A rectangular coil with 45 turns, area 0.0021 m² per turn, rotates at 11 Hz in a uniform magnetic field 0.21 T. At t = 0 its area normal is parallel to the field. Define positive flux linkage as NAB cos(ωt), with ω = 2πf, and signed induced emf ε = −d(NΦ)/dt. Find the peak emf, the signed emf at t = 0.007575758 s, and the new peak emf if rotation frequency doubles with the same coil and field. Explain why the induced emf is zero when flux linkage is greatest.

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Peak emf 1.372 V; ε at stated time +0.6858 V; doubled-frequency peak emf 2.743 V. The angular speed is ω = 2πf = 69.12 rad s⁻¹. Multiplying field, area and turn count gives the peak flux linkage NAB, not the emf itself. Faraday's law takes the time rate of change: ε = −d[NAB cos(ωt)]/dt = NABω sin(ωt). Therefore the emf amplitude is ε₀ = NABω. Substitution gives ε₀ = 1.372 V. The supplied time equals T/12, so ωt = π/6 and ε = ε₀ sin(π/6) = +0.6858 V under the defined sign convention. Doubling rotation frequency doubles ω, so the new peak emf is 2ε₀ = 2.743 V. The maximum flux linkage NAB is unchanged, but the same flux change occurs in half the time. When flux linkage is at a maximum its instantaneous rate of change is zero, so the emf is zero. Emf becomes greatest as flux passes through zero most rapidly; the two quantities are a quarter-cycle out of phase.

A rotating coil and alternating emf · set 1

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