EBK PHYSICS FOR SCIENTISTS AND ENGINEER
EBK PHYSICS FOR SCIENTISTS AND ENGINEER
10th Edition
ISBN: 8220106740163
Author: SERWAY
Publisher: CENGAGE L
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Chapter 28, Problem 18P

A particle in the cyclotron shown in Figure 28.16a gains energy qΔV from the alternating power supply each time it passes from one dee to the other. The time interval for each full orbit is

T = 2 π ω = 2 π m q B

so the particle’s average rate of increase in energy’ is

2 q Δ V T = q 2 B Δ V π m

Notice that this power input is constant in time. On the other hand, the rate of increase in the radius r of its path is not constant. (a) Show that the rate of increase in the radius r of the panicle’s path is given by

d r d t = 1 r Δ V π B

(b) Describe how the path of the particles in Figure 28.16a is consistent with the result of part (a). (c) At what rate is the radial position of the protons in a cyclotron increasing immediately before the protons leave the cyclotron? Assume the cyclotron has an outer radius of 0.350 m, an accelerating voltage of ΔV = 600 V, and a magnetic field of magnitude 0.800 T. (d) By how much does the radius of the protons’ path increase during their last full revolution?

Figure 28.16 (a) A cyclotron consists of an ion source at P, two does D1 and D2 across which an alternating potential difference is applied, and a uniform magnetic field. (The south pole of the magnet is not shown.) (b) The first cyclotron, invented by E. O. Lawrence and M. S. Livingston in 1934.

Chapter 28, Problem 18P, A particle in the cyclotron shown in Figure 28.16a gains energy qV from the alternating power supply

(a)

Expert Solution
Check Mark
To determine

To prove: The rate of increase in the radius r of the particles path is given by drdt=1rΔVπB .

Answer to Problem 18P

The rate of increase in the radius r of the particles path is given by drdt=1rΔVπB .

Explanation of Solution

Given Info: The time interval of full orbit is 2πmqB and the rate of increase in energy is 2qΔVT=q2BΔVπm .

The formula for the energy is,

E=12mv2 (1)

Here,

m is the mass of proton.

v is the velocity.

Differentiating equation (1) with respect to time

dEdt=mvdvdt

The above equation can be rewritten as,

dvdt=1mvdEdt

Substitute q2BΔVπm for dEdt in the above equation

dvdt=1mvq2BΔVπm (2)

The formula for the centripetal force is,

F=mv2r=qvB

The above equation can be rewritten as,

r=mvqB (3)

Differentiating equation (3) with respect to time

drdt=mqBdvdt (4)

Deducing from equation (2) and equation (4),

drdt=qmVΔVπdrdt=1rBΔVπ

Conclusion:

Therefore, the rate of increase in the radius r of the particles path is given by drdt=1rΔVπB

(b)

Expert Solution
Check Mark
To determine
The path of the particle is consistent.

Answer to Problem 18P

The path of the particle is consistent.

Explanation of Solution

Given Info: The time interval of full orbit is 2πmqB and the rate of increase in energy is 2qΔVT=q2BΔVπm .

The formula of change of radius with time is,

drdt=1rΔVπB

The value of the path of the particle is consistent with respect to time as according to the above formula the path is dependent on the radius of circle and the magnitude of the magnetic field which remains constant for a path.

Thus, the path of the particle is consistent.

Conclusion:

Therefore, the path of the particle is consistent.

(c)

Expert Solution
Check Mark
To determine
The rate of increase of the radial direction of proton.

Answer to Problem 18P

The rate of increase of the radial direction of proton. is 686.1m/s .

Explanation of Solution

Given Info: The time interval of full orbit is 2πmqB and the rate of increase in energy is 2qΔVT=q2BΔVπm . The outer radius is 0.350m , the accelerating voltage is 600V and the magnitude of magnetic field is 0.800T .

The formula for the change of radius with time is,

drdt=1rΔVπB

Substitute 0.350m for r , 0.800T for B and 600V for ΔV in above equation to find drdt .

drdt=1π(600V)(0.350m)(0.800T)=686.1m/s

Thus, the rate of increase of the radial direction of proton is 686.1m/s .

Conclusion:

Therefore, the rate of increase of the radial direction of proton is 686.1m/s .

(d)

Expert Solution
Check Mark
To determine
The increase in the radius of the path of proton.

Answer to Problem 18P

The increase in the radius of the path of proton is 100μm .

Explanation of Solution

Given Info: The time interval of full orbit is 2πmqB and the rate of increase in energy is 2qΔVT=q2BΔVπm . The outer radius is 0.350m , the accelerating voltage is 600V and the magnitude of magnetic field is 0.800T .

The formula for the velocity is,

v=rqBm

Substitute 0.350m for r , 1.6×1019C for q , 1.6×1027kg for m and 0.800T for B in above equation to find v .

v=(0.350m)(1.6×1019C)(0.800T)(1.6×1027kg)=2.8×107m/s

Thus, the velocity of proton is 2.8×107m/s .

The formula for the energy is,

E=12mv2

Substitute 1.6×1027kg for m and 2.8×107m/s for v in above equation to find E .

E=12(1.6×1027kg)(2.8×107m/s)2=6.272×1013J

The formula for the energy at the end is,

Eend=EΔErev=E2qΔV

Substitute 6.272×1013J for E , 1.6×1019C for q and 600V for ΔV in above equation to find Eend .

Eend=EΔErev=6.272×1013J2(1.6×1019C)(600V)=6.272×1013J0.002×1013J=6.270×1013J

The formula for the radius at the end is,

rend=2EendmqB

Substitute 6.27×1013J for Eend , 1.6×1027kg for m , 1.6×1019C for q and 0.800T for B in above equation to find rend .

rend=2(6.27×1013J)(1.6×1027kg)(1.6×1019C)(0.800T)=0.3499m

The formula for the increase in the radius is,

Δr=rrend

Substitute 0.350m for r and 0.3499m for rend in above equation to find Δr

Δr=0.350m0.3499m=0.0001m×(106μm1m)=100μm

Thus the increase in the radius of the path of proton is 100μm .

Conclusion:

Therefore, increase in the radius of the path of proton is 100μm .

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Chapter 28 Solutions

EBK PHYSICS FOR SCIENTISTS AND ENGINEER

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