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Chapter 19: Problem 60

A proton enters a uniform magnetic field that is at a right angle to its velocity. The field strength is \(0.80 \mathrm{~T}\) and the proton follows a circular path with a radius of \(4.6 \mathrm{~cm} .\) What are (a) the magnitude of its linear momentum and (b) its kinetic energy? (c) If its speed were doubled, what would then be the radius, momentum, and kinetic energy?

Short Answer

Expert verified
Momentum is \(5.888 \times 10^{-21} \text{ kg m/s}\), KE is \(1.037 \times 10^{-13} \text{ J}\). Doubled speed gives radius \(9.20 \text{ cm}\), momentum \(1.1776 \times 10^{-20} \text{ kg m/s}\), KE \(4.148 \times 10^{-13} \text{ J}\).

Step by step solution

01

Identify Given Values and Formulas

We have a proton in a magnetic field, with a field strength of \( B = 0.80 \text{ T} \) and the radius of the circular path given as \( r = 4.6 \times 10^{-2} \text{ m} \). The proton charge is \( q = 1.6 \times 10^{-19} \text{ C} \) and the mass is \( m = 1.67 \times 10^{-27} \text{ kg} \). The formula for the radius of a charged particle's path in a magnetic field is \( r = \frac{mv}{qB} \). We will use this formula to find the momentum \( p = mv \), as \( p = qBr \).
02

Calculate the Linear Momentum

We rearrange the radius formula to get momentum: \( p = qBr \). Substituting the known values, we have \[ p = (1.6 \times 10^{-19} \text{ C})(0.80 \text{ T})(4.6 \times 10^{-2} \text{ m}) = 5.888 \times 10^{-21} \text{ kg m/s}. \]
03

Calculate the Kinetic Energy

The kinetic energy \( KE \) of the proton can be calculated using the equation: \[ KE = \frac{p^2}{2m}. \] Substituting \( p = 5.888 \times 10^{-21} \text{ kg m/s} \) and the proton mass \( m = 1.67 \times 10^{-27} \text{ kg} \), we find: \[ KE = \frac{(5.888 \times 10^{-21})^2}{2 \times 1.67 \times 10^{-27}} = 1.037 \times 10^{-13} \text{ J}. \]
04

Analyze Dobled Speed Impact

When the velocity doubles, the linear momentum \( p \) also doubles. Thus, \( p' = 2 \times 5.888 \times 10^{-21} = 1.1776 \times 10^{-20} \text{ kg m/s} \). The new kinetic energy becomes four times the original since \( KE \propto v^2 \). Thus, \( KE' = 4 \times 1.037 \times 10^{-13} = 4.148 \times 10^{-13} \text{ J} \).
05

Calculate New Radius at Doubled Speed

With doubled speed, the radius is determined by \( r' = \frac{p'}{qB} \). Substituting \( p' \), we find: \[ r' = \frac{1.1776 \times 10^{-20}}{1.6 \times 10^{-19} \times 0.8}\approx 9.20 \times 10^{-2} \text{ m} = 9.20 \text{ cm}. \]

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Key Concepts

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

Linear Momentum
Linear momentum is a fundamental concept in physics that describes the motion of an object. It is defined as the product of an object's mass and velocity. For charged particles like protons moving in a magnetic field, calculating linear momentum can be straightforward thanks to the relationship between radius, charge, magnetic field strength, and velocity.

In the context of a proton moving through a magnetic field, the linear momentum is expressed as:
  • \( p = qBr \)
where:
  • \( p \) is the linear momentum
  • \( q \) is the charge of the proton
  • \( B \) is the magnetic field strength
  • \( r \) is the radius of the circular path
The magnetic field applies a perpendicular force to the moving proton, causing it to move in a circular path. By rearranging the formula \( r = \frac{mv}{qB} \), we see that the linear momentum \( p \) is simply the product of the three quantities: charge, magnetic field, and radius. Thus, the proton's momentum is directly proportional to the radius and the magnetic field strength.
Kinetic Energy
Kinetic energy is the energy possessed by an object due to its motion and is a crucial concept in analyzing dynamics in fields like electromagnetism. For a moving proton in a magnetic field, the kinetic energy can be calculated using its linear momentum.
  • Formula: \( KE = \frac{p^2}{2m} \)
where:
  • \( KE \) is the kinetic energy
  • \( p \) is the linear momentum
  • \( m \) is the mass of the proton
The kinetic energy depends on the square of the momentum, meaning that any increase in momentum results in a larger increase in kinetic energy due to the squared relationship. Thus, doubling the speed of the proton results in four times the kinetic energy. This quadratic relationship is crucial in understanding how changes in velocity impact the energy of moving particles.
Circular Motion in Magnetic Field
Protons and other charged particles exhibit circular motion when they move through a magnetic field that is perpendicular to their velocity. This motion arises from the Lorentz force, which acts as a centripetal force, keeping the particle in a circular path.

Some key points about circular motion in a magnetic field:
  • The centripetal force required for circular motion is provided by the magnetic force.
  • The radius of the circular path is determined by the relationship \( r = \frac{mv}{qB} \).
  • This radius will increase if the particle's speed increases, given by \( r' \), highlighting how velocity and magnetic field strength influence the path.
When the speed of the proton is doubled in the given problem, the radius doubles as well, resulting in a new path that is characteristic of higher energy states. Such interactions demonstrate the interplay between magnetic forces and motion, which is foundational in fields like particle physics and electromagnetism.

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Most popular questions from this chapter

A solenoid is \(0.20 \mathrm{~m}\) long and consists of 100 turns of wire. At its center, the solenoid produces a magnetic field with a strength of \(1.5 \mathrm{mT}\). Find the current in the coil.

In a mass spectrometer, a singly charged ion having a particular velocity is selected by using a magnetic field of 0.10 T perpendicular to an electric field of \(1.0 \times 10^{3} \mathrm{~V} / \mathrm{m} .\) A magnetic field of this same magnitude is then used to deflect the ion, which moves in a circular path with a radius of \(1.2 \mathrm{~cm} .\) What is the mass of the ion?

In a physics lab, a student discovers that the magnitude of the magnetic field at a certain distance from a long wire is \(4.0 \mu \mathrm{T}\). If the wire carries a current of \(5.0 \mathrm{~A}\), what is the distance of the magnetic field from the wire?

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 \(B_{y}=-0.050 \mathrm{~T}\)

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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