In a region of uniform magnetic induction of B = 10^-2 T, a circular coil of radius 30 cm having resistance of π^2 Ω is rotated about an axis which is perpendicular to the direction of coil which forms the diameter of the coil. . .

Question : In a region of uniform magnetic induction of B = 10^-2 T, a circular coil of radius 30 cm having resistance of π² Ω is rotated about an axis which is perpendicular to the direction of coil which forms the diameter of the coil. If the coil is rotated at 200 rpm, the amplitude of the alternating current induced in the coil is 

(A) 4π² mA

(B) 30 mA

(C) 6 mA

(D) 200 mA

Doubt by Chinmay

Solution : 

Here 
B = 10^-2 T
r = 30 cm
r = 30/100 = 0.3 m
R = π² Ω
f = 200 rpm
f = 200/60 rps
f = 10/3 rps
ω = 2πf 
    = 2π(10/3) rad/s
    = 20π/3 rad/s

We know, 
ε₀ = NABω
ε₀ = 1×π(0.3)²×(10^-2)(20π/3)
ε₀ =6π² 10^-3 V

Also 
I₀ = ε₀/R
I₀ = 6π² ×10^-3 /π²

I₀ = 6 × 10^-3
I₀ = 6 mA

Option (c) 6 mA is correct.