Faraday's Law: a Simple Generator

A coil turns in a magnetic field. Drive it at a fixed speed, or with a fixed torque, and connect a load or leave it open. The page follows the flux, the voltage, the current and the push-back through each turn, and the table under the plots does the arithmetic.

Worked example — Faraday's Law: a Simple Generator

Angular speed	314.2 rad/s
Frequency	50.00 Hz
Peak flux linkage	0.03000 Wb
Peak EMF	9.425 V
RMS EMF	6.664 V
Peak current	0.6283 A
Peak back-torque	0.01885 N·m
Mean back-torque	0.009425 N·m

These are the values the calculator shows when the page opens. Change any input above to recompute them.

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Lesson

The theory — Faraday's Law: a Simple Generator

Faraday's law: a changing magnetic flux through a coil drives an EMF round it, ε = −dΦ ÷ dt. For a coil of N turns and area A turning at angular speed ω in a field B, Φ = NBA cos ωt and ε = NBAω sin ωt. Lenz's law gives the sign: the induced current flows so that its own field opposes the change in flux. In a generator that shows up as a force opposing the motion. With a load R and a coil resistance r, the current is i = ε ÷ (R + r), and the power the back-torque absorbs, τ_eω, equals the electrical power delivered to both; a real driver supplies friction on top.

What each symbol means

Φ: the flux linkage, N B A cos ωt, in webers. Largest when the coil faces the field
ε: the EMF, in volts. Largest when the flux is changing fastest, a quarter-turn later
ω: the angular speed, in rad/s. The EMF is proportional to it
r: the coil's own resistance. Every current through it wastes i²r
τ_e: the back-torque from the current in the field. Zero on open circuit

Breaks when: The best load equals r only when the speed is held fixed. Drive the rotor with a fixed torque instead and a heavier load slows it down, which lowers the voltage; the best load is then larger than r, 48 Ω rather than 3 Ω at the constant-torque preset, and the page searches for it rather than assuming it. Real dynamos also limit their own current through the coil's inductance, which this model leaves out.

Nothing connected, nothing to push against 🖖

Press Open circuit. The coil turns at 3,000 rpm and makes 9.425 V at its peak, and the back-torque is zero: in this model nothing resists the turning until a current flows. Press Bicycle dynamo and connect a 12 Ω lamp. Now a current flows, the current sits in a magnetic field, and the field pushes on it, always against the rotation. That push is Lenz's law, and on average it is 0.009425 N·m, which the rider has to supply. Switch the drive to Constant-torque drive and the effect is visible as speed: the rotor spins at 1,000 rad/s with nothing connected and settles to 250 rad/s within about a second of the load going on, with a time constant of 0.25 s. This is why a bicycle dynamo gets heavier to pedal when the light is switched on.

Every watt of light is a watt from the hand 🖖

The power plot draws the mechanical power the back-torque absorbs, τ_eω, over the heat in the load and the coil, and at constant speed the two are the same at every instant, not only on average. At Bicycle dynamo the hand supplies 2.961 W; 2.369 W reaches the lamp and 0.5922 W warms the coil. The load decides the split. Press Heavy load: with 0.5 Ω the hand supplies 12.69 W, more than four times as much, and the lamp gets less than before, 1.813 W, because most of it heats the coil. The load that takes the most is the one equal to the coil's resistance, 3 Ω in Matched load, and there exactly half is wasted. That is the internal-resistance page's result, arrived at from the crank.

Twice the speed is twice the voltage and four times the power 🖖

The voltage comes from how fast the flux through the coil changes, ε = N B A ω sin ωt, so it grows in proportion to the speed. Press Twice the speed: at 6,000 rpm the peak goes from 9.425 to 18.85 V. The same lamp draws twice the current at twice the voltage, so it takes four times the power, 9.475 W against 2.369. The back-torque doubles too, and turning it twice as fast against twice the torque is four times the power, 11.84 W from the hand. A generator gives more for more effort, never more for nothing.

Problem solved in full

What a bicycle dynamo costs the rider 5 steps
A coil of 250 turns and area 4 cm² turns at 3,000 rpm in a field of 0.3 T. Its resistance is 3 Ω and it lights a 12 Ω lamp. Find the peak EMF, the mean power into the lamp, the mean power the rider supplies, and the mean back-torque. Press Bicycle dynamo to check each figure.

Open this problem in the tool
1. The angular speed and the peak flux linkage:
  
  $ω = \frac{2 π \cdot 3000}{60} = 314.16 rad/s, N B A = 250 \cdot 0.3 \cdot 4 \times 1 0^{- 4} = 0.03 Wb$
2. The peak EMF is the flux linkage times the angular speed:
  
  $ε_{m a x} = N B A ω = 0.03 \cdot 314.16 = 9.425 V$
3. The EMF is sinusoidal, so the mean power into a resistance R is half its peak, because the mean of sin² over a turn is ½:
  
  $⟨ P_{R} ⟩ = \frac{ε _{m a x}^{2} R}{2 ( R + r ) ^{2}} = \frac{9 . 42 5 ^{2} \cdot 12}{2 \cdot 1 5 ^{2}} = 2.369 W$
4. The rider supplies what the lamp and the coil take together:
  
  $⟨ P_{mech} ⟩ = \frac{ε _{m a x}^{2}}{2 ( R + r )} = \frac{9 . 42 5 ^{2}}{30} = 2.961 W$
5. Power is torque times angular speed, so the mean back-torque is:
  
  $⟨ τ_{e} ⟩ = \frac{⟨ P _{mech} ⟩}{ω} = \frac{2 . 961}{314 . 16} = 0.009425 N m$
Answer

The coil makes 9.425 V at its peak and the lamp gets 2.369 W on average. The rider supplies 2.961 W, 80% of which reaches the lamp, by pushing against a mean back-torque of 0.009425 N·m. Unplug the lamp and the torque and the power both fall to zero while the EMF stays.

$ε_{m a x} = 9.425 V, ⟨ P_{R} ⟩ = 2.369 W, ⟨ τ_{e} ⟩ = 0.009425 N m$

By the 42Math team · Reviewed against published sources. Corrections welcome.