EMF in a Rotating Coil Revision Notes | A Level AQA Physics

1. Why a Rotating Coil Generates Alternating EMF

Every generator in every power station — from a wind turbine to a nuclear plant — works on one principle: a coil rotating inside a magnetic field. Understanding this is understanding where mains electricity comes from.

You already know from Faraday's law that a changing flux linkage induces an EMF. The rotating coil is the most elegant way to produce that continuous change — and the mathematics reveals exactly why the output is sinusoidal.

2. Flux Linkage in a Rotating Coil

Picture a rectangular coil of $N$ turns and cross-sectional area $A$, sitting inside a uniform magnetic field of flux density $B$. At time $t = 0$, the plane of the coil is perpendicular to the field, so the normal to the coil is parallel to the field and the flux linkage is at its maximum.

As the coil rotates with angular velocity $\omega$, the angle between the field and the normal to the coil increases. The flux linkage at any instant is given by $N\Phi = BAN\cos(\omega t)$. When the coil has rotated so its plane is parallel to the field, the flux linkage drops to zero — because no field lines thread through the coil at that moment.

The cosine function tells you the flux linkage oscillates between $+BAN$ and $-BAN$ as the coil spins. That oscillation is the engine behind the alternating EMF.

3. Deriving the EMF Equation

Faraday's law states that the induced EMF equals the negative rate of change of flux linkage: $\varepsilon = -\frac{d(N\Phi)}{dt}$. To find the EMF in the rotating coil, differentiate the flux linkage expression with respect to time.

Worked example: Deriving $\varepsilon = BAN\omega\sin(\omega t)$

Starting from $N\Phi = BAN\cos(\omega t)$, derive the expression for the induced EMF.

1. $\varepsilon = -\dfrac{d(N\Phi)}{dt}$ — The induced EMF equals the negative rate of change of flux linkage. The negative sign reflects Lenz's law — the induced EMF opposes the change causing it.
2. $\varepsilon = -\dfrac{d}{dt}\left[BAN\cos(\omega t)\right]$ — $B$, $A$, $N$, and $\omega$ are all constants for a coil rotating at steady speed in a uniform field, so they come outside the derivative.
3. $\dfrac{d}{dt}\left[\cos(\omega t)\right] = -\omega\sin(\omega t)$ — Using the chain rule: the derivative of $\cos(u)$ is $-\sin(u)$, and $\frac{du}{dt} = \omega$. The result carries a factor of $\omega$.
4. $\varepsilon = -BAN \times (-\omega\sin(\omega t)) = BAN\omega\sin(\omega t)$ — The two negatives cancel. The result is a positive sine function — the EMF is sinusoidal and 90° out of phase with the flux linkage.
5. $\varepsilon_0 = BAN\omega$, so $\varepsilon = \varepsilon_0\sin(\omega t)$ — The amplitude $BAN\omega$ is the peak (maximum) EMF. It is reached whenever $\sin(\omega t) = 1$, i.e. when the coil plane is parallel to the field.

Answer: $\varepsilon = BAN\omega\sin(\omega t)$, with peak EMF $\varepsilon_0 = BAN\omega$

Notice the crucial phase relationship: the flux linkage follows a cosine, but the EMF follows a sine. They are 90° out of phase. When flux linkage is at its maximum (coil perpendicular to field), the EMF is zero. When flux linkage is zero (coil parallel to field), the EMF is at its peak.

4. The Sinusoidal Output and What It Means

The equation $\varepsilon = BAN\omega\sin(\omega t)$ tells you the EMF at every instant during the rotation. The output is a perfect sine wave — this is why generators produce alternating current (AC).

Reading the chart: at $t = 0$, flux linkage is maximum and EMF is zero. A quarter period later ($t = T/4$), flux linkage has fallen to zero and EMF is at its positive peak. This 90° phase difference is a direct consequence of differentiation — the EMF is the gradient of the flux linkage curve.

At $t = T/2$, the coil has rotated 180° and is back perpendicular to the field, but with the opposite face towards the field source. Flux linkage is at its negative maximum, and EMF is again zero. The cycle then repeats in the negative direction before returning to the start.

5. Calculating Peak and Instantaneous EMF

With the equation established, you can now tackle the two types of calculation that appear in AQA exam questions: finding the peak EMF from coil parameters, and finding the instantaneous EMF at a specific time or angle.

Worked example: Calculating Peak and Instantaneous EMF

A coil has 200 turns, area $4.0 \times 10^{-3}$ m$^2$, and rotates at 50 Hz in a uniform magnetic field of flux density 0.15 T. Calculate (a) the peak EMF and (b) the instantaneous EMF at $t = 3.0 \times 10^{-3}$ s.

1. $\omega = 2\pi f = 2 \times \pi \times 50 = 100\pi \approx 314$ rad s$^{-1}$ — Angular velocity links the rotation frequency to the sine function argument. Always convert frequency to $\omega$ before substituting.
2. $\varepsilon_0 = BAN\omega = 0.15 \times 4.0 \times 10^{-3} \times 200 \times 100\pi$ — Multiply all four quantities together. Keep $\pi$ symbolic until the final step to avoid rounding errors mid-calculation.
3. $0.15 \times 4.0 \times 10^{-3} = 6.0 \times 10^{-4}$ $6.0 \times 10^{-4} \times 200 = 0.12$ $0.12 \times 100\pi = 12\pi \approx 37.7$ V — Working step by step: $0.15 \times 4.0 \times 10^{-3} = 6.0 \times 10^{-4}$; then $6.0 \times 10^{-4} \times 200 = 0.12$; then $0.12 \times 314 = 37.7$ V. [M1 for correct formula, A1 for correct answer]
4. $\omega t = 100\pi \times 3.0 \times 10^{-3} = 0.300\pi = 0.942$ rad — The argument of the sine function must be in radians. $100\pi \times 3.0 \times 10^{-3} = 0.3\pi$ rad.
5. $\varepsilon = 37.7 \times \sin(0.942) = 37.7 \times 0.809 \approx 30.5$ V — $\sin(0.942 \text{ rad}) = 0.809$ (ensure your calculator is in radian mode). [A1 for correct substitution and answer]

Answer: (a) Peak EMF $\varepsilon_0 \approx 37.7$ V; (b) Instantaneous EMF $\varepsilon \approx 30.5$ V at $t = 3.0 \times 10^{-3}$ s

6. Effect of Coil Parameters on Peak EMF

The equation $\varepsilon_0 = BAN\omega$ is a powerful design tool. Each factor independently scales the peak output, which is why real generators are engineered with large coils, strong magnets, many turns, and high rotation speeds.

Notice that all four parameters have the same proportional relationship with $\varepsilon_0$. Doubling any one of them doubles the peak EMF. This linearity is a direct consequence of the derivative relationship between EMF and flux linkage.

7. Exam Technique and Common Mistakes

AQA questions on this topic tend to test three things: correct substitution into $\varepsilon = BAN\omega\sin(\omega t)$, the phase relationship between EMF and flux linkage, and the ability to explain changes in peak EMF in terms of the equation. Here is how to handle each one cleanly.

The core of this topic comes down to one derivation and one equation. If you can explain why differentiating $BAN\cos(\omega t)$ gives $BAN\omega\sin(\omega t)$, and if you can interpret the physical meaning of each factor, you have everything the AQA mark scheme is looking for.

This model also explains why the induced emf is zero when the coil momentarily presents maximum flux linkage and greatest when the flux linkage is changing most rapidly. In exam questions, always connect the shape of the emf graph to the rate of change of magnetic flux linkage rather than to flux linkage alone.