Faraday's Law and Electromagnetic Induction
For two centuries electricity and magnetism were separate sciences. Faraday's law is the hinge that joined them: move a magnet, and an electric field appears out of nowhere. A changing magnetic field is itself a source of electric field — and this is the first of Maxwell's equations to put time, and motion, into the picture.
- Faraday's law: the induced emf equals the negative rate of change of flux.
- Lenz's law and what the minus sign means physically.
- Motional emf in a conductor moving through a steady field.
- Transformer emf from a time-varying field in a stationary circuit.
- The differential form \(\nabla\times\vec{E}=-\partial\vec{B}/\partial t\) — a Maxwell equation.
- Eddy currents and the induction at work in generators and transformers.
Faraday's Law
Faraday's experiments revealed something the static chapters could not: a changing magnetic flux through a circuit drives a current, even with no battery present. The driving voltage is the induced emf, equal to the negative rate of change of the flux \(\Phi=\int_S\vec{B}\cdot d\vec{S}\) linking the loop. For \(N\) turns:
The emf is the work done per unit charge around the loop — the line integral of an electric field. So the deepest reading of Faraday's law is this: a changing magnetic field creates an electric field. Unlike the electrostatic field of Part 2, this induced \(\vec{E}\) is non-conservative — its closed-loop integral is not zero, and it has no potential.
In statics \(d\Phi/dt=0\), so \(\oint\vec{E}\cdot d\vec{l}=0\) and we recover the conservative field of Chapters 4–6. The moment the flux moves, that guarantee breaks — and the induced field can push charge around a closed loop indefinitely.
Lenz's Law
The minus sign is Lenz's law, and it is a statement of energy conservation. The induced current always flows in the direction whose own magnetic field opposes the change in flux that produced it. Push a magnet toward a loop and the loop pushes back; pull it away and the loop tries to follow.
If induction reinforced the change instead, the smallest disturbance would feed itself and energy would appear from nothing. Lenz's law fixes the direction; Faraday's law fixes the magnitude. Together they say the mechanical work you do against the opposing force is exactly the electrical energy that appears in the circuit.
Motional EMF
There are two ways to change the flux. The first is to move the conductor. A charge carried along with a wire of velocity \(\vec{u}\) through a field \(\vec{B}\) feels the magnetic force \(Q\vec{u}\times\vec{B}\) from Chapter 13 — and that force pushes charge along the wire, acting as a built-in battery. The motional emf is its line integral:
This is the "flux-cutting" picture, and it is the working principle of every generator: spin a coil in a field and its sides sweep through flux, producing a continuous emf. No changing field is needed — only relative motion.
Transformer EMF
The second way to change the flux is to change the field while the circuit sits still. Then the emf comes entirely from the time derivative of \(\vec{B}\) over the fixed surface — the transformer emf:
In the most general case the loop both moves and sits in a changing field, and the two effects simply add — the transformer term plus the motional term:
The transformer term alone runs the transformer: an alternating current in the primary makes an alternating flux, and that flux — shared through a high-\(\mu\) core, exactly the magnetic circuit of Chapter 13 — induces a voltage in the secondary set by the turns ratio \(V_2/V_1=N_2/N_1\).
The Differential Form
Apply Stokes' theorem to the integral law over a fixed surface, turning the closed line integral of \(\vec{E}\) into a surface integral of its curl. Since the surfaces are arbitrary, the integrands must match point by point:
This is one of the four Maxwell equations, and the first to couple the fields in time. Where electrostatics gave \(\nabla\times\vec{E}=0\), we now see that curl is sourced by \(-\partial\vec{B}/\partial t\). The induced \(\vec{E}\) forms closed loops with no start or end — there are no charges at its feet. Chapter 16 will complete the symmetry by showing that a changing \(\vec{E}\) likewise curls up a \(\vec{B}\).
Eddy Currents & Uses
A changing flux through a solid conductor induces swirling loops of current called eddy currents. They obey Lenz's law — opposing the change — and dissipate energy as \(I^2R\) heat. That loss is harnessed in induction heating and eddy-current brakes, but it is a parasite in transformer and motor cores, which is why those cores are built from thin laminations or powdered ferrite to break up the current paths.
| Device | Which emf | Principle |
|---|---|---|
| Generator / alternator | Motional | Coil rotates in a field; \(\text{emf}=NBA\omega\sin\omega t\) |
| Transformer | Transformer | AC flux shared through a core; \(V_2/V_1=N_2/N_1\) |
| Induction motor | Both | Rotating flux drags a conducting rotor |
| Betatron | Transformer | Induced loop \(\vec{E}\) accelerates electrons |
| Eddy-current brake | Motional | Induced currents oppose motion, dissipating energy |
Worked Examples
Problem. The flux through an \(80\)-turn coil rises uniformly from \(0\) to \(6\ \text{mWb}\) in \(0.02\ \text{s}\). Find the induced emf.
Solution. Apply \(\text{emf}=N\,\Delta\Phi/\Delta t\):
Problem. A \(0.5\ \text{m}\) rod slides at \(4\ \text{m/s}\) perpendicular to \(\vec{B}=0.3\ \text{T}\). Find the emf.
Solution. Use \(\text{emf}=B\ell u\):
Problem. A \(500\)-turn coil of area \(8\ \text{cm}^2\) spins at \(50\ \text{Hz}\) in \(\vec{B}=0.6\ \text{T}\). Find the peak emf.
Solution. Peak emf \(=NBA\omega\) with \(\omega=2\pi f\):
Problem. A flux \(\Phi=5\times10^{-3}\sin(314t)\ \text{Wb}\) links a \(300\)-turn winding. Find the peak induced emf.
Solution. \(\text{emf}=-N\,d\Phi/dt\), peak \(=N\Phi_0\omega\):
Problem. Inside a cylinder of radius \(0.1\ \text{m}\), \(\vec{B}=0.2\sin(377t)\ \hat{a}_z\ \text{T}\). Find the peak induced \(E\) at radius \(r=0.05\ \text{m}\).
Solution. By symmetry \(E(2\pi r)=-\pi r^2\,dB/dt\), so \(E=-(r/2)\,dB/dt\):
Problem. A circular loop of radius \(4\ \text{cm}\) sits in a uniform field (into the page) decreasing at \(0.5\ \text{T/s}\). Find the emf magnitude and the current direction.
Solution. \(|\text{emf}|=\pi r^2\,|dB/dt|\); the loop acts to maintain the vanishing inward flux:
Chapter Summary
\(\text{emf}=-N\,d\Phi/dt\); a changing flux drives an emf and an induced \(\vec{E}\).
The minus sign; induced current opposes the change; energy is conserved.
\(\text{emf}=\oint(\vec{u}\times\vec{B})\cdot d\vec{l}\); rod gives \(B\ell u\); runs the generator.
\(\text{emf}=-\int(\partial\vec{B}/\partial t)\cdot d\vec{S}\); fixed coil, AC field; the transformer.
\(\nabla\times\vec{E}=-\partial\vec{B}/\partial t\); a Maxwell equation; non-conservative \(\vec{E}\).
Eddy currents (laminations), generators, transformers, induction motors, betatron.
Problems
For each item, decide first which emf is at work — transformer, motional, or both — then apply the matching relation and use Lenz's law for the sign. Difficulty rises down the list.
- The flux through a \(120\)-turn coil changes from \(2\ \text{mWb}\) to \(8\ \text{mWb}\) in \(0.05\ \text{s}\). Find the average induced emf.
- A \(0.8\ \text{m}\) rod moves at \(6\ \text{m/s}\) perpendicular to \(\vec{B}=0.25\ \text{T}\). Find the motional emf.
- A flux \(\Phi=4\times10^{-3}\cos(377t)\ \text{Wb}\) links \(200\) turns. Write the induced emf and give its peak value.
- A \(300\)-turn coil of area \(5\ \text{cm}^2\) rotates at \(60\ \text{Hz}\) in \(\vec{B}=0.4\ \text{T}\). Find the peak emf.
- State the direction of the induced current in a loop when a north pole is pulled away from it, and justify with Lenz's law.
- Inside a solenoid the field rises at \(dB/dt=2\ \text{T/s}\). Find the induced \(E\) at radius \(r=3\ \text{cm}\) from the axis.
- An ideal transformer has \(N_1=400\), \(N_2=100\), primary \(230\ \text{V}\) rms. Find the secondary voltage.
- A square loop \(0.2\ \text{m}\) on a side, resistance \(2\ \Omega\), enters a \(0.5\ \text{T}\) field at \(3\ \text{m/s}\). Find the induced current while one side is in the field.
- Show that the mechanical power needed to push the rod in Problem 8 equals the electrical power dissipated.
- A \(10\ \text{cm}\) radius loop lies in a field \(B=0.3e^{-t/\tau}\ \text{T}\) with \(\tau=0.1\ \text{s}\). Find the emf at \(t=0\).
- Explain why laminating a transformer core reduces eddy-current loss, and why the laminations are insulated from one another.
- From \(\nabla\times\vec{E}=-\partial\vec{B}/\partial t\), argue that the induced electric field lines must form closed loops, unlike the field of a point charge.