The armature of an AC generator is rotating at a constant speed of 30 revolutions/second in a horizontal field of flux density 1.0Wb/m^2. The diameter of the cylindrical armature is 24cm and its length is 40cm. What is the maximum emf induced in the armature having 30 turns?
emf=-magnetic flux rate although this problem is more about magnitude
The Attempt at a Solution
The generator is rotating at 0.033seconds/period. The area facing the magnetic field follows a sin curve. The full area is (0.12m)^2*pi=0.045m^2 The sine curve has angular frequency b given by 2*pi/b=0.033seconds/period. So b=188.5 Hence the area is given by 0.045m^2*sin(188.5t)
The magnetic flux rate is dA/dt * B
max magnetic flux rate = 8.48*30 = 254.5V which includes the 30 turns. The answer stated 543V. Why?
When doing the problem I have not used the length of the cylindrical armature which stands at 0.4m. I don't know where I could use it.