Preview

Generator

Convert mechanical energy to electrical energy

Two labs ago you pulled a magnet through a coil and watched the EMF price the speed of the plunge — the faster the motion, the greater the kick. This lab asks what happens when the motion never stops. Bend the path into a circle: mount the coil on an axle between two magnetic poles and spin it. Now the flux doesn't surge once and die — it breathes, swelling and collapsing twice every turn, and the EMF it gives birth to is the most important waveform on Earth: the sine. Faraday built the first dynamo in 1831, and every power plant since — coal, nuclear, hydro, wind — is the same spinning coil with a bigger crank. The physics fits in one line: ε = NBAω·sin(ωt). Four levers set the peak — turns, field, area, spin — and the rotation itself writes the shape. Watch the spin rate work twice: it makes each turn stronger, because ω is a factor of the peak itself, and it makes the turns come faster, because the frequency IS ω/2π. That double duty is why the US grid hums at exactly 60 Hz — somewhere, a coil's flux cycle is repeating exactly sixty times a second: a two-pole machine spinning sixty turns, or a many-pole machine turning slower for the same beat. Watch the coil glow brighten and fade as the EMF swings, watch the trace unspool below the scene, and keep your eye on the strangest instant of all: the moment the flux is at its maximum and the voltage is exactly zero. Your first job is to predict what happens to the peak voltage AND the frequency when you double the spin — most people get one right and one wrong.

What you'll be able to do

  • Close the generator equation on the panel's own rows at the default state (ω = 5.0 rad/s, B = 1.0 T, A = 100 cm² = 0.01 m², N = 50): ε₀ = NABω = 50 × 1.0 × 0.01 × 5 = 2.5 V exactly → row '2.500' (OpenStax Eq. 13.18 'ε₀ = NABω is the peak emf'); T = 2π/ω = 1.2566 s → row '1.257' and f = ω/2π = 0.7958 Hz → row '0.80' (§13.6: 'f = ω/2π and the period is T = 1/f = 2π/ω'); ε_rms = ε₀/√2 = 1.7678 → row '1.768' (§15.2 Eq. 15.5); I_peak = ε₀/R = 0.250 A through the fixed 10 Ω load (the Example 13.1 I = ε/R book) — four closed-form rows, one state, all machine-certified
  • Price the spin rate twice: holding B, A, N fixed and dragging Angular Speed 5 → 10 rad/s flips the ε_max row 2.500 → 5.000 V AND the Frequency row 0.80 → 1.59 Hz in the same drag — 'the faster the generator is spun (greater ω), the greater the emf' (§13.6 verbatim) because ω is a multiplicative factor of the PEAK (Eq. 13.18), and f = ω/2π makes the same lever the frequency. Machine-certified by the full-grid ω monotonic sweep (0.500 → 10.000 V over 1–20 rad/s) and the IEEE-exact doubling-ratio bounds; the HTML's own quiz Q3 ('Both peak EMF and frequency double') is the in-scene sibling. The US grid exhibit: the Power Plant preset spins at exactly ω = 2π·60 = 120π rad/s → f = 60.00 Hz (§15.1's household ω = 120π rad/s)
  • Read the phase, not just the peak: with Φ = NBA·cos(ωt) the EMF ε = −dΦ/dt = NBAω·sin(ωt) is a quarter-cycle shifted — exactly ZERO at the flux-maximum orientation and at its ±NBAω peak where the flux itself is zero (the sim's own sidebar: 'Maximum EMF occurs when the coil plane is parallel to B… zero EMF when the coil is perpendicular to B (flux is at its peak — no change)'; machine gate: ε at the flux-maximum orientation ≡ 0 within 1e-9). Distinguish the three voltage numbers the panel shows at once: instantaneous ε(t) (signed, swings ±ε₀), peak ε₀, and ε_rms = ε₀/√2 — 'if we average out the values of current or voltage, these values are zero. Therefore, we often use a second convention called the root mean square value' (§15.2 verbatim). HONEST SCOPE: the 3D scene's rotation axis is parallel to B so the scene's geometric flux is identically zero (registered quirk) — the phase lesson anchors the engine's convention, the panel rows, and the ε-t trace, never the scene's flux geometry; the HTML quiz Q2's keyed orientation contradicts the sidebar and is registered as a bug, never anchored

Formulas

E(t)=NBAωsin(ωt)\mathcal{E}(t) = NBA\omega\sin(\omega t)
Generator EMF
ω=2πf\omega = 2\pi f
Angular frequency
Emax=NBAω\mathcal{E}_{max} = NBA\omega
Peak EMF

Make a prediction

Default state: Angular Speed ω = 5.0 rad/s, B = 1.0 T, Coil Area = 100 cm², N = 50 — the ε_max row reads 2.500 V and the Frequency row reads 0.80 Hz. You drag Angular Speed to 10.0 rad/s, exactly doubling the spin, with B, A, and N untouched. What do the ε_max and Frequency rows read now?

No grading here — commit to a guess, then scroll down and test it yourself.

This is a Pro experiment

Upgrade to Pro to access this experiment — or keep learning with one of our free labs.

Scroll for the debrief ↓

Check

Did the lab agree with you?

Your prediction

You skipped the prediction — jump back to the preview and commit to a guess first; the comparison is the whole point.

Answer: ε_max = 5.000 V and f = 1.59 Hz — both double: the peak EMF is NBAω, so ω is a factor of the peak itself, and the frequency is ω/2π — one slider prices both

The peak EMF is ε₀ = NABω (OpenStax Eq. 13.18, verified): ω sits inside the product, so doubling the spin from 5 to 10 rad/s doubles the peak from 2.5 V to exactly 5.0 V — the panel shows '5.000'. And the frequency is f = ω/2π (§13.6, verified), so the same drag doubles 0.7958 Hz to 1.5915 Hz — the panel shows '1.59'. The textbook states the voltage half flat: 'the faster the generator is spun (greater ω), the greater the emf' (§13.6). Option B correctly doubles the voltage but invents a geometry-owned frequency — turns and area price the peak, never the period; only ω sets the clock. Option C is the documented flux-for-flux-rate conflation in its rotating-frame form: it reads the law as ε₀ = NBA and concludes that unchanged flux means unchanged voltage — but Faraday's law prices a RATE of change, and spinning faster is precisely changing the flux faster; the same instinct the translating-magnet lab refuted with a plunging magnet, refuted here with a spinning coil. The lab's machine gates bracket the doubling ratio at exactly 2 and sweep the whole 1–20 rad/s grid monotonic from 0.500 to 10.000 V.

Quiz (0/3)

A coil with N=100, B=0.5T, A=0.01m² spins at 60Hz. What is the peak EMF?

Why does the EMF vary as sin(ωt) and not cos(ωt)?

Why do wind turbines use permanent magnet generators instead of electromagnets?

You can now

  • Close the generator equation on the panel's own rows at the default state (ω = 5.0 rad/s, B = 1.0 T, A = 100 cm² = 0.01 m², N = 50): ε₀ = NABω = 50 × 1.0 × 0.01 × 5 = 2.5 V exactly → row '2.500' (OpenStax Eq. 13.18 'ε₀ = NABω is the peak emf'); T = 2π/ω = 1.2566 s → row '1.257' and f = ω/2π = 0.7958 Hz → row '0.80' (§13.6: 'f = ω/2π and the period is T = 1/f = 2π/ω'); ε_rms = ε₀/√2 = 1.7678 → row '1.768' (§15.2 Eq. 15.5); I_peak = ε₀/R = 0.250 A through the fixed 10 Ω load (the Example 13.1 I = ε/R book) — four closed-form rows, one state, all machine-certified
  • Price the spin rate twice: holding B, A, N fixed and dragging Angular Speed 5 → 10 rad/s flips the ε_max row 2.500 → 5.000 V AND the Frequency row 0.80 → 1.59 Hz in the same drag — 'the faster the generator is spun (greater ω), the greater the emf' (§13.6 verbatim) because ω is a multiplicative factor of the PEAK (Eq. 13.18), and f = ω/2π makes the same lever the frequency. Machine-certified by the full-grid ω monotonic sweep (0.500 → 10.000 V over 1–20 rad/s) and the IEEE-exact doubling-ratio bounds; the HTML's own quiz Q3 ('Both peak EMF and frequency double') is the in-scene sibling. The US grid exhibit: the Power Plant preset spins at exactly ω = 2π·60 = 120π rad/s → f = 60.00 Hz (§15.1's household ω = 120π rad/s)
  • Read the phase, not just the peak: with Φ = NBA·cos(ωt) the EMF ε = −dΦ/dt = NBAω·sin(ωt) is a quarter-cycle shifted — exactly ZERO at the flux-maximum orientation and at its ±NBAω peak where the flux itself is zero (the sim's own sidebar: 'Maximum EMF occurs when the coil plane is parallel to B… zero EMF when the coil is perpendicular to B (flux is at its peak — no change)'; machine gate: ε at the flux-maximum orientation ≡ 0 within 1e-9). Distinguish the three voltage numbers the panel shows at once: instantaneous ε(t) (signed, swings ±ε₀), peak ε₀, and ε_rms = ε₀/√2 — 'if we average out the values of current or voltage, these values are zero. Therefore, we often use a second convention called the root mean square value' (§15.2 verbatim). HONEST SCOPE: the 3D scene's rotation axis is parallel to B so the scene's geometric flux is identically zero (registered quirk) — the phase lesson anchors the engine's convention, the panel rows, and the ε-t trace, never the scene's flux geometry; the HTML quiz Q2's keyed orientation contradicts the sidebar and is registered as a bug, never anchored

Step-by-step

  1. Drag the four sliders and watch the data panel reprice live: Angular Speed ω (1–20 rad/s) scales both peak EMF and frequency; B Field, Coil Area, and N Turns each scale the peak linearly.
  2. The EMF trace at lower left draws ε(t) in real time — pause at a zero crossing and note the flux is at its maximum right then.
  3. Try the presets: Hand Crank (0.24 V), Power Plant (ω jumps to 377 rad/s — beyond the slider, its label reads the true value — delivering 60.00 Hz), and High Field (80 V).
  4. Peak Current is priced against a fixed 10 Ω load.
  5. The panel stays live — sliders reprice every row even while paused (the legacy build froze its panel when paused, a registered quirk not ported; 组装线 mha Fire 先例).

Key formulas

  • E(t)=NBAωsin(ωt)\mathcal{E}(t) = NBA\omega\sin(\omega t)Generator EMF
  • ω=2πf\omega = 2\pi fAngular frequency
  • Emax=NBAω\mathcal{E}_{max} = NBA\omegaPeak EMF

Frequently asked questions

A coil with N=100, B=0.5T, A=0.01m² spins at 60Hz. What is the peak EMF?
Ε_max = NBAω = 100 × 0.5 × 0.01 × 2π×60 ≈ 188 V.
Why does the EMF vary as sin(ωt) and not cos(ωt)?
It depends on the starting orientation — at t=0 the coil plane is perpendicular to B (flux at maximum), so ε(0) = 0 and the EMF follows sin. Starting from the plane parallel to B (zero flux, fastest change) would give cos.
Why do wind turbines use permanent magnet generators instead of electromagnets?
Permanent magnets need no excitation power and work even when grid is down.