Electromagnetic Induction: Study Guide (Physics C: E&M 13.2)

Course note. This guide is for the calculus-based Physics C: Electricity and Magnetism course, Topic 13.2. Physics 2 also covers Faraday’s and Lenz’s laws; this course adds time-dependent fluxes that you must differentiate, and the induced electric field.

Constants used: μ₀ = 4π × 10⁻⁷ T·m/A and ε₀ = 8.85 × 10⁻¹² C²/(N·m²).

Faraday’s law

In Topic 13.1 you learned to calculate magnetic flux, Φ_B = ∫B·dA. Faraday’s law says what happens when that flux changes: an emf appears around the edge of the surface.

ℰ = −dΦ_B/dt (one loop)

ℰ = −N dΦ_B/dt (a coil of N identical turns)

• The emf depends on the rate of change of flux, not on the flux itself. A huge steady flux induces nothing.
• In a coil or a long solenoid, each turn has the same flux change, so the emfs of the turns add: the total is N times the emf of one turn.
• Units: Wb/s = T·m²/s = V.
• If the loop is a closed conductor of resistance R, the induced current is I = |ℰ|/R.

Sign convention. Pick a direction for the area vector A. Curl the fingers of your right hand around the loop with your thumb along A. That way round the loop is the positive direction for ℰ and for the current. The minus sign then gives the direction automatically. Most students find it quicker to use Lenz’s law (below) for direction and Faraday’s law for size.

Three ways to change the flux

For a flat loop in a uniform field, Φ_B = B A cos θ. Any of the three factors can change.

What changes	Size of the emf (one loop)	Example
The field, B (area fixed)	ℰ = A dB⊥/dt	a coil inside an electromagnet whose current is ramped
The area, A (field fixed)	ℰ = B dA⊥/dt	a rod sliding along rails; a loop being stretched or squeezed
The angle, θ	ℰ = BAω sin ωt for θ = ωt	a coil spinning in a field: a generator

Here B⊥ is the component of B perpendicular to the loop, and A⊥ is the area perpendicular to B.

For a rotating coil with N turns and constant angular speed ω, the flux per turn is BA cos ωt, so ℰ = −N d(BA cos ωt)/dt = NBAω sin ωt. The peak emf, NBAω, grows with every factor, and the emf alternates with the frequency of rotation. The emf is largest when the plane of the coil lies along the field (the flux is zero but changing fastest) and zero when the coil faces the field (the flux is greatest but momentarily not changing).

Lenz’s law: the direction

Lenz’s law: the induced current flows in the direction that makes its own magnetic field oppose the change in flux. It opposes the change, not the field.

Use four steps:

1. What is the direction of the external field through the loop?
2. Is the flux in that direction increasing or decreasing?
3. The induced field inside the loop points opposite to the external field if the flux is increasing, and the same way if it is decreasing.
4. Use the right-hand rule: thumb along the induced field inside the loop, fingers curl in the direction of the induced current.

Lenz’s law is energy conservation at work. If the induced current helped the change, the flux would grow faster, giving a bigger current, and so on, creating energy from nothing.

Reading flux graphs

Since ℰ = −N dΦ_B/dt, the emf graph is −N times the gradient of the flux graph. Straight sections of Φ_B(t) give constant emf; flat sections give zero emf.

To go the other way, from an emf graph to the flux, use the area under the graph: ΔΦ_B = −(1/N)∫ℰ dt.

Worked example 1: a coil in a growing field

Question. A flat circular coil of 200 turns and radius 0.040 m has its axis along a uniform field that points towards you. The field grows as B(t) = 0.10 T + (0.030 T/s²)t². The coil’s ends are joined, and its total resistance is 4.0 Ω. At t = 2.0 s, find (a) the size of the induced emf, (b) the current and its direction as you see it.

1. Area: A = π(0.040 m)² = 5.03 × 10⁻³ m². It is constant, so ℰ = NA dB/dt.
2. dB/dt = 2(0.030 T/s²)t = 0.12 T/s at t = 2.0 s.
3. (a) |ℰ| = (200)(5.03 × 10⁻³ m²)(0.12 T/s) = 0.121 V.
4. (b) I = 0.121 V ÷ 4.0 Ω = 0.030 A.
5. Direction: the flux towards you is increasing, so the induced field inside the coil points away from you. Thumb away from you: fingers curl clockwise as you see it.

Check. Units: m² × T/s = Wb/s = V. Because dB/dt ∝ t, the emf at t = 4.0 s would be twice as big (0.241 V). The emf depends on how fast B changes, not on B itself (B = 0.22 T at t = 2.0 s does not appear in the answer).

Worked example 2: a rod sliding on rails

Question. Two parallel horizontal rails 0.30 m apart are joined at the left by a 1.5 Ω resistor. A uniform 0.50 T field points vertically down through the circuit. A metal rod lies across the rails and moves to the right at a steady 2.0 m/s. Viewed from above, find the emf, the current and its direction in the rod.

1. Let x be the distance from the resistor to the rod. The circuit’s area is A = Lx, with L = 0.30 m. So dA/dt = L dx/dt = Lv = (0.30 m)(2.0 m/s) = 0.60 m²/s.
2. B is constant, so |ℰ| = B dA/dt = BLv = (0.50 T)(0.60 m²/s) = 0.30 V.
3. I = 0.30 V ÷ 1.5 Ω = 0.20 A.
4. Direction: the downward flux is increasing, so the induced field inside the circuit points up. Draw the circuit as seen from above, with the resistor on the left and the rails running across the page. Thumb up (out of your drawing): the current goes anticlockwise, so in the rod it flows up the page, from the lower rail to the upper rail.

Check. You can also find the direction from the magnetic force on positive charges in the moving rod, qv × B; it gives the same answer. Now that the rod carries a current in a field, a force acts on it. That force, and the motion it causes, are the subject of Topic 13.3.

Induced electric fields

What pushes the charges round a loop when only B changes and nothing moves? A changing magnetic field creates an electric field. This is Maxwell’s third equation, Faraday’s law in field form:

∮ E·dl = −dΦ_B/dt

The left side is the emf around a closed path. The induced field E forms closed loops around the region where B changes. It is non-conservative: its line integral around a closed path is not zero, so you cannot describe it with an electric potential.

Worked example 3: the field around a changing solenoid

Question. A long solenoid of radius R = 0.025 m has a uniform field along its axis that is increasing at 0.40 T/s. Find the size of the induced electric field at (a) r = 0.010 m and (b) r = 0.050 m from the axis. (c) Find the emf around a circle of radius 0.050 m.

By symmetry, E is tangent to circles centred on the axis and has the same size all round each circle, so ∮E·dl = E(2πr).

1. (a) Inside (r < R): the flux through the circle is Bπr². E(2πr) = πr² dB/dt, so E = (r/2) dB/dt = (0.010 m ÷ 2)(0.40 T/s) = 2.0 × 10⁻³ V/m.
2. (b) Outside (r > R): only the solenoid’s cross-section has flux, πR²B. E(2πr) = πR² dB/dt, so E = (R²/2r) dB/dt = (0.025)² ÷ (2 × 0.050) × 0.40 = 2.5 × 10⁻³ V/m.
3. (c) ℰ = πR² dB/dt = π(0.025)²(0.40) = 7.85 × 10⁻⁴ V. Check: E(2πr) = (2.5 × 10⁻³)(2π × 0.050) gives the same value.

Interpretation. There is an electric field outside the solenoid, where B is almost zero. E rises linearly inside, peaks at r = R (5.0 × 10⁻³ V/m) and falls as 1/r outside. Viewed from the end where the field points towards you, E circulates clockwise, by Lenz’s law. A wire loop placed anywhere around the solenoid would carry an induced current.

Maxwell’s equations and light

You have now met all four of Maxwell’s equations:

Equation	Meaning
∮E·dA = q_enc/ε₀	Gauss’s law: charges create electric fields (Topic 8.6)
∮B·dA = 0	No magnetic monopoles; field lines form closed loops (Topic 12.1)
∮E·dl = −dΦ_B/dt	Faraday: a changing magnetic flux creates an electric field (this topic)
∮B·dl = μ₀I + μ₀ε₀ dΦ_E/dt	Ampère–Maxwell: currents and changing electric flux create magnetic fields (Topic 12.4)

Together they predict that changing E and B fields can sustain each other and travel as an electromagnetic wave, at the speed c = 1/√(μ₀ε₀) = 1/√[(4π × 10⁻⁷)(8.85 × 10⁻¹²)] = 3.00 × 10⁸ m/s, the speed of light. The course does not expect you to derive this.

Measuring an induced emf

A typical investigation: place a small flat search coil inside a long solenoid, with the axes lined up. Drive the solenoid with a current that rises at a steady, known rate, so that dB/dt = μ₀n dI/dt is known. Record the coil’s emf with a voltmeter or data logger. Repeat for several ramp rates and plot ℰ against dB/dt. Faraday’s law predicts a straight line through the origin with gradient NA, so the graph gives the number of turns or the area. Hold the angle and position of the coil fixed, and check that the emf drops to zero when the current is steady.

Common misconceptions

• “A strong field induces a big emf.” Only a changing flux induces an emf. A coil at rest in a strong steady field has ℰ = 0.
• “The induced field always opposes the external field.” It opposes the change. When the flux is decreasing, the induced field points the same way as the external field.
• Forgetting N. Every turn of a coil adds its own emf.
• Ignoring a sign change. If the flux goes from +Φ to −Φ, the change is 2Φ.
• “Maximum flux means maximum emf.” For a rotating coil the emf is largest when the flux is zero.
• “Induced electric fields need a wire.” The field exists in empty space; the wire only lets a current flow.
• Using V = −∫E·dl round a loop. The induced field is non-conservative, so “potential” around a closed loop has no single value.

Where this leads

Next, in Topic 13.3, Induced Currents and Magnetic Forces, you will find the forces on induced currents and use Newton’s second law for moving loops and rods. Later, inductance (Topic 13.4) applies Faraday’s law to a coil’s own changing current. Practise now with the practice questions, then use the revision notes and the topic checklist.