Emf induced in a coil rotating uniformly in a magnetic field: ε = BANωsin ωt

The idea

A flat coil of N turns and area A rotates at constant angular velocity ω in a uniform magnetic flux density B. The flux linkage through the coil is NΦ = BAN cos ωt, because the component of B perpendicular to the coil face varies as the cosine of the angle between the field and the normal. Faraday's law states that the induced emf equals the negative rate of change of flux linkage, so differentiating gives ε = BANω sin ωt. The emf is therefore sinusoidal, with peak value ε₀ = BANω reached when the coil plane is parallel to B, where flux linkage is momentarily zero but changing fastest. At the peaks of flux linkage the emf is zero. Increasing B, A, N or ω raises the peak emf and, for ω, also the frequency.

What earns the marks

• States that the flux linkage through the rotating coil is NΦ = BAN cos ωt, identifying the cosine variation with the angle between the field and the coil normal.
• Applies Faraday's law as the rate of change of flux linkage, differentiating BAN cos ωt to obtain ε = BANω sin ωt.
• Identifies the peak emf as ε₀ = BANω and explains that it occurs when the coil plane is parallel to the magnetic field, where flux linkage is zero but changing at its greatest rate.
• Explains that the emf is zero at maximum flux linkage, when the coil plane is perpendicular to the field, because the rate of change of flux linkage is momentarily zero.
• Relates the angular frequency ω to the rotation frequency f using ω = 2πf, and links the sinusoidal emf to the sinusoidal variation of flux linkage.
• Describes how increasing B, A, N or ω increases the peak induced emf, and that changing ω also changes the frequency of the output.

Common mistakes and how to put them right

• Treating the induced emf as proportional to the flux linkage rather than to its rate of change.

  Put it right: Differentiate the flux linkage expression, giving a sine term from the cosine term.

• Using degrees in the differentiation or in ωt.

  Put it right: Work in radians, since ω is in rad s⁻¹ and the derivative of cos ωt is −ω sin ωt.

• Assuming the emf is maximum when flux linkage is maximum.

  Put it right: Maximum emf occurs when flux linkage is zero and changing fastest, and zero emf occurs at maximum flux linkage.

Exam tips

• Sketch the flux linkage and emf graphs on the same time axis, labelling the quarter-cycle phase difference, to show clearly how the sine and cosine variations are related.
• When substituting values, convert a rotation rate in revolutions per second to angular frequency using ω = 2πf before calculating BANω.
• State explicitly that the derivation uses Faraday's law and the chain rule, so the examiner can see the physics behind each mathematical step.