Navjyot Kalra
Last Activity: 10 Years ago
Sol. A = 5 cm^2 = 5 x 10^–4 m^2
B = B0 sin ωt = 0.2 sin(300 t)
θ = 60°
a) Max emf induced in the coil
E = dϕ/dt = d/dt (BA cosθ)
= d/dt (B base 0 sin ω t x 5 x 10^-4 x 1/2)
= B base 0 x 5/2 x 10^-4 d/dt (sin ωt) = B base 0 5/2 x 10^-4 cosωt . ω
= 0.2x5/2 x 300 x 10^-4 x cos ωt = 15 x 10^-3 cos ωt
E base max = 15 x 10^-3 = 0.015 V
b) Induced emf at t = (π/900) s
E = 15 x 10^–3 x cos ωt
= 15 x 10^–3 x cos (300 x π/900) = 15 x 10^–3 x ½
= 0.015/2 = 0.0075 = 7.5 x 10^–3 V
c) Induced emf at t = π/600 s
E = 15 x 10^–3 x cos (300 x π/600)
= 15 x 10^–3 x 0 = 0 V.