A bicycle generator rotates at 1,975 rad/s, producing a 20.0 V peak emf. It has

Question

A bicycle generator rotates at 1,975 rad/s, producing a 20.0 V peak emf. It has a 53-turn, 1.00 by 3.00 cm rectangular coil in a 0.640 T field. It is driven by a 1.64 cm diameter wheel that rolls on the outside rim of the bicycle tire. (a) What is the velocity of the bicycle? (Enter the magnitude in m/s.) m/s (b) What is the maximum emf (in V) of the generator when the bicycle moves at 10.0 m/s, noting that it was 20.0 V under the original conditions? V (c) If the sophisticated generator can vary its own magnetic field, what field strength (in T) will it need at 5.00 m/s to produce a 9.00 V maximum emf? T

Explanation / Answer

Part A:

Velocity of bicycle will be

V = w*r

w = 1975 rad/sec

r = 1.64 cm/2 = 0.82 cm

V = 1975*0.82*10^-2 = 16.20 m/sec

Part B

Max Emf is given by:

EMF)max = N*A*B*w

w = V/r

EMF)max = N*A*B*V/r

A = 1 cm* 3 cm = 1*3*10^-4 m^2

N = number of turns = 53

EMF)max = 53*1*3*10^-4*0.640*10/(0.82*10^-2)

EMF)max = 12.41 V

Part C

EMF)max = N*A*B*V/r

B = r*EMF)max/(N*A*V)

Using given values:

B = 0.82*10^-2*9/(53*1*3*10^-4*5)

B = 0.928 T

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