Practice question

**Back emf** in motor opposes applied voltage, e_b = N B A ω sin ωt, reduces net current, at start ω=0 e_b=0 current large, as speed increases e_b increases limiting current, power conversion mechanical, principle of motor and generator reciprocity. Initial flux: Φ = B A = 0.2 × π × (0.11)² = 0.0076 Wb . Final flux = 0. ε = (Δ Φ/Δ t) = (0.0076/0.5) = 0.0152 V ≈ 0.015 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)