Question

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 37 times as great as the angular speed of the tire itself. Each tire has a radius of 0.303 m. The coil consists of 137 turns, has an area of 3.92 10-3 m2, and rotates in a 0.0981-T magnetic field. The bicycle starts from rest and has an acceleration of +0.570 m/s2. What is the peak emf produced by the generator at the end of 5.39 s? V

Answer 1

We know,

$v=r\omega$ where v=linear velocity, $\omega$ is the angular velocity, and r is the radius

So,

$a=r\alpha$ where a= tangential acceleration of the tyre=linear acceleration of the bicycle, $\alpha$ is the angular acceleration.

r=0.303 m., a= 0.57 m/s² from which,

$\alpha =1.88 rad/s^{2}$

We know, $\dot{\omega }=\alpha =1.88 rad/s^{2}$

Hence,integrate wrt t,

$\omega =1.88t$

We are given that, angular velocity of coil is 37 times that of tyre.So

$\omega_{c} =37(1.88t)=69.6 t$

Now,

EMF induced by a coil, $\epsilon =-N\frac{\mathrm{d\phi } }{\mathrm{d}t }$ where N=no. of turns, $\phi$ is the magnetic flux passing through the coil.

No.of turns N = 137

$\phi =\int \vec{B}.\hat{n}ds=(A)(Bcos\theta)=(A)(Bcos(\omega _{c}t))$

where we consider the face of the coil to be parallel to the face of the magnetic pole pieces at t=0.

The cos factor comes as the coil is rotating and hence the flux through the coil varies sinusoidally.So,

$\phi =(3.92)(10^{-3})(0.0981)cos(69.6t)$

Subst. the values into the equation for induced emf,

$\epsilon =\frac{137}{69.6}(0.384)(10^{-3})sin(69.6t)$

$\epsilon =(0.75)(10^{-3})sin(69.6t)$

This is the general equation for induced EMF at a time t.

We have been given time t=5.39 s.Therefore,

$\epsilon =0.75(10^{-3})sin(375.14)$

And so, $\epsilon =(-0.720)(10^{-3})$

Hence the instantaneous induced EMF at time = 5.39 s is ,$\epsilon =(-0.720)(10^{-3})V$

For the peak EMF induced we consider only the amplitude.

So Peak EMF induced = 0.75 x 10^-3 V