Practice question

**Motional emf** for rod length l moving with velocity v perpendicular to uniform field B, e = B l v (V), B in T, l in m, v in m/s, direction given by right-hand rule, positive end where positive charges accumulate due to q v×B force. For l=0.9 m, v=1.2 m/s, B=0.5 T, e=0.5×0.9×1.2=0.54 V. ε = B l v , l = 0.14 m . ε = 0.7 × 0.14 × 1.2 = 0.1176 V ≈ 0.118 V . Using Φ = B A cosθ, e = -N dΦ/dt = -N A dB/dt = B l v = N B A ω sinωt, L