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A proton of charge \(+e\) and mass \(m\) enters a uniform magnetic field \(\vec{B}=B \hat{\mathrm{i}}\) with an initial velocity \(\vec{v}=v_{0 x} \hat{\mathrm{i}}+v_{0 y} \hat{\mathrm{j}} .\) Find an expression in unit-vector notation for its velocity \(\vec{v}\) at any later time \(t\).

Short Answer

Expert verified
The velocity \( \vec{v}(t) = v_{0x}\hat{\mathrm{i}} + v_{0y}\cos \left(\frac{eB}{m} t\right) \hat{\mathrm{j}} - v_{0y}\sin \left(\frac{eB}{m} t\right) \hat{\mathrm{k}} \).

Step by step solution

01

Understanding the Problem

The proton enters a region with a magnetic field pointing along the x-direction, given by \( \vec{B} = B \hat{\mathrm{i}} \). The initial velocity of the proton is \( \vec{v} = v_{0x} \hat{\mathrm{i}} + v_{0y} \hat{\mathrm{j}} \). We need to determine how the velocity vector \( \vec{v} \) evolves over time given the magnetic field.
02

Identify the Force

The force on a charged particle in a magnetic field is given by the Lorentz force: \( \vec{F} = q(\vec{v} \times \vec{B}) \). Here, the charge \( q \) of the proton is \( +e \). The cross product \( \vec{v} \times \vec{B} = (v_{0x} \hat{\mathrm{i}} + v_{0y} \hat{\mathrm{j}}) \times (B \hat{\mathrm{i}}) \) simplifies to \(-v_{0y}B \hat{\mathrm{k}}\). Thus, \( \vec{F} = -ev_{0y}B \hat{\mathrm{k}} \).
03

Calculate the Acceleration

From the force, we determine the acceleration by using Newton's second law: \( \vec{F} = m \vec{a} \). Therefore, \( \vec{a} = \frac{\vec{F}}{m} = \frac{-ev_{0y}B}{m} \hat{\mathrm{k}} \). The acceleration is along the \( z \)-axis, rotating the velocity around \( \hat{\mathrm{i}} \).
04

Velocity Evolution

Consider the motion as circular in the plane perpendicular to \( \hat{\mathrm{i}} \). The components of the velocity in the \( yz \)-plane, affected by the field, result in \( v_y(t) = v_{0y} \cos(\omega t) \) and \( v_z(t) = -v_{0y} \sin(\omega t) \), where \( \omega = \frac{eB}{m} \) is the cyclotron frequency.
05

Express Final Velocity in Unit-Vector Notation

The velocity vector in unit-vector notation becomes: \( \vec{v}(t) = v_{0x} \hat{\mathrm{i}} + v_{0y}\cos \left(\frac{eB}{m} t\right) \hat{\mathrm{j}} - v_{0y}\sin \left(\frac{eB}{m} t\right) \hat{\mathrm{k}} \). This expression represents the velocity at any time \( t \), including the effects of the magnetic field.

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Key Concepts

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

Proton Motion
When a proton enters a magnetic field, its motion is governed by specific principles which determine how it behaves over time. Protons are positively charged particles, and this charge plays a critical role in their interaction with magnetic fields. Consider a proton that initially moves with a velocity made up of two components: one in the x-direction and the other in the y-direction. As it enters a magnetic field that is uniform and oriented along the x-axis, we need to understand how this affects its motion.

The magnetic force acts perpendicular to both the velocity and the magnetic field. This means, for the proton moving initially in the xy-plane, the force results in a change in velocity not in the x-direction but rather in the plane perpendicular to the field and the initial velocity – in this case, affecting movements in the y and z directions. This leads to a circular motion of the proton in a plane perpendicular to the magnetic field line.
Magnetic Field Effects
The effects of a magnetic field on a charged particle, such as a proton, are deeply tied to the Lorentz force. This force tells us how the velocity of the particle is influenced when it moves through the magnetic field. The formula for the Lorentz force is given by \( \vec{F} = q(\vec{v} \times \vec{B}) \), where \( q \) is the charge, \( \vec{v} \) is the velocity, and \( \vec{B} \) is the magnetic field.

In our case, the proton with positive charge \( +e \) experiences this force. The cross product calculation shows that the force acts in the z-direction. This means that the proton experiences acceleration along this axis, causing it to rotate around the magnetic field direction, but not move through it directly. The magnetic field doesn’t do work on the particle, as the force is always perpendicular to the velocity. Hence, the speed of the proton remains constant; only its direction changes, resulting in circular motion.
Cyclotron Frequency
Cyclotron frequency is a crucial concept that helps us understand how charged particles move in a magnetic field. It is the frequency with which a charged particle orbits in a magnetic field. For a proton, this frequency is calculated using \( \omega = \frac{eB}{m} \), where \( B \) is the magnetic field strength, \( e \) is the charge of the proton, and \( m \) is its mass.

This frequency tells us how quickly the proton spins around a central path due to the magnetic influence. In the context of the problem, the cyclotron frequency \( \omega \) is used to express how the velocity components in the plane perpendicular to the field change over time. Specifically, the y-velocity becomes \( v_{0y} \cos(\omega t) \) and the z-velocity becomes \(-v_{0y} \sin(\omega t) \).

Cyclotron frequency is fundamental in fields like mass spectrometry and particle accelerators, where understanding the movement of charged particles is crucial. It helps explain not only the behavior of individual particles but also has practical applications in controlling their paths in various scientific and technological applications.

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Most popular questions from this chapter

An electric field of \(1.50 \mathrm{kV} / \mathrm{m}\) and a perpendicular magnetic field of \(0.400 \mathrm{~T}\) act on a moving electron to produce no net force. What is the electron's speed?

A \(5.0 \mu \mathrm{C}\) particle moves through a region containing the uniform magnetic field \(-20 \hat{\mathrm{i}} \mathrm{m} \mathrm{T}\) and the uniform electric field \(300 \hat{\mathrm{j}} \mathrm{V} / \mathrm{m} .\) At a certain instant the velocity of the particle is \((17 \hat{i}-11 \hat{j}+7.0 \hat{k}) \mathrm{km} / \mathrm{s}\). At that instant and in unit-vector notation, what is the net electromagnetic force (the sum of the electric and magnetic forces) on the particle?

An alpha particle can be produced in certain radioactive decays of nuclei and consists of two protons and two neutrons. The particle has a charge of \(q=+2 e\) and a mass of \(4.00 \mathrm{u}\), where \(\mathrm{u}\) is the atomic mass unit, with \(1 \mathrm{u}=1.661 \times 10^{-27} \mathrm{~kg} .\) Suppose an alpha particle travels in a circular path of radius \(4.50 \mathrm{~cm}\) in a uniform magnetic field with \(B=1.20 \mathrm{~T}\). Calculate (a) its speed, (b) its period of revolution, (c) its kinetic energy, and (d) the potential difference through which it would have to be accelerated to achieve this energy.

An electron has velocity \(\vec{v}=(32 \hat{\mathrm{i}}+40 \hat{\mathrm{j}}) \mathrm{km} / \mathrm{s}\) as it enters a uniform magnetic field \(\vec{B}=60 \hat{\mathrm{i}} \mu \mathrm{T}\). What are (a) the radius of the helical path taken by the electron and (b) the pitch of that path? (c) To an observer looking into the magnetic field region from the entrance point of the electron, does the electron spiral clockwise or counterclockwise as it moves?

A wire of length \(25.0 \mathrm{~cm}\) carrying a current of \(4.51 \mathrm{~mA}\) is to be formed into a circular coil and placed in a uniform magnetic field \(\vec{B}\) of magnitude \(5.71 \mathrm{mT}\). If the torque on the coil from the field is maximized, what are (a) the angle between \(\vec{B}\) and the coil's magnetic dipole moment and (b) the number of turns in the coil? (c) What is the magnitude of that maximum torque?

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