Physics for Scientists and Engineers
Physics for Scientists and Engineers
10th Edition
ISBN: 9781337553278
Author: Raymond A. Serway, John W. Jewett
Publisher: Cengage Learning
bartleby

Concept explainers

bartleby

Videos

Textbook Question
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 .

Want to see more full solutions like this?

Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!
Students have asked these similar questions
A positron with kinetic energy 2.00 keV is projected into a uniform magnetic field of magnitude 0.100 T, with its velocity vector making an angle of 89.0° with . Find (a) the period, (b) the pitch p, and (c) the radius r of its helical path.
A positron with kinetic energy 1.80 keV is projected into a uniform magnetic field of magnitude 0.140 T, with its velocity vector making an angle of 88.0° with the field. Find (a) the period, (b) the pitch p, and (c) the radius r of its helical path.
What is the radius of the path of an electron (mass 9×10-31 kg and charge 1.6×10-19 C) moving at a speed of 3x107 m/s in a magnetic field of 6 × 10-4 T perpendicular to it? What is its frequency? Calculate its energy in keV. ( 1 eV = 1.6 × 10-19 J).

Chapter 28 Solutions

Physics for Scientists and Engineers

Ch. 28 - A proton moves perpendicular to a uniform magnetic...Ch. 28 - An accelerating voltage of 2.50103 V is applied to...Ch. 28 - A proton (charge + e, mass mp), a deuteron (charge...Ch. 28 - Review. A 30.0-g metal hall having net charge Q =...Ch. 28 - Review. One electron collides elastically with a...Ch. 28 - Review. One electron collides elastically with a...Ch. 28 - Review. An electron moves in a circular path...Ch. 28 - A cyclotron designed to accelerate protons has a...Ch. 28 - Prob. 15PCh. 28 - Singly charged uranium-238 ions are accelerated...Ch. 28 - A cyclotron (Fig. 28.16) designed to accelerate...Ch. 28 - A particle in the cyclotron shown in Figure 28.16a...Ch. 28 - Prob. 19PCh. 28 - A straight wire earning a 3.00-A current is placed...Ch. 28 - A wire carries a steady current of 2.40 A. A...Ch. 28 - Why is the following situation impossible? Imagine...Ch. 28 - Review. A rod of mass 0.720 kg and radius 6.00 cm...Ch. 28 - Review. A rod of mass m and radius R rests on two...Ch. 28 - A wire having a mass per unit length of 0.500 g/cm...Ch. 28 - Consider the system pictured in Figure P28.26. A...Ch. 28 - A strong magnet is placed under a horizontal...Ch. 28 - In Figure P28.28, the cube is 40.0 cm on each...Ch. 28 - A magnetized sewing needle has a magnetic moment...Ch. 28 - A 50.0-turn circular coil of radius 5.00 cm can be...Ch. 28 - You are in charge of planning a physics magic show...Ch. 28 - You are working in your dream job: an assistant...Ch. 28 - A rectangular coil consists of N = 100 closely...Ch. 28 - A rectangular loop of wire has dimensions 0.500 m...Ch. 28 - A wire is formed into a circle having a diameter...Ch. 28 - A Hall-effect probe operates with a 120-mA...Ch. 28 - Prob. 37APCh. 28 - Figure 28.11 shows a charged particle traveling in...Ch. 28 - Within a cylindrical region of space of radius 100...Ch. 28 - Prob. 40APCh. 28 - Prob. 41APCh. 28 - (a) A proton moving with velocity v=ii experiences...Ch. 28 - A proton having an initial velocity of 20.0iMm/s...Ch. 28 - You have been called in as an expert witness in a...Ch. 28 - Prob. 45APCh. 28 - Why is the following situation impossible? Figure...Ch. 28 - A heart surgeon monitors the flow rate of blood...Ch. 28 - Review. (a) Show that a magnetic dipole in a...Ch. 28 - Consider an electron orbiting a proton and...Ch. 28 - Protons having a kinetic energy of 5.00 MeV (1 eV...Ch. 28 - Review. A wire having a linear mass density of...
Knowledge Booster
Physics
Learn more about
Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, physics and related others by exploring similar questions and additional content below.
Similar questions
    • SEE MORE QUESTIONS
    Recommended textbooks for you
  • Physics for Scientists and Engineers, Technology ...
    Physics
    ISBN:9781305116399
    Author:Raymond A. Serway, John W. Jewett
    Publisher:Cengage Learning
    Modern Physics
    Physics
    ISBN:9781111794378
    Author:Raymond A. Serway, Clement J. Moses, Curt A. Moyer
    Publisher:Cengage Learning
    University Physics Volume 2
    Physics
    ISBN:9781938168161
    Author:OpenStax
    Publisher:OpenStax
  • Principles of Physics: A Calculus-Based Text
    Physics
    ISBN:9781133104261
    Author:Raymond A. Serway, John W. Jewett
    Publisher:Cengage Learning
    Physics for Scientists and Engineers with Modern ...
    Physics
    ISBN:9781337553292
    Author:Raymond A. Serway, John W. Jewett
    Publisher:Cengage Learning
  • Physics for Scientists and Engineers, Technology ...
    Physics
    ISBN:9781305116399
    Author:Raymond A. Serway, John W. Jewett
    Publisher:Cengage Learning
    Modern Physics
    Physics
    ISBN:9781111794378
    Author:Raymond A. Serway, Clement J. Moses, Curt A. Moyer
    Publisher:Cengage Learning
    University Physics Volume 2
    Physics
    ISBN:9781938168161
    Author:OpenStax
    Publisher:OpenStax
    Principles of Physics: A Calculus-Based Text
    Physics
    ISBN:9781133104261
    Author:Raymond A. Serway, John W. Jewett
    Publisher:Cengage Learning
    Physics for Scientists and Engineers with Modern ...
    Physics
    ISBN:9781337553292
    Author:Raymond A. Serway, John W. Jewett
    Publisher:Cengage Learning
    Magnets and Magnetic Fields; Author: Professor Dave explains;https://www.youtube.com/watch?v=IgtIdttfGVw;License: Standard YouTube License, CC-BY