Question

The armature of a small generator consists of a flat, square coil with 130 turns and sides with a length of 1.10 cm . The coil rotates in a magnetic field of 7.65×10⁻² T .

Part A

What is the angular speed of the coil if the maximum emf produced is 2.00×10⁻² V ?

(answer in rad/s)

Answer 1

flux = phi = BA cos wt
A = area of crossection = 0.018*0.018 m^2
w = angular speed, t =time
instantaneous emf = e(t) = - N d(phi)/dt = - w (BA) [- sin wt]
e (max) = e0, when sin wt = 1 or angle wt = 90 degree
----------------------
eo = BANw
w = e0/BAN = 2.00×10−2/(7.65×10−2*0.011*0.011*130)
w = 16.62 rad/sec ANSWER