Question

A cyclotron (Fig. 29.16) designed to accelerate protons has an outer radius of 0.350 m . The protons are emitted nearly at rest from a source at the center and are accelerated through 600 V each time they cross the gap between the dees. The dees are between the poles of an electromagnet where the field is 0.800 T . (e) For what time interval does the proton accelerate?

Answer 1

The time interval for which the proton accelerated can be calculated using -

$t=\frac{N}{f}=$   $\frac{3.3\times 10^{25} }{f}$

Using the above formula you can find the time interval for any frequency.

We have a Cyclotron designed to accelerate protons.

We have to determine for what time interval does the proton accelerate.

What is a Cyclotron? On what principle it works?

Cyclotron is a device used to accelerate charged particles to high energies. It works on the principle that a charged particle moving normal to a magnetic field experiences magnetic Lorentz force due to which the particle moves in a circular path. The magnetic Lorentz force is given by-

F = qvB sinθ

According to the question -

In a cyclotron the force by a magnetic field is equal to the centrifugal force.

Now, for r(max) -                                     ( Since v = rω)

q v(max) B sin (90) =  $\frac{m\times v(max)^{2} }{r(max)}$

$v(max)=\frac{qBr(max)}{m}$

Now -

r(max) = 0.35 m

B = 0.8 Tesla

Therefore -

v(max) =  $\frac{1.6\times 10^{-19} \times 0.8 \times 0.35}{1.6\times 10^{-27} }$

v(max) = 0.28 x  $10^{8}$  m/sec

Therefore, the Maximum kinetic energy = Kinetic energy at which they leave the cyclotron-

E(max) =  $\frac{1}{2} m\times v(max)^{2}$

E(max) = $\frac{1}{2} \times 1.6\times 10^{-27} \times 0.28\times 10^{8} \times 0.28\times 10^{8}$  

E(max) = 0.063 x  $10^{11}$  joules.

Now -

The kinetic energy K is built up from 2N passes through Voltage V = 600 volts. Therefore -

$N = \frac{E(max)}{2qV}$ = $\frac{0.063\times 10^{11} }{2\times 1.6\times 10^{-19} \times 600}$  = 3.3 x  $10^{25}$ revolutions.

The total number of revolutions are 3.3 x $10^{25}$ revolutions.

Now - The time interval for which the proton accelerated can be calculated using -

$t=\frac{N}{f}=$   $\frac{3.3\times 10^{25} }{f}$

Using the above formula you can find the time interval for any frequency.

To solve more questions on Cyclotron, visit the link below-

brainly.com/question/14555284

#SPJ4