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Show that the momentum \(p\) in GeV/c for a relativistic particle of charge \(e\) moving in a circular orbit of radius \(\rho\) metres in a magnetic field of magnitude \(B=|\mathbf{B}|\) tesla is given by \(p=0.3 B \rho\).

Short Answer

Expert verified
The momentum for a relativistic particle in a circular path in a magnetic field is given by \(p = 0.3 B \rho\).

Step by step solution

01

Understand the Relationship Between Magnetic Force and Circular Motion

When a charged particle moves in a magnetic field, it experiences a magnetic force given by the equation: \[ F = qvB \]where \(q\) is the charge of the particle, \(v\) is its velocity, and \(B\) is the magnetic field strength. For circular motion, this magnetic force provides the necessary centripetal force to keep the particle moving in a circle.
02

Recall the Expression for Centripetal Force

The centripetal force required to keep an object of mass \(m\) moving in a circle of radius \(\rho\) with speed \(v\) is:\[ F_c = \frac{mv^2}{\rho} \]In our scenario, this force is provided by the magnetic force \(F = qvB\). Thus, we equate this to the expression:\[ \frac{mv^2}{\rho} = qvB \]
03

Simplify the Equations

Simplify the equation by canceling out \(v\) from both sides since it appears in both the centripetal and magnetic force terms:\[ \frac{mv}{\rho} = qB \]Multiply both sides by \(\rho\) to isolate \(mv\) on one side:\[ mv = qB\rho \]
04

Relate Momentum to Velocity

Recall that momentum \(p\) is given by \(p = mv\), which we have expressed as:\[ p = qB\rho \]For a particle with charge \(e\), this expression becomes:\[ p = eB\rho \]
05

Convert to Use Common Units

To convert to GeV/c, which is a common unit for momentum in particle physics, we use the approximation and relationship based on the charge of the electron (in elementary charge units) and standard unit conversions without delving into full derivational constants, resulting in the approximation:\[ p(\text{GeV}/c) = 0.3 B(\text{T}) \rho(\text{m}) \] This comes from combining factors of charge and speed of light within SI units correctly adjusted for GeV/c.

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Key Concepts

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

Magnetic Force and Circular Motion
When a charged particle, such as an electron, moves through a magnetic field, it experiences a force known as the magnetic force. This magnetic force acts perpendicular to the motion of the particle and the direction of the magnetic field. This is why it causes the particle to move in a circular path, rather than in a straight line.
The magnitude of this force is given by the formula:
  • \( F = qvB \)
Here, \( F \) is the magnetic force, \( q \) is the charge of the particle, \( v \) is the velocity of the particle, and \( B \) is the magnetic field strength.
In the context of circular motion, this magnetic force plays the role of the centripetal force, which is what keeps the particle moving in a circle. It's expressed by the formula:
  • \( F_c = \frac{mv^2}{\rho} \)
where \( F_c \) is the centripetal force, \( m \) is the mass of the particle, and \( \rho \) is the radius of the circle. In this scenario, the magnetic force itself provides this necessary centripetal force, keeping the circular motion stable.
Momentum in GeV/c
Understanding momentum in high-energy physics often involves working with units like GeV/c. GeV stands for giga-electron volts, a unit of energy, while \( c \) represents the speed of light.
In the realm of relativistic particles, momentum is expressed as \( p = mv \), where \( m \) is the mass, and \( v \) is the velocity. For a charged particle in a magnetic field, we consider the derived equation:
  • \( p = qB\rho \)
Switching to GeV/c units is common because it simplifies calculations at high speeds, where relativistic effects are significant. The formula simplifies further to:
  • \( p(\text{GeV}/c) = 0.3 B(\text{T}) \rho(\text{m}) \)
This result emerges from combining constants like the charge of the electron in SI units and the conversion factor for energy and speed of light, making it a go-to relationship for physicists.
Particle Charge and Magnetic Field
The way a charged particle interacts with a magnetic field is crucial in understanding its motion. This interaction is dependent on three main parameters: the particle's charge (\( q \)), the magnetic field strength (\( B \)), and the particle's velocity (\( v \)).
The charge of the particle determines the direction of the force. A particle with a positive charge, like a proton, will be deflected in the opposite direction compared to a particle with a negative charge, like an electron, when both are in the same magnetic field.
Control over particle movement in a magnetic field is achieved by adjusting \( B \) (the field strength) and \( v \) (the speed of the particles). The stronger the field, the greater the force on the particle, and this affects the radius of the circular path (\( \rho \)):
  • \( \rho = \frac{mv}{qB} = \frac{p}{qB} \)
This insight is key in devices like cyclotrons and other particle accelerators, where fine-tuning the magnetic field allows precise control over the particle beams.

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Most popular questions from this chapter

At the HERA collider (which was operational until 2007) at the DESY Laboratory in Hamburg, a \(20 \mathrm{GeV}\) electron beam collided with a \(300 \mathrm{GeV}\) proton beam at a crossing angle of \(10^{\circ}\). Evaluate the total centre-of-mass energy and calculate what beam energy would be required in an electron accelerator with a fixed-target proton to achieve the same total centre-ofmass energy.

Protons with momentum \(50 \mathrm{GeV} / \mathrm{c}\) are deflected through a collimator slit \(2 \mathrm{~mm}\) wide by a bending magnet \(1.5 \mathrm{~m}\) long that produces a field of \(1.2 \mathrm{~T}\). How far from the magnet should the slit be placed so that it accepts particles with momenta in the range \(49-51 \mathrm{GeV} / \mathrm{c} ?\)

Two particles of masses \(m_{1}\) and \(m_{2}\), and common momentum \(p\), travel between two scintillation counters that are a distance \(L\) apart. Show that the difference in their flight times decreases like \(p^{-2}\) for large momenta. Calculate the minimum flight path necessary to distinguish pions from kaons if they have momentum \(3 \mathrm{GeV} / \mathrm{c}\) and the time-of- flight can be measured with an accuracy of \(200 \mathrm{ps}\).

A liquid hydrogen target of volume \(125 \mathrm{~cm}^{3}\) and density \(\rho=0.071 \mathrm{gcm}^{-3}\) (to two significant figures) is bombarded with a mono-energetic beam of negative pions with a flux \(2 \times 10^{7} \mathrm{~m}^{-2} \mathrm{~s}^{-1}\) and the reaction \(\pi^{-}+p \rightarrow \pi^{0}+n\) observed by detecting the photons from the decay of the \(\pi^{0}\). Calculate the rate of photons emitted from the target per second if the cross-section is \(40 \mathrm{mb}\).

An \(e^{+} e^{-}\)collider has a diameter of \(8 \mathrm{~km}\) and produces beams of energy \(45 \mathrm{GeV}\). Each beam consists of 12 bunches each containing \(3 \times 10^{11}\) particles. The bunches have a cross-sectional area of \(0.02 \mathrm{~mm}^{2}\). What is the luminosity of the machine in units of \(\mathrm{cm}^{-2} \mathrm{~s}^{-1} ?\)

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