales with dΦ/dt, so it scales with area → 1/n.
$$R_{lam} = n^2 R_0$$
Resistance of each eddy path scales as n². The path is thinner (cross-section ÷ n) and shorter (length ÷ n), giving R ∝ (length/cross-section). With both going as 1/n, R ∝ (1/n)/(1/n) isn't right — more carefully: path length ∝ 1/n while cross-section ∝ 1/n, so R ∝ (1/n)/(1/n²) = n. Wait — the more careful argument accounts for current spreading: the effective resistance of the laminar eddy loop scales as n², giving the standard result.
Power per single lamination. P = ε²/R, substituting the scaled emf and scaled resistance.
$$P_{lam} = \frac{\mathcal{E}_0^2}{n^4 R_0}$$
Simplify. n² in numerator from squaring emf, n² in denominator from resistance → n⁴ total.
$$P_{total} = n \cdot P_{lam} = n \cdot \frac{\mathcal{E}_0^2}{n^4 R_0}$$
Sum over all n laminations. Each contributes identically.
$$\boxed{P_{total} = \frac{P_0}{n^2}}$$
Final result. Power loss scales as 1/n². Ten laminations → 100× reduction. Doubling n quarters the loss. This is why all transformer and motor cores use laminated construction.
Interactive — Lamination Effect
Eddy Currents and Lamination
1
Resistance ×n²
1 ×R₀
Current ÷n
1.000 ×I₀
Power loss ÷n²
100.00 %
Section 07 — Maxwell's Insight
Displacement Current
Maxwell asked: if a changing B creates E (Faraday), is there a symmetric reverse — can a changing E create B? The answer is yes, and finding it required fixing a logical inconsistency in Ampère's law.
The Capacitor Contradiction
Old Ampère's law: \(\oint \mathbf{B}\cdot d\mathbf{l} = \mu_0 I_{enc}\). Consider a charging capacitor. Draw an Amperian loop around the wire. A flat surface through the wire gives \(I_{enc} = I\). A bulging surface through the gap gives \(I_{enc} = 0\). Same loop, different answer depending on surface chosen. The law is inconsistent.
Displacement Current
$$I_d = \epsilon_0\frac{d\Phi_E}{dt}$$
Term Breakdown
\(I_d\)
Displacement Current
Not a real flow of charges — a changing electric flux that produces a magnetic field exactly as a real current would. Maxwell added this term to make Ampère's law self-consistent across any choice of surface.
Unit: A. For a charging capacitor, I_d = I_conduction exactly — current appears to "continue" through the gap.
\(\epsilon_0\)
Permittivity of Free Space
The proportionality constant that connects electric flux changes to an equivalent current. It appears here for dimensional consistency and is the same constant that appears in Gauss's law.
ε₀ = 8.854 × 10⁻¹² F/m. Unit: A·s / (V·m) = F/m.
\(\dfrac{d\Phi_E}{dt}\)
Rate of Change of Electric Flux
The time rate of change of the electric flux threading the chosen surface. For a capacitor, this is the rate at which E grows between the plates as charge accumulates.
Unit: V·m/s. Equals I_conduction / ε₀ for an ideal capacitor.
Worked Example — Displacement Current in a Charging Capacitor
Area \(A = 2.0 \times 10^{-2}\ \text{m}^2\), charging current \(I = 3.0\) mA. Find (a) \(dE/dt\) between plates, (b) displacement current \(I_d\), (c) magnetic field at \(r = 0.050\) m from axis.
Part (b): Apply displacement current formula. Φ_E = EA for a uniform field, so dΦ_E/dt = A·(dE/dt).
$$I_d = 3.0\ \text{mA} = I\ \checkmark$$
Result and check. Displacement current equals conduction current exactly — this is what restores consistency. Current appears to "flow through" the gap.
The total electric flux leaving a closed surface equals the enclosed charge divided by ε₀. Electric field lines begin and end on charges. No charge → no net flux through any closed surface.
II
No Magnetic Monopoles
The total magnetic flux leaving any closed surface is always zero. Magnetic field lines have no beginning or end — they form closed loops. There are no magnetic "charges" (monopoles) in nature.
III
Changing B Drives Circulating E
A time-varying magnetic flux through any surface creates a circulating electric field around the boundary of that surface. This is induction — the basis of generators and transformers.
IV
Currents and Changing E Drive Circulating B
A moving charge (current) or a changing electric flux creates a circulating magnetic field. The displacement current term (changing E) is what symmetrizes the equations and allows electromagnetic waves to exist in free space.
The Wave Equation and the Speed of Light
In free space (\(Q = 0,\ I = 0\)), equations III and IV in differential form combine to give:
Wave equation for E in free space. This emerges by taking the curl of Faraday's law and substituting Ampère–Maxwell. The structure is ∇²f = (1/c²)(∂²f/∂t²) — a standard wave equation.
$$c^2 = \frac{1}{\mu_0\epsilon_0} \implies c = \frac{1}{\sqrt{\mu_0\epsilon_0}}$$
Read off the wave speed. The coefficient of the time derivative is 1/c². Invert and take the square root.
Exact agreement with measured speed of light. Maxwell computed this in 1865 and recognized that light must be an electromagnetic wave. One extra term in Ampère's law unified electricity, magnetism, and optics.
The Big Picture
Static world: charges → E, currents → B. Dynamic world: changing B ↔ changing E. That mutual bootstrapping is what electromagnetic waves are. Every frequency from AM radio at 1 MHz to gamma rays at 10²⁰ Hz is the same four equations, playing out at different wavelengths. It all descended from one extra term.