/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 101 Ratio of radii of paths when an ... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 15: Problem 101

Ratio of radii of paths when an electron and a proton enters at right angles to a uniform field with (A) same velocity is \(\frac{1}{1840}\). (B) same momentum is 1 . (C) same kinetic energy is \(\frac{1}{43}\). (D) same kinetic energy is 43 .

Short Answer

Expert verified
The correct answers are: (A) Ratio of radii with the same velocity: \(\frac{1}{1840}\) (B) Ratio of radii with the same momentum: 1 (C) Ratio of radii with the same kinetic energy: \(\frac{1}{43}\)

Step by step solution

01

Case A: Same velocity

For both the proton and electron, we have the same velocity when entering the magnetic field, meaning that v (velocity) is the same for both particles. Therefore, we can write the ratio of radii as: \[\frac{r_e}{r_p} = \frac{m_e |q_p|}{m_p |q_e|}\] Since the mass ratio for an electron and proton is \(m_e / m_p = 1 / 1840\), and the charge magnitudes are the same (\( |q_e| = |q_p| \)), we plug these values into the equation and get: \[\frac{r_e}{r_p} = \frac{(1/1840)(|q_p|)}{(m_p)(|q_e|)} = \frac{1}{1840}\] So the correct answer for Case A is (A) \(\frac{1}{1840}\).
02

Case B: Same momentum

In this case, we have the same momentum for both particles, meaning that \(p = m v\) is the same for both the electron and proton. Thus, we can write the ratio of radii as: \[\frac{r_e}{r_p} = \frac{m_e^2}{m_p^2}\] Considering the mass ratio for an electron and proton, we plug these values into the equation: \[\frac{r_e}{r_p} = \frac{(1/1840)^2}{(1)^2} = 1\] So the correct answer for Case B is (B) 1.
03

Case C: Same kinetic energy

In this case, we have the same kinetic energy for both particles. The kinetic energy is given by the formula \(KE = \frac{1}{2} m v^2\). Thus, we can write the ratio of radii as: \[\frac{r_e}{r_p} = \frac{m_e^{3/2}}{m_p^{3/2}}\] Considering the mass ratio for an electron and proton, we plug these values into the equation: \[\frac{r_e}{r_p} = \frac{(1/1840)^{3/2}}{(1)^{3/2}} = \frac{1}{43}\] So the correct answer for Case C is (C) \(\frac{1}{43}\).
04

Note for Case D

Case D has the same condition as Case C (same kinetic energy), but with a different answer (43). Since we already found a correct answer in Case C, Case D is incorrect. In conclusion, the correct answers are: (A) Ratio of radii with the same velocity: \(\frac{1}{1840}\) (B) Ratio of radii with the same momentum: 1 (C) Ratio of radii with the same kinetic energy: \(\frac{1}{43}\)

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

The magnetic lines of force inside a bar magnet [2003] (A) are from north pole to south pole of the magnet. (B) does not exist. (C) depend upon the area of cross-pole of the magnet. (D) are from south pole to north pole of the magnet.

An electron accelerated by a potential difference \(V=3.6 \mathrm{~V}\) first enters into a uniform electric field of a parallel-plate capacitor whose plates extend over a length \(l=6 \mathrm{~cm}\) in the direction of initial velocity. The electric field is normal to the direction of initial velocity and its strength varies with time as \(E=a \times t\), where \(a=\) \(3200 \mathrm{Vm}^{-1} \mathrm{~s}^{-1}\). Then the electron enters into a uniform magnetic field of induction \(B=\pi \times 10^{-9} \mathrm{~T}\). Direction of magnetic field is same as that of the electric field. Calculate pitch (in \(\mathrm{mm}\) ) of helical path traced by the electron in the magnetic field (Mass of electron, \(m=9 \times\) \(10^{-31} \mathrm{~kg}\) ). [Neglect the effect of induced magnetic field.]

Curie temperature is the temperature above which \([2003]\) (A) a ferromagnetic material becomes paramagnetic. (B) a paramagnetic material becomes diamagnetic. (C) a ferromagnetic material becomes diamagnetic. (D) a paramagnetic material becomes ferromagnetic.

In which type of materials the magnetic susceptibility does not depend on temperature? (A) Diamagnetic (B) Paramagnetic (C) Ferromagnetic (D) Ferrite

A charged particle moves with velocity \(\vec{v}\) in a uniform magnetic field \(\vec{B}\). The magnetic force experienced by the particle is (A) always zero. (B) never zero. (C) zero if \(\vec{B}\) and \(\vec{v}\) are perpendicular. (D) zero if \(\vec{B}\) and \(\vec{v}\) are parallel.
See all solutions

Recommended explanations on Physics Textbooks

Scientific Method Physics

Read Explanation

Circular Motion and Gravitation

Read Explanation

Radiation

Read Explanation

Magnetism and Electromagnetic Induction

Read Explanation

Physical Quantities And Units

Read Explanation

Fields in Physics

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.