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Chapter 6: Problem 29

Motion in a B field \(*\) A particle of charge \(q\) and rest mass \(m\) is moving with velocity \(\mathbf{v}\) where the magnetic field is \(\mathbf{B}\). Here \(\mathbf{B}\) is perpendicular to \(\mathbf{v}\), and there is no electric field. Show that the path of the particle is a curve with radius of curvature \(R\) given by \(R=p / q B\), where \(p\) is the momentum of the particle, \(\gamma m v\). (Hint: Note that the force \(q \mathbf{v} \times \mathbf{B}\) can only change the direction of the momentum, not the magnitude. By what angle \(\Delta \theta\) is the direction of \(\mathbf{p}\) changed in a short time \(\Delta t ?\) ) If \(\mathbf{B}\) is the same everywhere, the particle will follow a circular path. Find the time required to complete one revolution.

Short Answer

Expert verified
The radius of curvature of the particle's path is given by \( R = p/qB \), and the time required to complete one revolution is \( T = 2\pi p/ (qvB) \). Here, \( p \) is the relativistic momentum of the particle.

Step by step solution

01

Understanding the problem

The charged particle moving in the magnetic field \( \mathbf{B} \) perpendicular to its velocity \( \mathbf{v} \) experiences a Lorentz force \( F = q \mathbf{v} \times \mathbf{B} \). This force, being perpendicular to the velocity at all times, changes the direction of the particle's motion without affecting its speed. As a result, the particle will move along a circular path, and the magnitude of the momentum \( p = \gamma m v \) will remain constant.
02

Calculating the radius of curvature

For a particle moving in a circle, the centripetal force is equal to \( F = mv^2/R \). Since the Lorentz force is the centripetal force in this case we get \( qvB = mv^2/R \). By cross-multiplication and re-arrangement we can find the radius \( R = p/qB \) where \( p = \gamma m v \) is the relativistic momentum of the particle.
03

Calculating the time for one revolution

The time required to complete one revolution, also called the period \( T \), can be obtained by dividing the circumference of the circular path \( 2\pi R \) by the speed of the particle \( v \). So. \( T = 2\pi R/v = 2\pi p/ (qvB) \), as \( R = p/qB \).

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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 plays a crucial role when a charged particle moves in a magnetic field. This force is defined as the cross product of the charge's velocity vector \( \mathbf{v} \) and the magnetic field vector \( \mathbf{B} \), giving us \( F = q \mathbf{v} \times \mathbf{B} \).

The unique aspect of the Lorentz force is that it is always perpendicular to the velocity of the particle. This means it can alter the direction of the particle but not its speed.

This property ensures that a charged particle moving perpendicularly to a uniform magnetic field will follow a circular path. The magnetic field doesn't do work on the particle; it merely changes the trajectory.

In summary, the Lorentz force maintains the constant speed of the particle while continuously changing its direction, thus tracing a circular trajectory.
Relativistic momentum
When dealing with high-speed particles, such as those approaching the speed of light, we must consider relativistic effects. Relativistic momentum comes into play here.

The formula for relativistic momentum is \( p = \gamma mv \), where \( \gamma \), the Lorentz factor, is defined as \( \gamma = \frac{1}{\sqrt{1 - \frac{v^2}{c^2}}} \). \( m \) is the rest mass of the particle and \( v \) the velocity.

As a particle's velocity approaches the speed of light, relativistic momentum increases significantly. This momentum differs from classical momentum because it accounts for the effects of relativity, which are essential at high speeds.

Understanding relativistic momentum is crucial in calculating the radius of curvature in a magnetic field, as shown by the formula \( R = \frac{p}{qB} \). This formula highlights the particle's momentum alongside the charge \( q \) and magnetic field \( B \), clearly showing the interplay between momentum and the magnetic forces experienced by the charge.
Circular motion
Charged particles undergoing circular motion in a magnetic field is a fascinating phenomenon resulting from the interaction between the charge and the magnetic field.

The radius of this circular path, known as the radius of curvature, is given by \( R = \frac{p}{qB} \), where \( p \) is the relativistic momentum of the particle. This relationship indicates that higher momentum or weaker magnetic fields result in larger circular paths.

The concept of centripetal force is vital here; it refers to the inward force necessary for circular motion. The Lorentz force acts as this centripetal force, maintaining the circular path. This is explained by the equation \( qvB = \frac{mv^2}{R} \), linking the charge, velocity, and radius.

Moreover, the time taken for one complete revolution, called the period \( T \), can be determined by \( T = \frac{2\pi R}{v} \). This period highlights how both the path radius and speed determine how long the particle takes to complete a loop.
  • Large momentum means a bigger circle and longer period.
  • Faster speeds result in shorter periods.
Understanding these principles is essential for grasping how charged particles behave in magnetic fields.

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Most popular questions from this chapter

Field at the center of an orbit \(*\) An electron is moving at a speed \(0.01 c\) on a circular orbit of radius \(10^{-10} \mathrm{~m}\). What is the strength of the resulting magnetic field at the center of the orbit? (The numbers given are typical, in order of magnitude, for an electron in an atom.)

Vector potential inside a wire ** A round wire of radius \(r_{0}\) carries a current \(I\) distributed uniformly over the cross section of the wire. Let the axis of the wire be the \(z\) axis, with \(\hat{z}\) the direction of the current. Show that a vector potential of the form \(\mathbf{A}=A_{0} \hat{\mathbf{z}}\left(x^{2}+y^{2}\right)\) will correctly give the magnetic field \(\mathbf{B}\) of this current at all points inside the wire. What is the value of the constant, \(A_{0}\) ?

Force in three frames *** A charge \(q\) moves with speed \(v\) parallel to a wire with linear charge density \(\lambda\) (as measured in the lab frame). The charges in the wire move with speed \(u\) in the opposite direction, as shown in Fig. 6.38. If the charge \(q\) is a distance \(r\) from the wire, find the force on it in (a) the given lab frame, (b) its own rest frame, (c) the rest frame of the charges in the wire. Do this by calculating the electric and magnetic forces in the various frames. Then check that the force in the charge's rest frame relates properly to the forces in the other two frames. You can use the fact that the \(\gamma\) factor associated with the relativistic addition of \(u\) and \(v\) is \(\gamma_{u} \gamma_{v}\left(1+\beta_{u} \beta_{v}\right)\)

Field at the tip of a cone \(* *\) A hollow cone (like a party hat) has vertex angle \(2 \theta\), slant height \(L\), and surface charge density \(\sigma\). It spins around its symmetry axis with angular frequency \(\omega\). What is the magnetic field at the tip?

Integral of \(A\), flux of \(B\) Show that the line integral of the vector potential \(\mathbf{A}\) around a closed curve \(C\) equals the magnetic flux \(\Phi\) through a surface \(S\) bounded by the curve. This result is very similar to Amp猫re's law, which says that the line integral of the magnetic field \(\mathbf{B}\) around a closed curve \(C\) equals (up to a factor of \(\mu_{0}\) ) the current flux \(I\) through a surface \(S\) bounded by the curve.
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