Physics · Electromagnetic Induction · NEET
The EMF is the rate of change of flux, ε = −N dΦ/dt. The flux is Φ = BA cos(ωt). When you differentiate cos(ωt) you get −ω sin(ωt). So ε = −N × BA × (−ω sin ωt) = NBAω sin(ωt). The two minus signs cancel and you are left with a positive sine. Physically, EMF is largest when the flux is changing fastest — that happens when the flux itself is passing through zero (the steepest part of a cosine), which is exactly where a sine peaks.
It comes from differentiating cos(ωt). By the chain rule, d/dt of cos(ωt) is −ω sin(ωt). That extra ω is the angular speed of the coil. This is why a coil that spins faster gives a bigger peak EMF — ε₀ = NBAω grows with ω. In problems, remember ω = 2πf, where f is revolutions per second (frequency).
ε (lowercase, no subscript) is the instantaneous EMF — the value at one particular instant of time t, given by ε = ε₀ sin(ωt). ε₀ (with subscript zero) is the peak or maximum EMF, a fixed number equal to NBAω. It is the tallest value the sine wave ever reaches (when sin ωt = 1). NEET usually asks for ε₀, the maximum value.
It does not disappear — it cancels. Faraday's law has a minus sign: ε = −N dΦ/dt. But differentiating cos(ωt) also produces a minus sign: d(cos ωt)/dt = −ω sin(ωt). Minus times minus is plus, so the final equation ε = NBAω sin(ωt) is positive. The Lenz's-law minus sign only tells us direction; the magnitude of the peak EMF is simply NBAω.
The coil rotates at a constant angular speed ω. Angle turned = angular speed × time, so θ = ωt (taking θ = 0 at t = 0). The flux depends on the angle between the field B and the coil's area vector A through Φ = BA cos θ. Substituting θ = ωt turns the static formula into a time-dependent one, Φ = BA cos(ωt), which is what we differentiate.
An emf is generated by an ac generator having a 100 turn coil of loop area 1 m². The coil rotates at a speed of one revolution per second and is placed in a uniform magnetic field of 0.05 T perpendicular to the axis of rotation. The maximum value of emf is:
A big circular coil of 1000 turns and average radius 10 m is rotating about its horizontal diameter at 2 rad/s. If the vertical component of earth's magnetic field is 2 × 10⁻⁵ T and the coil resistance is 12.56 Ω, the maximum induced current in the coil will be:
Try the real previous-year questions from this chapter — each with the answer and a full solution.
ε = ε₀ sin(ωt), where ε₀ = NBAω is the peak EMF. N is the number of turns, B the magnetic field, A the coil area, and ω the angular speed of rotation.
Start with flux Φ = BA cos(ωt). Apply Faraday's law ε = −N dΦ/dt. Differentiating cos(ωt) gives −ω sin(ωt), so ε = NBAω sin(ωt). Writing ε₀ = NBAω gives ε = ε₀ sin(ωt).
ε₀ is the peak (maximum) EMF, equal to NBAω. It is the highest value the EMF reaches, when sin(ωt) = 1. It grows if you increase turns, field, area, or spin speed.
EMF depends on how fast flux changes. When the coil plane is parallel to B (area vector perpendicular to B), the flux is momentarily zero but changing fastest, so the EMF peaks. When the coil is perpendicular to B, flux is maximum but changing slowest, so EMF is zero.
Use ω = 2πf, where f is the frequency in revolutions per second (Hz). For example, 1 rev/s gives ω = 2π ≈ 6.28 rad/s. Always convert before using ε₀ = NBAω unless ω is already given in rad/s.
