A coil of 190 turns and area 0.03 m² is rotated at 40 Hz in a 0.09 T field. What is the maximum emf?
**Maximum emf** e₀ = N B A ω, for 180 turns A=0.02 m² B=0.03 T ω=60 rad/s, e₀=180×0.03×0.02×60=6.48 V, illustrating dependence on N, B, A, ω, used in generator design, increasing N or B or A or ω raises output. ω = 2π v = 2π × 40 = 80π rad/s . ε₀ = N B A ω = 190 × 0.09 × 0.03 × 80π = 128.91 V ≈ 128.9 V . Using Φ = B A cosθ, e = -N dΦ/dt = -N A dB/dt = B l v = N B A ω sinωt, L = μ₀ N²A/l, M = e/(dI/dt) and
Ref: NCERT > Physics Book > Electromagnetic Induction > Rotational EMF and AC Generator