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:
revolution per second
uniform (constant)
constant

ɛ = − BA dN/dt
= − BπR2N

|ɛ| = πNBR2

Answer: C

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