Question

A bar 1.91 m long pivots in a vertical plane about one end. The first 0.37 m of this rod is made of nonconducting material, but the outer part are made of iron (see Figure). The apparatus is within a uniform 1.2 T magnetic field oriented at right angles to the plane in which the bar rotates. At what angular speed would you need to rotate this bar to generate a potential difference of 11.8 V between the ends of the iron segment? Express the answer with two decimal places.

Answer 1

area dA = pi*(L^2-l^2)

time dt = 2*pi/w

dA/dt = (L^2 - l^2)*w/2

magnetic flux = B*A*cos0 = B*A

emf = rate of change of flux emf = (d/dt)*(B*A) =

emf = B*(dA/dt)

emf = B*(L^2 - l^2)*w/2

given emf = 11.8 V

L = 1.91

l = 0.37 m

B = 1.2

11.8 = 1.2*(1.91^2 - 0.37^2)*w/2

w = 5.6 rad/s

w = 5.6*(60/(2pi))

w = 53.5 revolutions per minute