Faraday’s law of induction
The rate of change of flux
Faraday’s experiments showed that a changing magnetic flux induces a current. But how big is the effect? Faraday found a remarkably simple answer.
The magnitude of the induced emf in a circuit is equal to the time rate of change of magnetic flux through the circuit. In symbols: ε equals minus dΦ_B by dt.
The emf does not depend on how much flux there is, only on how fast it changes. A huge steady flux gives zero emf. A small flux that changes quickly can give a large one.
The minus sign tells us the direction of the emf. We will meet it properly as Lenz’s law in the next lesson.
Most coils have many turns, N, wound closely together. The same flux passes through every turn, and each turn adds its own emf. So ε equals minus N dΦ_B by dt.
The product N Φ_B is called the flux linkage. A coil of 100 turns gives 100 times the emf of a single loop.
Read it on a graph. Here is the flux through a coil against time. The emf is minus the slope. A steady rise gives a steady emf. A flat stretch gives zero. A steep fall gives a large emf of the opposite sign.
If the flux changes by ΔΦ_B in a time Δt, the average emf is N ΔΦ_B over Δt. Make the same change in half the time, and you get twice the emf.
And remember the three ways to change the flux B A cos θ: change B, change A, or change θ. Each of them induces an emf.
Two worked examples
A square loop of side 10 centimetres and resistance 0.5 ohm stands vertically in the east–west plane. A uniform field of 0.10 tesla points north-east, across the loop.
The field is reduced to zero at a steady rate in 0.70 seconds. Find the induced emf and current.
The loop lies east–west, so its normal points north. The field points north-east, at 45° to the normal. So Φ_B equals B A cos 45°, which is 0.10 × 10⁻² divided by √2: about 7.1 × 10⁻⁴ weber.
The flux falls to zero in 0.70 seconds. So ε equals 7.1 × 10⁻⁴ divided by 0.70, which is 1.0 × 10⁻³ volt, or 1.0 millivolt. The current is ε by R, 1.0 millivolt over 0.5 ohm: 2 milliamperes.
Second example. A circular coil of radius 10 centimetres has 500 turns and a resistance of 2 ohm. Its plane is perpendicular to the horizontal component of the Earth’s field, 3.0 × 10⁻⁵ tesla.
The coil is turned about its vertical diameter through 180° in 0.25 seconds. At the start, the flux through each turn is B A, which is 3.0 × 10⁻⁵ × π × 10⁻², about 9.4 × 10⁻⁷ weber.
After half a turn, the field enters from the other side, so the flux is minus B A. The change is 2 B A, about 1.9 × 10⁻⁶ weber per turn, not zero!
The average emf is N ΔΦ_B over Δt: 500 × 1.9 × 10⁻⁶ divided by 0.25, about 3.8 × 10⁻³ volt. The average current is 3.8 millivolts over 2 ohm, about 1.9 milliamperes.
These are average values. While the coil turns, the emf changes from moment to moment. That idea is the heart of the AC generator, which we will meet at the end of this chapter.
Try it: change the flux
- This is the first worked example: B = 0.10 T at 45°, side 10 cm, R = 0.5 Ω. Run the change and check the average emf, 1.0 mV.
- The lit-up marks in the front view are the field lines through the loop. Tilt the loop with θ and watch Φ_B = BA cos θ shrink.
- Halve Δt: the same change of flux in half the time gives twice the emf.
- Try “Flip loop 180°”. The flux changes by 2BA, and the emf is not steady while the loop turns.
Reading a flux graph
The flux through a single loop rises steadily from 0 to 4 mWb in 2 s, stays at 4 mWb for 1 s, then falls steadily to zero in 1 s. What is the induced emf in these three stages?
- 2 mV, then 0, then 4 mV in the opposite direction
- Small at first, largest during the 1 s when the flux is steady at 4 mWb
- 2 mV, then 4 mV, then 4 mV
- 1 mV all the time, since 4 mWb changes over 4 s
Answer
A. The emf is minus the slope of the Φ–t graph, not the flux itself. Stage 1: 4 mWb / 2 s = 2 mV. Stage 2: the flux is constant, so ε = 0. Stage 3: −4 mWb / 1 s, so 4 mV with the opposite sign. Averaging over the whole 4 s hides all of this.
Switch it off faster
In the first example (square loop of side 10 cm, R = 0.5 Ω, B = 0.10 T at 45° to the normal), the field is now reduced to zero in 0.35 s instead of 0.70 s. What is the induced current?
- 1 mA
- 2 mA
- 4 mA
- 8 mA
Answer
C. The change of flux is the same, 7.1 × 10⁻⁴ Wb, but it happens in half the time. ε = 7.1 × 10⁻⁴ / 0.35 = 2.0 mV, and I = 2.0 mV / 0.5 Ω = 4 mA, twice the original 2 mA.
Summary
- Faraday’s law (one loop):
ε = −dΦ_B/dt - Coil of N turns:
ε = −N dΦ_B/dt - Average emf:
|ε| = N |ΔΦ_B| / Δt - Induced current:
I = ε/R
- The induced emf equals the rate of change of magnetic flux, not the flux itself.
- A coil of N turns has flux linkage NΦ_B and N times the emf of one turn.
- On a Φ–t graph, the emf is minus the slope: steady rise → steady emf, flat → zero.
- Turning a loop through 180° changes its flux by 2BA, not zero.
- The minus sign gives the direction of the emf: Lenz’s law.