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. ε = (1/2) B ω R² . ε = (1/2) × 0.3 × 25 × (0.8)² = 2.4 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 U = ½ L I², result 2.4 V follows, reflecting Faraday's law and Lenz's opposition.