/*! 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 7 A beam of protons is accelerated... [FREE SOLUTION] | 91Ó°ÊÓ

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A beam of protons is accelerated to a speed of \(5.0 \times 10^{6} \mathrm{~m} / \mathrm{s}\) in a particle accelerator and emerges horizontally from the accelerator into a uniform magnetic field. What magnetic field (give its direction and magnitude) oriented perpendicularly to the velocity of the proton would cancel the force of gravity and keep the beam moving exactly horizontally?

Short Answer

Expert verified
The magnetic field is \(2.05 \times 10^{-7}\, \text{T}\) and is directed out of the page.

Step by step solution

01

Identify Forces Acting on the Protons

The protons experience gravitational force acting downward and magnetic force acting perpendicular to their velocity. For the beam to continue moving horizontally, these two forces must balance each other out.
02

Write the Expression for Gravitational Force

The gravitational force on each proton is given by the equation:\[ F_{ ext{gravity}} = m imes g \]where \(m\) is the mass of a proton \( (1.67 \times 10^{-27} \, ext{kg}) \) and \(g\) is the acceleration due to gravity \(9.81 \, ext{m/s}^2\).
03

Write the Expression for Magnetic Force

The magnetic force acting on the proton is given by the equation:\[ F_{ ext{magnetic}} = q imes v imes B \]where \(q\) is the charge of a proton \((1.6 \times 10^{-19} \, ext{C})\), \(v\) is the velocity of the protons \((5.0 \times 10^{6} \, ext{m/s})\), and \(B\) is the magnetic field strength.
04

Set Forces Equal to Solve for Magnetic Field

To keep the beam moving horizontally, set the gravitational force equal to the magnetic force:\[ m imes g = q imes v imes B \]Rearrange to solve for \(B\):\[ B = \frac{m imes g}{q imes v} \]
05

Calculate Magnetic Field Magnitude

Plug in the values:\[ B = \frac{(1.67 \times 10^{-27} \, ext{kg}) \times 9.81 \, ext{m/s}^2}{(1.6 \times 10^{-19} \, ext{C}) \times (5.0 \times 10^{6} \, ext{m/s})} \]Calculate the result to find \(B = 2.05 \times 10^{-7} \, ext{T}\).
06

Determine Direction of Magnetic Field

Using the right-hand rule, with the velocity of the protons pointing horizontally and gravitational force pointing downward, the magnetic field should come out of the page to exert an upward magnetic force to balance gravity.

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Key Concepts

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

Proton Beam
A proton beam is a stream of protons, which are positively charged particles found in the nucleus of an atom. These beams are fascinating because protons have specific properties like mass and charge that allow them to be manipulated using electric and magnetic fields. br> Protons in a beam typically travel at very high speeds, often close to a significant fraction of the speed of light. This high velocity is crucial because it allows the protons to acquire significant kinetic energy, making them useful in fields such as medicine and physics research. br> In the context of the given problem, the proton beam moves with a velocity of \(5.0 \times 10^{6} \text{ m/s}\). Understanding this property helps us see how it interacts with magnetic fields to achieve desired outcomes, such as maintaining a horizontal path despite gravitational forces.
Particle Accelerator
Particle accelerators are specialized devices that propel charged particles, like protons and electrons, to very high speeds. They use electric and magnetic fields to increase the velocity of particles and direct their paths. br> These machines are fundamental in experimental physics, helping scientists understand the nature of matter and forces at a subatomic level. br> Two well-known types of particle accelerators include linear accelerators and synchrotrons. In linear accelerators, particles are accelerated along a straight path. In synchrotrons, particles are accelerated in a circular path using both magnetic and electric fields. br> In the current exercise, the particle accelerator serves as the starting point for the proton beam, accelerating it to a significant speed before it enters the magnetic field. Understanding how these machines work is key to appreciate their role in modern physics experiments.
Gravitational Force vs Magnetic Force
In the scenario presented, the gravitational force and magnetic force play crucial roles. The objective is to balance these forces to maintain the proton beam's horizontal path. br> **Gravitational Force:** This force is a result of the Earth's gravity acting on the protons, calculated using the formula \( F_{\text{gravity}} = m \times g \), where \( m \) is the mass of one proton, and \( g \) is the acceleration due to gravity. This force acts downward. br> **Magnetic Force:** When charged particles move through a magnetic field, they experience a force described by \( F_{\text{magnetic}} = q \times v \times B \), where \( q \) is the charge of the proton, \( v \) is its velocity, and \( B \) is the magnetic field. This force can be adjusted to act in any direction based on the orientation of the magnetic field. br> By equating these forces, we can determine the magnetic field necessary to cancel out gravity's effect, ensuring that the beam continues horizontally without deviation.
Right-Hand Rule
The right-hand rule is a handy mnemonic in physics for understanding the directions of vectors in three dimensions, especially in electromagnetism. This rule is crucial for determining the direction of the magnetic force on a charged particle moving in a magnetic field. br> To use the right-hand rule: br>
  • Point your thumb in the direction of the velocity (\( v \)) of the charged particle.
  • Point your fingers in the direction of the magnetic field (\( B \)).
  • Your palm faces in the direction of the force (\( F_{\text{magnetic}} \)) exerted on a positive charge.
By applying the right-hand rule, we can visualize the magnetic force direction needed for the proton beam to balance the downward gravitational force. In this exercise, since the gravitational force acts downward, we orient the magnetic field such that it emerges out of the page, thereby providing the necessary upward force to maintain the beam's horizontal movement.

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Most popular questions from this chapter

In a mass spectrometer, a beam of protons enters a magnetic field. Some protons make exactly a one-quarter circular arc of radius \(0.50 \mathrm{~m}\). If the field is always perpendicular to the proton's velocity, (a) what is the field's magnitude if exiting protons have a kinetic energy of \(10 \mathrm{keV} ?\) (b) How long does it take the proton to complete the quarter circle? (c) Find the net force (magnitude) on a proton while it is in the field.

A nearly horizontal dc power line in the midlatitudes of North America carries a current of 1000 A directly eastward. If the Earth's magnetic field at the location of the power line is northward with a magnitude of \(5.0 \times 10^{-5} \mathrm{~T}\) at an angle of \(45^{\circ}\) below the horizontal, what are the magnitude and direction of the magnetic force on a \(15-\mathrm{m}\) section of the line?

A beam of protons exits from a particle accelerator due east at a speed of \(3.0 \times 10^{5} \mathrm{~m} / \mathrm{s}\). The protons then enter a uniform magnetic field of magnitude \(0.50 \mathrm{~T}\) that is oriented at \(37^{\circ}\) above the horizontal relative to the beam direction. (a) What is the initial acceleration of a proton as it enters the field? (b) What if the magnetic field were angled at \(37^{\circ}\) below the horizontal instead? (c) If the beam were made of electrons traveling at the same speed rather than protons and the field were angled upward at \(37^{\circ},\) would there be any difference in the force on the electrons compared to the protons? Explain. (d) In part (c), what would be the ratio of the acceleration of an electron to that of a proton?

The magnetic field \(7.5 \mathrm{~cm}\) directly above the center of a 25 -turn coil of radius \(15 \mathrm{~cm}\) is \(0.80 \mathrm{mT}\). Find the current in the coil.

A current-carrying solenoid is \(10 \mathrm{~cm}\) long and is wound with 1000 turns of wire. It produces a magnetic field of \(4.0 \times 10^{-4} \mathrm{~T}\) at the solenoid's center. (a) How long would you make the solenoid in order to produce a field of \(6.0 \times 10^{-4} \mathrm{~T}\) at its center? (b) Adjusting only the windings, what number would be needed to produce a field of \(8.0 \times 10^{-4} \mathrm{~T}\) at the center? (c) What current in the solenoid would be needed to produce a field of \(9.0 \times 10^{-4} \mathrm{~T}\) but in the opposite direction?

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