/*! 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 29 An electron and a proton move in... [FREE SOLUTION] | 91Ó°ÊÓ

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An electron and a proton move in circular orbits in a plane perpendicular to a uniform magnetic field \(\overrightarrow{\mathbf{B}}\). Find the ratio of the radii of their circular orbits when the electron and the proton have (a) the same momentum and (b) the same kinetic energy.

Short Answer

Expert verified
(a) Radii are equal when momenta are equal; (b) ratio is \(\sqrt{m_e/m_p}\) for equal kinetic energy.

Step by step solution

01

Understanding Motion in a Magnetic Field

Both the electron and the proton are charged particles moving in a magnetic field which applies a force on them, keeping them in a circular path. The force is given by the Lorentz force formula: \[ F = qvB = \frac{mv^2}{r} \]where \( q \) is the charge, \( v \) is the velocity, \( m \) is the mass of the particle, and \( r \) is the radius of the orbit.This can be rearranged to find the radius: \[ r = \frac{mv}{qB} \].This will be essential in both scenarios.
02

Calculate Radius for Same Momentum

For particles with the same momentum \( p = mv \), substitute \( p = mv \) into the radius expression:\[ r = \frac{p}{qB} \].Now, since \( p \) and \( B \) are the same for both particles:- The charge of the electron is \( -e \), and for the proton it is \( +e \).- The radius ratio is:\[ \text{Ratio}(r_e/r_p) = \frac{p/eB}{p/eB} = 1 \].Thus, when they have the same momentum, their orbits will have the same radius.
03

Calculate Radius for Same Kinetic Energy

For particles with the same kinetic energy \( KE = \frac{1}{2}mv^2 \), express momentum using kinetic energy:\[ p = \sqrt{2m \cdot KE} \].Substitute \( p \) into the expression for radius:\[ r = \frac{\sqrt{2m \cdot KE}}{qB}. \]Now, the radius ratio for the same kinetic energy is:\[ \text{Ratio}(r_e/r_p) = \frac{\sqrt{2m_e \cdot KE}/eB}{\sqrt{2m_p \cdot KE}/eB} = \sqrt{\frac{m_e}{m_p}}. \]Given \( m_p \gg m_e \), this ratio will be much smaller than 1.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Lorentz Force
The Lorentz Force is a fundamental concept in electromagnetism, named after the Dutch physicist Hendrik Lorentz. It describes the force experienced by a charged particle moving in a magnetic field. This force is what keeps particles like electrons and protons in circular motion when they enter a magnetic field perpendicular to their velocity vector. The Lorentz Force is given by the equation:\[ F = qvB \]where:
  • \( F \) is the force exerted on the particle,
  • \( q \) is the electric charge of the particle,
  • \( v \) is the velocity of the particle, and
  • \( B \) is the magnetic field strength.
This force acts perpendicular to both the velocity of the particle and the magnetic field direction, resulting in a circular motion of the particle.
The particle's path radius can be expressed as:\[ r = \frac{mv}{qB} \],indicating that the larger the mass or velocity, or smaller the charge or magnetic field, the larger the radius of its path. This relationship is crucial in understanding how different conditions affect the trajectory of charged particles like electrons and protons.
Electron and Proton Orbits
The behavior of electrons and protons in a magnetic field is fascinating due to their charge differences and masses. When both particles have the same momentum in a magnetic field, their orbits can be compared easily.For a given momentum, the radius of the path can be calculated as:\[ r = \frac{p}{qB} \],where \( p \) stands for momentum, \( q \) represents charge, and \( B \) is the magnetic field. Since the proton and electron have equal magnitude but opposite charges,
  • their paths maintain the same radius if their momentum is identical.
However, when treating them with equal kinetic energy, we use the relationship:\[ p = \sqrt{2m \cdot KE} \].Here, the radius of each particle's orbit becomes:\[ r = \frac{\sqrt{2m \cdot KE}}{qB} \].Given that an electron has much less mass than a proton (\( m_e \ll m_p \)), the radius of the electron's orbit will be significantly smaller compared to that of the proton under the scenario of equal kinetic energies.
This illustrates the critical role of mass in determining orbital radius when kinetic energy is the same.
Momentum and Kinetic Energy
Momentum and kinetic energy are key concepts in physics that often determine the behavior of particles in motion. These concepts greatly influence the radii of orbits for electrons and protons in a magnetic field.

Momentum

Momentum, denoted \( p \), is a measure of the amount of motion a particle has, and is calculated by the expression:\[ p = mv \].It depends on both the mass \( m \) and velocity \( v \) of the particle. Given the same momentum, an electron and a proton will orbit a magnetic field path with the same radius, despite differing masses, due to the charge factor.

Kinetic Energy

Kinetic energy, \( KE \), represents the energy of motion and is given by:\[ KE = \frac{1}{2}mv^2 \].When two particles possess the same kinetic energy, their velocities differ because of their differing masses. For example,
  • a lighter electron moves faster than a heavier proton with the same kinetic energy.
When considering orbit radii under constant kinetic energy, the profound mass difference significantly affects the outcomes, leading to the electron's orbit being smaller compared to the proton's. This comparison underscores the importance of mass when energy is conserved, explaining why smaller particles have more compact orbits at constant kinetic forces.

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Most popular questions from this chapter

The astronomical object \(4 \mathrm{U} 014+61\) has the distinction of creating the most powerful magnetic field ever observed. This object is referred to as a "magnetar" (a subclass of pulsars), and its magnetic field is \(1.3 \times 10^{15}\) times greater than the Earth's magnetic field. (a) Suppose a \(2.5-\mathrm{m}\) straight wire carrying a current of \(1.1 \mathrm{A}\) is placed in this magnetic field at an angle of \(65^{\circ}\) to the field lines. What force does this wire experience? (b) A field this strong can significantly change the behavior of an a tom. To see this, consider an electron moving with a speed of \(2.2 \times 10^{6} \mathrm{m} / \mathrm{s} .\) Compare the maximum magnetic force exerted on the electron to the electric force a proton exerts on an electron in a hydrogen atom. The radius of the hydrogen atom is \(5.29 \times 10^{-11} \mathrm{m}\).

Two power lines, each \(270 \mathrm{m}\) in length, run parallel to each other with a separation of \(25 \mathrm{cm}\). If the lines carry parallel currents of \(110 \mathrm{A},\) what are the magnitude and direction of the magnetic force each exerts on the other?

A charged particle moves in a horizontal plane with a speed of \(8.70 \times 10^{6} \mathrm{m} / \mathrm{s} .\) When this particle encounters a uniform magnetic field in the vertical direction it begins to move on a circular path of radius \(15.9 \mathrm{cm}\). (a) If the magnitude of the magnetic field is \(1.21 \mathrm{T}\), what is the charge-to-mass ratio \((q / m)\) of this particle? (b) If the radius of the circular path were greater than \(15.9 \mathrm{cm},\) would the corresponding charge-to-mass ratio be greater than, less than, or the same as that found in part (a)? Explain. (Assume that the magnetic field remains the same.)

Medical X-rays An electron in a medical X-ray machine is accelerated from rest through a voltage of \(10.0 \mathrm{kV}\). (a) Find the maximum force a magnetic field of \(0.957 \mathrm{T}\) can exert on this electron. (b) If the voltage of the \(X\) -ray machine is increased, does the maximum force found in part (a) increase, decrease, or stay the same? Explain. (c) Repeat part (a) for an electron accelerated through a potential of \(25.0 \mathrm{kV}\).

A high-voltage power line carries a current of \(110 \mathrm{A}\) at a location where the Earth's magnetic field has a magnitude of \(0.59 \mathrm{G}\) and points to the north, \(72^{\circ}\) below the horizontal. Find the direction and magnitude of the magnetic force exerted on a \(250-\mathrm{m}\) length of wire if the current in the wire flows (a) horizontally toward the east or (b) horizontally toward the south.

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