Lessons · circuits

The AC generator: the induced EMF is the slope of the flux

Regime 2 — calculus does what algebra cannot. The acceleration isn't constant, so the algebra formulas don't apply; calculus is the only road in, and SymPy proves the closed form solves the equation of motion.

Regime 2 · calculus does more Math machine-derived & checked by SymPy 2 modeling assumptions (author-asserted)

A coil of area $A = 0.1$ m² spins at $\omega = 10$ rad/s in a uniform magnetic field $B = 0.5$ T. The flux through it varies as $\Phi(t) = BA\cos(\omega t)$ , and the induced EMF is its rate of change, $\mathcal{E} = -d\Phi/dt = BA\omega\sin(\omega t)$ — a sinusoid that lags the flux by a quarter cycle (its peak comes a quarter turn after the flux's). This is how a generator turns rotation into alternating current. Plotted as a stacked $\Phi$ – $t$ over EMF– $t$ , the slope of the flux is the (negated) EMF at every instant. Drag $B$ or $\omega$ : when the flux is at its peak the EMF is zero, and when the flux crosses zero the EMF is at its peak.

Peak EMF

\mathcal{E}_0 = BA\omega

0.5 V

Peak flux

\Phi_0 = BA

0.05 Wb

Period

T = 2\pi/\omega

0.6283 s

The instantaneous EMF is the derivative of the flux

\mathcal{E}(t) = -\frac{d\Phi}{dt}, \quad \Phi(t) = BA\cos(\omega t)

Faraday's law in its exact form uses the instantaneous rate of change — a derivative, not a ratio of finite changes. For a coil rotating at constant $\omega$ , the flux is a cosine, $\Phi = BA\cos(\omega t)$ .

check $\mathcal{E} = -d\Phi/dt$ ; check $\int_0^t \mathcal{E}\,dt' = -\Delta\Phi$ ; peak $\mathcal{E}_0 = BA\omega$ ; $90^\circ$ phase; period $T = 2\pi/\omega$

✓ The EMF is the negative slope of the flux: $\mathcal{E} = -\dfrac{d\Phi}{dt}$ (Faraday's law). [structural]
✓ Integrating the EMF recovers the change in flux: $\int_0^t \mathcal{E}\,dt' = -\Delta\Phi$ — the area↔change pivot, one domain over. [structural]
✓ The peak EMF is $\mathcal{E}_0 = BA\omega$ , reached a quarter period in (when the flux crosses zero). [structural]
✓ The EMF is zero exactly when the flux is at its peak (here $t=0$ ) — the slope of a cosine at its crest is flat, so flux and EMF are $90^\circ$ out of phase. [structural]
✓ The generator repeats every period $T = 2\pi/\omega$ : $\Phi(t+T) = \Phi(t)$ . [structural]

Dimensional homogeneity: checked by SymPy (holds).

✗ Common misconception: “The flux and the EMF reach their maxima at the same time — they are in phase.”

No. The EMF is the rate of change of the flux, $\mathcal{E} = -d\Phi/dt$ , so it is $90^\circ$ out of phase. When the flux $\Phi$ is at its peak, its slope is zero, so the EMF is zero. When the flux passes through zero (its steepest point), $|d\Phi/dt|$ is largest, so the EMF is at its peak. Watch the two panels: each EMF (lower) peak falls a quarter cycle after the corresponding flux (upper) peak — a sine lagging the cosine by $90^\circ$ .

Modeling assumptions — author-asserted, disclosed not discharged

A uniform, constant magnetic field, and a single-turn coil rotating at constant angular velocity $\omega$ .
An ideal EMF source (no coil resistance), so the terminal voltage is the induced EMF $\mathcal{E} = -d\Phi/dt$ exactly.

The stacked graph, fully annotated

A static rendering (Matplotlib) at the default parameters — the interactive version is in the Graph tab above.

Formulas used

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