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PU-204 / L02 · lecture

Electromagnetic Induction and Motional EMF

~8 min read · Tier toggle changes depth (T1 plain · T2 standard · T3 rigour)

BEFORE YOU START

You should be comfortable with the Lorentz force law, the definition of magnetic flux, and basic vector calculus (line and surface integrals). Familiarity with the distinction between conservative and non-conservative electric fields is essential.

Why this matters

Electromagnetic induction is the principle behind generators, transformers, and wireless charging, making it the bridge between static electricity and dynamic field theory.

In the previous lecture, we established that time-varying magnetic fields generate electric fields, a concept that fundamentally breaks the static assumption that electric fields are always conservative. This lecture dissects the total electromotive force (EMF) in a circuit into two distinct physical mechanisms: the "transformer EMF" caused by a changing magnetic field in a stationary circuit, and the "motional EMF" caused by the movement of the circuit through a magnetic field. Understanding this separation is critical because it clarifies the source of energy transfer and resolves common paradoxes regarding the direction of induced currents.

The Two Faces of Induction

Imagine a metal ring sitting on a table. If you place a magnet above it and move the magnet up and down, a current flows in the ring. This is induction due to a changing magnetic field. Now imagine the ring is fixed, but you slide the ring itself across a uniform magnetic field. A current also flows. In both cases, the result looks the same: a voltage is induced. However, the physics is different. In the first case, the electric field lines form closed loops because the magnetic field is changing in time. In the second case, the magnetic field is static, but the charges in the wire are moving through that field, experiencing a magnetic force that pushes them along the wire. The total EMF is the sum of these two effects. This distinction is not just academic; it determines how we calculate energy in motors and generators.

Faraday’s Law and the Flux Rule

The fundamental law governing induction is Faraday’s Law. For any circuit \( C \) bounding a surface \( S \), the electromotive force \( \mathcal{E} \) is equal to the negative rate of change of the magnetic flux \( \Phi_B \) through that surface. This is often called the "flux rule."

\[ \mathcal{E} = -\frac{d\Phi_B}{dt} \]

Here, the magnetic flux is defined as \( \Phi_B = \int_S \mathbf{B} \cdot d\mathbf{A} \). The negative sign represents Lenz’s Law, which dictates that the induced current flows in a direction that opposes the change in flux. This is a direct consequence of the conservation of energy; if the induced current aided the change, it would create a runaway feedback loop, violating energy conservation.

Mathematically, the total EMF can be expressed as the line integral of the total force per unit charge around the circuit. If the circuit moves with velocity \( \mathbf{v} \), the force per unit charge is \( \mathbf{E} + \mathbf{v} \times \mathbf{B} \). Thus:

\[ \mathcal{E} = \oint_C (\mathbf{E} + \mathbf{v} \times \mathbf{B}) \cdot d\mathbf{l} \]

This equation reveals that the total EMF is the sum of two terms: the integral of \( \mathbf{E} \cdot d\mathbf{l} \) (the transformer EMF) and the integral of \( (\mathbf{v} \times \mathbf{B}) \cdot d\mathbf{l} \) (the motional EMF).

Transformer EMF: The Induced Electric Field

Consider a stationary circuit where the magnetic field \( \mathbf{B} \) changes with time. In this case, \( \mathbf{v} = 0 \), so the motional term vanishes. The EMF is purely due to the electric field:

\[ \mathcal{E} = \oint_C \mathbf{E} \cdot d\mathbf{l} = -\frac{d}{dt} \int_S \mathbf{B} \cdot d\mathbf{A} \]

Assuming the surface \( S \) is fixed, we can move the time derivative inside the integral:

\[ \oint_C \mathbf{E} \cdot d\mathbf{l} = -\int_S \frac{\partial \mathbf{B}}{\partial t} \cdot d\mathbf{A} \]

By applying Stokes' Theorem to the left-hand side, \( \oint_C \mathbf{E} \cdot d\mathbf{l} = \int_S (\nabla \times \mathbf{E}) \cdot d\mathbf{A} \), we arrive at the differential form of Faraday’s Law. Since this must hold for any arbitrary surface \( S \), the integrands must be equal:

\[ \nabla \times \mathbf{E} = -\frac{\partial \mathbf{B}}{\partial t} \]

This is a profound result. In electrostatics, \( \nabla \times \mathbf{E} = 0 \), meaning the electric field is conservative and can be described by a scalar potential. Here, the curl is non-zero, indicating that the electric field is non-conservative. The field lines of \( \mathbf{E} \) form closed loops, encircling the changing magnetic flux. This is the "transformer EMF," named after the device that utilizes this effect to transfer energy between circuits without physical contact.

Motional EMF: The Lorentz Force Mechanism

Now consider a circuit moving with velocity \( \mathbf{v} \) through a static magnetic field \( \mathbf{B} \). Here, \( \partial \mathbf{B} / \partial t = 0 \), so the transformer EMF is zero. The EMF arises entirely from the magnetic force on the charges in the conductor. The force per unit charge is \( \mathbf{v} \times \mathbf{B} \).

\[ \mathcal{E} = \oint_C (\mathbf{v} \times \mathbf{B}) \cdot d\mathbf{l} \]

This is the "motional EMF." It is crucial to note that this effect is not due to a non-conservative electric field in the lab frame; rather, the magnetic force pushes the charges along the wire while an equal and opposite component slows the conductor, so the energy comes from whatever keeps the conductor moving; the magnetic force itself does no net work. In the frame of the moving conductor, this effect can be interpreted as an electric field \( \mathbf{E}' = \mathbf{v} \times \mathbf{B} \) (for non-relativistic speeds), which drives the current. The energy for this current comes from the external agent moving the conductor against the magnetic drag force, not from the magnetic field itself (which does no work on stationary charges).

The total EMF is the sum of these two contributions. In many practical scenarios, both effects are present simultaneously. For example, a coil rotating in a fixed magnet's field develops a motional EMF, while a fixed coil beside a rotating magnet develops a transformer EMF; the flux rule gives the same answer for both. The flux rule \( \mathcal{E} = -d\Phi_B/dt \) elegantly combines both effects into a single scalar calculation, provided the circuit is a single rigid body or the deformation is accounted for correctly.

The derivation

◆ DERIVATION EMBED · D-160 · not re-derived here

Faraday's Law and the Flux Rule

For any circuit C bounding a surface S, the electromotive force equals minus the rate of change of magnetic flux, ℰ = −dΦB/dt, and this single "flux rule" resolves into two physically distinct contributions — the induced curl of the electric field, giving the differential law ∇ × E = −∂B/∂t, and the motional term ∮ (v × B)·dl arising from the Lorentz force on charges carried by a moving or deforming conductor.

∇ × E = −∂B/∂t   ⇔   ∮C E·dl = −∫S ∂B/∂t ·dA   ⇒   ℰ = −dΦB/dt
Open full derivation →
Tier 3 · watch out

A common pitfall is assuming that \( \mathbf{B} \) is always static in motional EMF problems. If the circuit moves through a non-uniform magnetic field, the flux changes due to both the change in area and the change in \( \mathbf{B} \) over that area. The flux rule remains valid, but the separation into "transformer" and "motional" parts becomes frame-dependent. In the lab frame, it is motional; in the circuit frame, it may appear as a transformer effect due to the time-varying field seen by the moving observer. Always specify your reference frame when decomposing the EMF. C

Worked example

Problem. A rectangular loop of wire with width \( w = 0.50\ \mathrm{m} \) and length \( l = 0.30\ \mathrm{m} \) moves with constant velocity \( v = 2.0\ \mathrm{m/s} \) into a region of uniform magnetic field \( B = 0.40\ \mathrm{T} \) directed into the page. The loop has a total resistance \( R = 0.10\ \Omega \). Calculate the induced EMF and the power dissipated while the loop is partially entering the field.

  1. Identify the type of EMF. The magnetic field is static, but the loop is moving. This is a motional EMF problem. The flux changes because the area of the loop inside the field changes.
  2. Calculate the rate of change of flux. The area \( A \) inside the field changes as \( A = w \cdot x \), where \( x \) is the length of the loop inside the field. The rate of change of area is \( dA/dt = w \cdot dx/dt = w \cdot v \).
  3. Apply Faraday’s Law. The magnitude of the EMF is \( |\mathcal{E}| = |d\Phi_B/dt| = B \cdot |dA/dt| = B \cdot w \cdot v \).
  4. Substitute the values. \( |\mathcal{E}| = 0.40\ \mathrm{T} \times 0.50\ \mathrm{m} \times 2.0\ \mathrm{m/s} \).
  5. Calculate the power. Power dissipated is \( P = \mathcal{E}^2 / R \).
\[ \mathcal{E} = 0.40 \times 0.50 \times 2.0 = 0.40\ \mathrm{V} \]
\[ P = \frac{(0.40)^2}{0.10} = 1.6\ \mathrm{W} \]

Units check. \( \mathrm{T} \cdot \mathrm{m} \cdot \mathrm{m/s} = (\mathrm{N}/(\mathrm{A}\cdot\mathrm{m})) \cdot \mathrm{m}^2/\mathrm{s} = \mathrm{N}\cdot\mathrm{m}/(\mathrm{A}\cdot\mathrm{s}) = \mathrm{J}/(\mathrm{C}) = \mathrm{V} \). Correct. Power: \( \mathrm{V}^2/\Omega = \mathrm{W} \). Correct.

CHECK YOURSELF

Three quick tests

  1. If a loop moves through a uniform magnetic field such that the flux through it remains constant, is there an induced EMF?
    AnswerNo. Since \( d\Phi_B/dt = 0 \), the net EMF is zero. The motional EMFs on opposite sides of the loop cancel each other out.
  2. Can a static magnetic field do work on a stationary charge?
    AnswerNo. The magnetic force \( \mathbf{F} = q\mathbf{v} \times \mathbf{B} \) is zero if \( \mathbf{v} = 0 \). Even if moving, the force is perpendicular to velocity, so no work is done.
  3. In the differential form \( \nabla \times \mathbf{E} = -\partial \mathbf{B}/\partial t \), what does the non-zero curl of \( \mathbf{E} \) imply about the electric potential?
    AnswerIt implies that the electric field is non-conservative, and thus a single-valued scalar potential \( V \) cannot be defined globally for the entire field.
PROBLEMS

Problem set for PU-204 →

Each derivation in this unit ends with five graded problems with full solutions.

NOTES

Lecture notes checked line by line by the editor; every number in the worked example was recomputed.

Derivations cited: D-160 · Threads: force, energy, light