Electromagnetism - Faraday’s law

A wire is being wound around a rotating wooden cylinder of radius R. One end of the wire is connected to the axis of the cylinder, as shown in the figure. The cylinder is placed in a uniform magnetic field of magnitude B parallel to its axis and rotates at N revolutions per second. What is the potential difference between the open ends of the wire?

A. 0
B. 2πNBR
C. πNBR²
D. BR²/N
E. πNBR³

(GR9677 #47)

Solution:

Faraday's Law: ɛ = − dΦ/dt
with  Φ = NBA  

Given:
N revolution per second
B uniform (constant)
A constant

ɛ = − BA dN/dt
= − BπR²N

|ɛ| = πNBR²

Answer: C