A square coil of side 625cm in ail ac generator rotates at 5

A square coil of side 6.25-cm in ail ac generator rotates at 50.0 Hz and develops an emf of 220 V (rms), (a) If the coil contains 812 turns, find the magnitude of the magnetic field in which it rotates, (b) If the magnetic field were 0.205 T and the generator were to rotate at 60.0 Hz, what would be the rms emf developed by the generator?

Solution

Emf = N A B w sin(wt)

Erms = N A B w / sqrt(2)

(A)
Erms = 220 Volt, N = 812 turns

A = 0.0625 * 0.0625 m^2 , w = 50 Hz

220 = (812)(0.0625 * 0.0625) ( B) ( 50) / sqrt(2)

B = 1.96 T .....Ans

(B) B = 0.205 T , w = 60 Hz , N = 812 , A = (0.0625)^2

Erms = (812)(0.0625^2)(0.205)(60)/sqrt(2)

Erms = 27.6 Volt