Q. 70

Expert-verifiedFound in: Page 834

Book edition
4th

Author(s)
Randall D. Knight

Pages
1240 pages

ISBN
9780133942651

A proton in a cyclotron gains of kinetic energy per revolution, where is the potential between the dees. Although the energy gain comes in small pulses, the proton makes so many revolutions that it is reasonable to model the energy as increasing at the constant rate , where is the period of the cyclotron motion. This is power input because it is a rate of increase of energy.

a. Find an expression for , the radius of a proton's orbit in a cyclotron, in terms of . Assume that at .

**Hint: **Start by finding an expression for the proton's kinetic energy in terms of .

b. A relatively small cyclotron is in diameter, uses a magnetic field, and has a potential difference between the dees. What is the power input to a proton, in ?

c. How long does it take a proton to spiral from the center out to the edge?

a) The expression for the radius of the proton's orbit in cyclotron is determined as

b) The power input to the proton is determined as

c) The time taken for a proton to travel spiral from the center out to the edge is determined as

The kinetic energy per revolution for the proton in a cyclotron is given as

The objecives are to find an expression for the radius of the proton in a cyclotron, The power input to the proton and the time taken for a proton to spiral from the center out to the edge

The cyclotron motion is defined as the uniform circular motion of a specific perpendicular to the magnetic field at a constant speed. The radius of a proton whose circular motion is caused by a magnetic field is related to the radius of the motion by equation.

Where is the proton's charge, is the particle's speed, is the magnetic field, and is the particle's mass.

We must first determine the proton's velocity in order to determine the kinetic energy, therefore we rearrange equation for to be

The kinetic energy is equal to half of the product of mass and velocity squared.

Let us differentiate both sides of equation using as the energy changes as varies.

By integrating equation , we may obtain an expression for in terms of .

b)

The power is given to us by

is the period, and it equals the frequency reciprocal. Equation gives the cyclotron's frequency . We may get the period by taking the reciprocal of this equation.

Because the change in energy is given by , we can utilise the expressions of and to get by plugging them into equation.

To find , we now insert the values for into equation.

In component of equation , we have the radius of the cyclotron, thus we solve this equation for to be in the form

To find t, we need to plug the values for into equation .

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