Question

A bicycle generator rotates at 1,825 rad/s, producing a 16.0 V peak emf. It has a 46-turn, 1.00 by 3.00 cm rectangular coil in a 0.640 T field. It is driven by a 1.44 cm diameter wheel that rolls on the outside rim of the bicycle tire.

What is the velocity of the bicycle? (Enter the magnitude in m/s.)

What is the maximum emf (in V) of the generator when the bicycle moves at 10.0 m/s, noting that it was 16.0 V under the original conditions?

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?

Answer 1

Part A:

linear velocity is given by:

V = w*r

w = angular velocity = 1825 rad/sec

r = radius of wheel = diameter/2 = 1.44 cm/2 = 0.72 cm = 0.0072 m

So,

V = (1825 rad/sec)*0.0072 m = 13.14 m/sec

Part B.

Emf is given by:

Emf = N*A*B*w = N*A*B*V/r

Since N, A and B and r are constant, So

E0/E0' = (w1/w2) = (v1/v2)

E0' = E0*(v2/v1)

E0 = 16.0 V

v1 = 13.14 m/sec

v2 = 10.0 m/sec

So,

E0' = 16.0*(10.0/13.14)

E0 = 12.2 V

Part C.

As we know

Emf = N*A*B*w = N*A*B*v/r

B = r*Emf/(N*A*v)

Given that

Emf = 9.00 V

N = Number of turns = 46 turns

v = speed of wheel = 5.00 m/sec

A = Area of coil = length*width = 1.00 cm*3.00 cm = 3.00 cm^2 = 3.00*10^-4 m^2

Using these values:

B = 0.0072*9.00/(46*3.00*10^-4*5.00)

B = 0.9391 T = 0.94 T

Let me know if you've any query.