The coil of a generator has a radius of 0.12 m. When this coil is unwound, the wire from which it is made has a length of 5.5 m. The magnetic field of the generator is 0.10 T, and the coil rotates at an angular speed of 25 rad/s. What is the peak emf of this generator?

Answer 1

Number of turns in the coil :
N = 5.5/(2 pi R) = 7.3
Coil area = A = pi R^2
B = 0.1 T
R = 0.12 m

Phi = magnetic flux = B A N cos theta
Induced EMF =d(Phi)/dt = - B A N sin theta * w
where w = d(theta) dt = 25 radians per second

Peak EMF = B A N w