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Chapter 19: Problem 51

(a) Exiting from a small linear accelerator near Washington D.C., a proton is moving parallel to the ground. What should the direction(s) of its velocity be in order to maximize the upward component of the Earth's magnetic force on it? (b) Using a value of \(0.05 \mathrm{mT}\) for the value of the horizontal component of the Earth's magnetic field at this location, what is the upward acceleration of the proton as it exits the accelerator at a speed of \(2000 \mathrm{~m} / \mathrm{s} ?\)

Short Answer

Expert verified
Velocity should be perpendicular to the horizontal magnetic field, with an upward acceleration of about \(9.58 \times 10^3 \, m/s^2\).

Step by step solution

01

Understand the Problem

We are asked to find the direction of a proton's velocity to maximize the upward magnetic force and then to calculate the upward acceleration given a specific velocity and magnetic field.
02

Apply the Right-Hand Rule

To maximize the upward component of the magnetic force on a charged particle moving in a magnetic field, use the right-hand rule. The magnetic force (\( F = qvB \sin \theta \) ) is maximized when the velocity is perpendicular to the magnetic field. Thus, the velocity should be perpendicular to the magnetic field lines, which are horizontal.
03

Calculate the Magnetic Force

Use \( F = qvB \sin \theta \) . Given that the motion is perpendicular to the magnetic field, \( \sin \theta = 1 \). Here: \( q = 1.6 \times 10^{-19} \, C \) (charge of proton), \( v = 2000 \, m/s \), \( B = 0.05 \, mT = 0.05 \times 10^{-3} \, T \). So \( F = 1.6 \times 10^{-19} \times 2000 \times 0.05 \times 10^{-3} \).
04

Calculate the Upward Acceleration

Use the formula \( a = \frac{F}{m} \) where \( m = 1.67 \times 10^{-27} \, kg \) is the mass of a proton. Plugging in the value of \( F \) from Step 2: \( a = \frac{1.6 \times 10^{-19} \times 2000 \times 0.05 \times 10^{-3}}{1.67 \times 10^{-27}} \). Simplify to find \( a \).
05

Solve for Acceleration Value

Calculate \( a = \frac{1.6 \times 10^{-19} \times 2000 \times 0.05 \times 10^{-3}}{1.67 \times 10^{-27}} \approx 9.58 \times 10^{3} \, m/s^2 \).

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Key Concepts

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

Magnetic Force
Magnetic force is a fundamental concept in physics that describes how charged particles like protons experience force when moving through a magnetic field. The magnitude of this force is given by the equation \( F = qvB \sin \theta \), where:
  • \( F \) is the magnetic force.
  • \( q \) is the charge of the particle.
  • \( v \) is the velocity of the particle.
  • \( B \) is the magnetic field strength.
  • \( \theta \) is the angle between the velocity vector and the magnetic field.
The force is maximum when the velocity is perpendicular to the magnetic field lines, making \( \sin \theta = 1 \). In such a situation, the magnetic force can significantly change the motion of the charged particle, which is crucial in controlling protons in accelerators.
Proton Motion
Understanding the motion of a proton in a magnetic field is pivotal in fields like particle physics. Protons, which are positively charged particles, move according to forces acting on them. When in motion through a magnetic field, they experience a force perpendicular to their velocity, causing them to change their path. The velocity of a proton is crucial because it determines how strong the force will be. In the discussed scenario, a proton moving parallel to the ground exiting a linear accelerator can have its path altered by the magnetic field of the Earth. By properly aligning the velocity with the magnetic field, the proton's path can be manipulated to achieve certain intended effects, such as maximizing lift or altering its trajectory for experimental purposes.
Linear Accelerator
A linear accelerator (often known as a linac) is an apparatus used to accelerate charged particles, such as protons, along a straight line. It uses electromagnetic waves to move the particles at high speeds in a controlled beam. This technology is central to many physics experiments and medical procedures. The linear motion helps in maintaining the energy of particles and is effective in research and applications like radiation therapy. As the proton exits the accelerator, knowledge of its velocity and direction becomes critical in studying its interaction with external magnetic fields, like that of the Earth, to influence their paths.
Right-Hand Rule
The right-hand rule is a simple rule that helps determine the direction of the magnetic force acting on a moving charged particle in a magnetic field. To apply it, imagine positioning your right hand so that your thumb points in the direction of the positive charge's velocity (such as a proton), and your fingers in the direction of the magnetic field. The force will then emerge from your palm, indicating its direction. Using this rule allows you to quickly ascertain how the forces act and helps in predicting the next step in controlling or manipulating the path of charged particles. Applying the right-hand rule correctly enables the maximum upward component of magnetic force when the velocity is perpendicular to the magnetic field, which is key in various applications, including the operation of particle accelerators.

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Most popular questions from this chapter

Two long, straight, parallel wires \(10 \mathrm{~cm}\) apart carry currents in opposite directions. (a) Use the right-hand source and force rules to determine whether the forces on the wires are (1) attractive or (2) repulsive. Show your reasoning. (b) If the wires carry equal currents of \(3.0 \mathrm{~A}\), what is the magnetic field magnitude that each produces at the other's location? (c) Use the result of part (b) to determine the magnitude of the force per unit length they exert on each other.

(a) What angle(s) does a particle's velocity have to make with the magnetic field direction for the particle to be subjected to half the maximum possible magnetic force, \(F_{\max } ?\) (b) Express the magnetic force on a charged particle in terms of \(F_{\max }\) if the angle between its velocity and the magnetic field direction is (i) \(10^{\circ},\) (ii) \(80^{\circ},\) and (iii) \(100^{\circ} .\) (c) If the particle's velocity makes an angle of \(50^{\circ}\) with respect to the magnetic field direction, at what other angle(s) would the magnetic force on it be the same? Would the direction be the same? Explain.

A long, straight wire carries a current of \(2.5 \mathrm{~A}\). Find the magnitude of the magnetic field \(25 \mathrm{~cm}\) from the wire.

An electron travels at a speed of \(2.0 \times 10^{4} \mathrm{~m} / \mathrm{s}\) through a uniform magnetic field whose magnitude is \(1.2 \times 10^{-3} \mathrm{~T}\). What is the magnitude of the magnetic force on the electron if its velocity and the magnetic field (a) are perpendicular, (b) make an angle of \(45^{\circ},(\mathrm{c})\) are parallel, and (d) are exactly opposite?

A wire carries a current of \(10 \mathrm{~A}\) in the \(+x\) -direction. (a) Find the force per unit length on the wire if it is in a magnetic field that has components of \(B_{x}=0.020 \mathrm{~T}\) \(B_{y}=0.040 \mathrm{~T},\) and \(B_{z}=0 \mathrm{~T} .\) (b) Find the force per unit length on the wire if only the field's \(x\) -component is changed to \(B_{x}=0.050 \mathrm{~T}\). (c) Find the force per unit length on the wire if only the field's \(y\) -component is changed to \(B_{y}=-0.050 \mathrm{~T}\)
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