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Hrishant Goswami Grade: 10
        A closed coil having 100 turns is rotated in a uniform magnetic field B = 4.0 ω 10^-4 T about a diameter which is perpendicular to the field. The angular velocity of rotation if 300 revolutions per minute. The area of the coil is 25 cm^2 and its resistance is 4.0 Ω. Find (a) the average emf developed in half a turn from a position where the coil is perpendicular to the magnetic field, (b) the average emf in a full turn and (c) the net charge displaced in part (a). 
3 years ago

Answers : (1)

Aditi Chauhan
askIITians Faculty
396 Points
										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.
3 years ago
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