A closed coil having November turns is rotated on a uniform magnetic field Bh about a diameter which is perpendicular to the field the angular velocity of rotation is. The average emf developed in a full turn is ?
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November turns kisme hota h
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Sol. n = 100 turns, B = 4 × 10^–4 T A = 25 cm^2 = 25 × 10^–4 m^2 a) When the coil is perpendicular to the field ϕ = nBA When coil goes through half a turn ϕ = BA cos 18° = 0 – Nba dϕ = 2nBA The coil undergoes 300 rev, in 1 min 300 × 2π rad/min = 10 π rad/sec 10π rad is swept in 1 sec. π/π rad is swept 1/10π × π = 1/10 sec E = dϕ/dt = 2nBA/dt = 2 x 100 x 4 x 10^-4 x 25 x 10^-4/ 1/10 = 2 x 10^-3 V b) ϕ1 = nBA, ϕ2 = nBA (ϕ = 360°) dϕ = 0 (c) i = E/R = 2x10^-3 / 4 = ½ x 10^-3 = 0.5 × 10^–3 = 5 × 10^–4 q = idt = 5 × 10^–4 × 1/10 = 5 × 10^–5 C.
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