Electromagnetic Induction and Faraday's Law: Study Guide (Physics 2 12.4)

The big idea

In Topic 12.3 a current made a magnetic field. This topic runs the other way. If the magnetic field passing through a loop of wire changes, an emf appears in the loop, and if the loop is a closed circuit, a current flows. No battery is needed. This is electromagnetic induction, and it is how generators, transformers and induction cooktops work.

The key word is changes. A strong magnet sitting still next to a loop induces nothing. To describe “how much field passes through a loop” we need one new quantity: magnetic flux.

Magnetic flux

Every flat surface has an area vector A: its size is the area, and it points at right angles to the surface. For a flat loop you choose which of the two perpendicular directions to use. For a closed surface, such as a box, each face’s area vector points outward.

The magnetic flux Φ through the surface depends only on the part of B that is perpendicular to the surface, that is, the part along A:

Φ = B · A = BA cos θ

where θ is the angle between B and the area vector (not the surface). The unit is the weber: 1 Wb = 1 T·m².

For a square loop of side 0.10 m (A = 0.010 m²) in a 0.30 T field:

θ (B to area vector)	cos θ	Φ (Wb)	Meaning
0°	1	+3.0 × 10⁻³	field straight through, along A
60°	0.5	+1.5 × 10⁻³	only part of B is perpendicular to the loop
90°	0	0	field lies in the plane of the loop
180°	−1	−3.0 × 10⁻³	field straight through, against A

The sign of Φ tells you whether B is along the area vector (positive) or against it (negative). The size tells you how much field passes through.

Three ways to change the flux

Since Φ = BA cos θ, the flux through a loop changes if any one of these changes:

1. B — move a magnet towards or away from the loop, or change the current in a nearby coil;
2. A — squash or stretch the loop, or slide part of the circuit (the rod on rails below);
3. θ — rotate the loop in the field. This is how a generator works.

For example, a loop of area 0.050 m² in a 0.40 T field starts with its area vector along B (Φ = 0.020 Wb). It is turned through 90° in 0.10 s, so the flux falls to zero. The average emf is 0.020 Wb ÷ 0.10 s = 0.20 V.

Faraday’s law

Faraday’s law links the emf to the rate of change of flux:

ε = −ΔΦ / Δt

• The emf depends on how fast the flux changes, not on how big the flux is.
• With ΔΦ/Δt over a finite time, you get the average emf over that time.
• The minus sign is Lenz’s law (next section). For sizes, use |ε| = |ΔΦ|/Δt and find the direction separately.

Coils. A coil of N identical turns has the same flux through each turn. Each turn gets the same emf, and the turns are connected in series, so the emfs add: |ε| = N|ΔΦ|/Δt, where Φ is the flux through one turn.

If the loop is part of a circuit with total resistance R, the induced current is I = ε/R, as in Unit 11.

Unit check: Wb/s = T·m²/s. Since T = N/(A·m), this is N·m/(A·s) = J/C = V.

Lenz’s law: which way does the current go?

Lenz’s law: the induced current flows in the direction that makes its own magnetic field oppose the change in flux that caused it.

Work through four steps every time:

1. External field. Which way does the outside field B point through the loop?
2. Change. Is the flux through the loop increasing or decreasing?
3. Induced field. If the flux is increasing, the induced field inside the loop points opposite to B. If it is decreasing, the induced field points the same way as B, trying to keep the flux up.
4. Current. Use the right-hand rule for a loop (Topic 12.3): point your thumb along the induced field inside the loop; your fingers curl in the direction of the induced current.

Lenz’s law opposes the change, not the field. A decreasing flux gets “topped up”, not opposed.

Why it must be so. Suppose the induced current helped the change. Pushing a magnet towards a loop would then pull the magnet in faster, which would make a bigger current, which would pull harder, and so on, creating energy from nothing. Lenz’s law is conservation of energy: whoever causes the change has to do work against the induced effect, and that work becomes the electrical energy in the circuit.

The rod on rails: ε = Bℓv

A common set-up: two parallel conducting rails a distance ℓ apart, joined at one end by a resistor R. A conducting rod rests across the rails and slides along them at speed v. A uniform field B points at right angles to the plane of the rails.

Derivation from Faraday’s law. In time Δt the rod moves vΔt, so the area of the circuit grows by ΔA = ℓvΔt. With B perpendicular to the loop, ΔΦ = BΔA = BℓvΔt. So:

|ε| = ΔΦ/Δt = Bℓv

Check from forces on charges. Each charge q in the rod moves with the rod at speed v, at right angles to B, so it feels a magnetic force qvB along the rod (Topic 12.2). Charges pile up at the ends until an electric field E stops them: qE = qvB. Across a rod of length ℓ that gives a potential difference Eℓ = Bℓv, the same answer.

Worked example 1: a coil in a growing field

Question. A flat rectangular coil of 200 turns, 4.0 cm by 5.0 cm, lies in the plane of the page. A uniform field out of the page increases steadily from 0.10 T to 0.50 T in 0.20 s. The coil’s total resistance is 4.0 Ω. (a) Find the average emf. (b) Find the current. (c) Find the direction of the current, as seen from the front of the page. (d) What would the emf be if the coil were tilted so its area vector made 60° with the field?

1. Area: A = 0.040 m × 0.050 m = 2.0 × 10⁻³ m².
2. Flux per turn. B is along the area vector (θ = 0), so Φ = BA. Φ goes from (0.10)(2.0 × 10⁻³) = 2.0 × 10⁻⁴ Wb to (0.50)(2.0 × 10⁻³) = 1.0 × 10⁻³ Wb. ΔΦ = 8.0 × 10⁻⁴ Wb.
3. (a) emf: |ε| = NΔΦ/Δt = 200 × 8.0 × 10⁻⁴ Wb ÷ 0.20 s = 0.80 V.
4. (b) Current: I = ε/R = 0.80 V ÷ 4.0 Ω = 0.20 A.
5. (c) Direction. The external field points out of the page and the flux is increasing. The induced field must point into the page inside the coil. Thumb into the page: the fingers curl clockwise as seen from the front.
6. (d) Tilted. Every flux value is multiplied by cos 60° = 0.5, so ΔΦ halves and |ε| = 0.40 V.

Check. Units: Wb/s = V. Over the 0.20 s, a charge of I × Δt = 0.040 C passes round the coil.

Worked example 2: rod on rails

Question. In Figure 2, ℓ = 0.30 m, B = 0.50 T into the page, R = 1.5 Ω, and the rod is pulled to the right at a steady 2.0 m/s. Ignore the resistance of the rod and rails, and friction. Find (a) the emf, (b) the current and its direction in the rod, (c) the magnetic force on the rod and (d) the power needed to keep the rod moving. Compare with the power in the resistor.

1. (a) emf: ε = Bℓv = (0.50 T)(0.30 m)(2.0 m/s) = 0.30 V.
2. (b) Current: I = ε/R = 0.30 V ÷ 1.5 Ω = 0.20 A. Direction: the loop’s area grows, so flux into the page increases. The induced field inside the loop points out of the page, so the current runs anticlockwise: up the rod in Figure 2. (Check: a positive charge moving right in a field into the page is pushed up the rod.)
3. (c) Force: F = IℓB = (0.20 A)(0.30 m)(0.50 T) = 0.030 N. Right-hand rule: fingers up the rod (current), curl into the page (B): the thumb points left, opposite to v. The field brakes the rod, as Lenz’s law predicts.
4. (d) Power: to keep v constant, the puller must supply 0.030 N to the right. P = Fv = (0.030 N)(2.0 m/s) = 0.060 W. In the resistor, I²R = (0.20 A)²(1.5 Ω) = 0.060 W.

Interpretation. The two powers are equal: the work you do on the rod ends up as thermal energy in the resistor. Double the speed and the emf, current and braking force all double, so the power needed rises by a factor of 4.

Sketching emf from a flux graph

Since ε = −ΔΦ/Δt, the emf at any moment is minus the slope of the Φ–t graph. Read the flux graph in pieces:

• straight rising section → constant emf of one sign;
• flat section → zero emf, however large Φ is;
• straight falling section → constant emf of the opposite sign; a steeper fall gives a larger emf.

Example: a single loop of area 0.040 m² sits in a field at right angles to it. The field rises steadily from 0 to 0.20 T over 2 s, stays at 0.20 T until t = 5 s, then falls steadily to zero by t = 6 s. Then Φ rises from 0 to 8.0 × 10⁻³ Wb, stays flat, then falls back. The emf is −4.0 × 10⁻³ V for 0–2 s, zero for 2–5 s, and +8.0 × 10⁻³ V for 5–6 s: twice the size, because the flux falls twice as fast as it rose.

Common misconceptions

• “A big flux means a big emf.” The emf depends on the rate of change of flux. A loop in a huge steady field has no emf.
• “Lenz’s law opposes the field.” It opposes the change in flux. When the flux is falling, the induced field points the same way as the external field.
• Using the angle to the surface. In Φ = BA cos θ, θ is measured from the area vector, which is perpendicular to the surface. A field lying in the plane of the loop gives zero flux.
• “No current means no emf.” An open loop still has an emf when its flux changes; it just has no complete path for a current.
• Forgetting N. For a coil, multiply the single-turn emf by the number of turns.
• “The induced current speeds the rod up.” The magnetic force on the induced current always opposes the motion that causes it. Otherwise energy would not be conserved.
• Quoting a rule without reasons. For a direction, write the chain: external field direction, increase or decrease, induced field direction, then current from the right-hand rule.

Where this leads

Induction closes Unit 12. Unit 13 begins with Topic 13.1, Reflection, where light (an electromagnetic wave) is the subject. First test yourself with the practice questions, then use the revision notes and the topic checklist to consolidate.