Question

Consult Multiple-Concept Example 11 for background material relating to this problem. A small rubber wheel on the shaft of a bicycle generator presses against the bike tire and turns the coil of the generator at an angular speed that is 40 times as great as the angular speed of the tire itself. Each tire has a radius of 0.330 m. The coil consists of 163 turns, has an area of 3.23 x 10-3 m2, and rotates in a 0.0948-T magnetic field. The bicycle starts from rest and has an acceleration of 0.601 m/s2. What is the peak emf produced by the generator at the end of 6.72 s

Answer 1

Answer:

$Emf=24.4V$

Explanation:

From the question we are told that:

Angular speed $\omega=40*\omega'$

Radius $r=0.33m$

No. Turns $N=163$

Area $A= 3.23 * 10^{-3} m^2$

Magnetic field. $B=0.0948T$

Acceleration $a=0.60m/s^2$

Time $t=6.72s$

Generally the equation for momentum is mathematically given by

$\omega'=\omega_o+\frac[a}{r}t$

$\omega'=0+\frac{0.60}{0.33}*6.72$

$W=12.22rad/s^2$

Therefore

$\omega=Aw$

$\omega=12.22*40$

$\omega=488.7rads/s$

Generally the equation for Peak emf is mathematically given by

$Emf=NBA \omega$

$Emf=163*0.0948* 3.23 * 10^{-3} m^2*488.7rads/s$

$Emf=24.4V$