Explanation 1) Concept: For a charge circulating in a magnetic field, the centripetal force is provided by magnetic force. By using this concept, we can find the radius of the circular path of each particle and then the required ratios. 2) Formula: i) Magnetic force F B = q v B ii) Centripetal force, F = m v 2 r 3) Given: i) Mass of proton, m p = 1 u ii) Charge of proton, q p = + e iii) Mass of deuteron, m d = 2 u iv) Charge of deuteron, q d = + e v) Mass of ∝ particle, m ∝ = 4 u vi) Charge of ∝ particle, q ∝ = + 2 e 4) Calculations: For a charge circulating in a magnetic field, centripetal force is provided by the magnetic force. Magnetic force F B = q v B … … … … … … … … … … … … … … … … … … … … … … … … … … … . ( 1 ) Centripetal force F = m v 2 r … … … … … … … … … … … … … … … … … … … … … … … … … … … . ( 2 ) Equating equation 1 and 2, q v B = m v 2 r Rearranging the equation for radius r, r = m v q B … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … ( 3 ) Kinetic energy of the circulation particle is given by K = 1 2 m v 2 v = 2 K m Using in equation 3, we get r = m q B × 2 K m r = m q × 2 K B … … … … … … … … … … … … … … … … … … … … … … … … … … … . . … … … ( 4 ) a) It is given that magnetic force and kinetic energy of each particle is the same

10th Edition

ISBN: 9781118230725

Author: David Halliday, Robert Resnick, Jearl Walker

Publisher: Wiley, John & Sons, Incorporated

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Magnetic Force

A magnetic force arises when a charged particle or a current-carrying conductor is placed in a magnetic field. It is created because of the movement of the charged particles.

Question

Chapter 28, Problem 75P

To determine

To find:

a) Ratio of the radius rd of the deuteron path to the radius rp of the proton path, rd/rp

b) Ratio of the radius r∝ of the alpha particle path to the radius rp of the proton path, r∝/rp

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Chapter 28, Problem 74P

Chapter 28, Problem 76P

Chapter 28 Solutions

Fundamentals of Physics Extended

Ch. 28 - Prob. 1Q Ch. 28 - Prob. 2Q Ch. 28 - Prob. 3Q Ch. 28 - Prob. 4Q Ch. 28 - In Module 28-2, we discussed a charged particle...Ch. 28 - Figure 28-26 shows crossed uniform electic and...Ch. 28 - Figure 28-27 shows the path of an electron that...Ch. 28 - Figure 28-28 shows the path of an electron in a...Ch. 28 - a In Checkpoint 5, if the dipole moment is rotated...Ch. 28 - Particle round about. Figure 28-29 shows 11 paths...

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a) Ratio of the radius r d of the deuteron path to the radius r p of the proton path, r d / r p b) Ratio of the radius r ∝ of the alpha particle path to the radius r p of the proton path, r ∝ / r p

Solution Summary: The author explains how to find the radius of the circular path of each particle and then the required ratios.

Concept explainers

To find:

a) Ratio of the radius rd of the deuteron path to the radius rp of the proton path, rd/rp

b) Ratio of the radius r∝ of the alpha particle path to the radius rp of the proton path, r∝/rp

Want to see the full answer?

Chapter 28 Solutions

a) Ratio of the radius r d of the deuteron path to the radius r p of the proton path, r d / r p b) Ratio of the radius r ∝ of the alpha particle path to the radius r p of the proton path, r ∝ / r p

Solution Summary: The author explains how to find the radius of the circular path of each particle and then the required ratios.

Concept explainers

To find: a) Ratio of the radius rd of the deuteron path to the radius rp of the proton path, rd/rp b) Ratio of the radius r∝ of the alpha particle path to the radius rp of the proton path, r∝/rp

Want to see the full answer?

Chapter 28 Solutions

To find:

a) Ratio of the radius rd of the deuteron path to the radius rp of the proton path, rd/rp

b) Ratio of the radius r∝ of the alpha particle path to the radius rp of the proton path, r∝/rp