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Proton beam \(* *\) A high-energy accelerator produces a beam of protons with kinetic energy \(2 \mathrm{GeV}\) (that is, \(2 \cdot 10^{9} \mathrm{eV}\) per proton). You may assume that the rest energy of a proton is \(1 \mathrm{GeV}\). The current is 1 milliamp, and the beam diameter is \(2 \mathrm{~mm}\). As measured in the laboratory frame: (a) what is the strength of the electric field caused by the beam \(1 \mathrm{~cm}\) from the central axis of the beam? (b) What is the strength of the magnetic field at the same distance? (c) Now consider a frame \(F^{\prime}\) that is moving along with the protons. What fields would be measured in \(F^{\prime}\) ?

Short Answer

Expert verified
The strength of the electric field caused by the beam, 1 cm away from its central axis, can be computed using the formula for the electric field of a long charged wire, and the strength of the magnetic field at the same distance, using Ampere's law. In a frame moving with the protons, the velocity is 0 and so there is no magnetic field (\(B' = 0\)). The electric charge density would be considered infinite and therefore, the electric field (\(E' = \infty\)).

Step by step solution

01

Determine Proton Velocity

First, we need to find the velocity of the proton beam. The total energy of a proton is the sum of its kinetic and rest energy. Here, total energy equals to rest energy, because they are moving in the same frame. Energy is given by the formula \(E = \sqrt{(pc)^2 + (m_0c^2)^2}\), where \(E\) is the total energy, \(p\) is the momentum, \(m_0\) is the rest mass and \(c\) is the speed of light. We are only given the total energy and the rest mass of the proton. However, the momentum can be replaced with \(mv\) for non-relativistic particles, where \(m\) is the mass and \(v\) is the velocity. Therefore, we can express velocity as \(v = \sqrt{[(E/c)^2 - m_0^2]}/m_0\).
02

Compute the Electric Field Strength

Next, we need to determine the strength of the electric field caused by the beam 1 cm away from its central axis. The beam can be considered as a long charged wire with a linear charge density of \(\lambda = I/v\), where \(I\) is the current. The electric field of a long charged wire is \(E = \lambda/(2\pi\epsilon_0r)\), where \(\epsilon_0\) is the permittivity of free space, and \(r\) is the distance. By substituting for \(\lambda\), we will find the electric field.
03

Compute the Magnetic Field Strength

Now, we can calculate the magnetic field strength at the same distance. Ampere’s law equates the line integral of magnetic field around a closed path to the total current enclosed by the path. If we take a circular path oriented perpendicular to the proton beam around the beam axis as the path, then Ampere's law gives \(B = \mu_0I/(2\pi r)\), where \(\mu_0\) is the permeability of free space. Substitute the given current and radius into the above equation to find the magnetic field.
04

Relativistic Frame Fields

Finally, since the frame \(F^\prime\) is moving with the protons, the velocity of the protons in \(F^\prime\) is zero. Therefore, the linear charge density is infinite and the electric field is also infinite. The magnetic field will be zero because it is generated by the moving charge, and in this frame, the charges are at rest. Thus, the fields in \(F^\prime\) would be \(E^\prime = \infty\) and \(B^\prime = 0\).

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Key Concepts

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

Kinetic Energy of Protons
To get to the bottom of what the kinetic energy of protons means, let's first unpack the concept of kinetic energy itself. It corresponds to the energy that an object possesses due to its motion. For protons, which are subatomic particles, the kinetic energy is crucial for understanding their behavior in processes such as those taking place in particle accelerators. In our exercise, each proton in the beam has a kinetic energy of 2 GeV, which is quite significant considering that the rest energy, or the energy due to its mass, is 1 GeV.

It's important to understand that in the realm of high-energy physics, protons with such energies can no longer be described accurately by classical mechanics, which is why we use relativistic mechanics instead. This framework provides us with the equation for energy that includes both the kinetic and rest components. The kinetic energy can thus be found by subtracting the rest energy from the total energy. With this understanding, one can see why the step-by-step solution begins with determining the proton's velocity. It's because velocity is directly linked to kinetic energy, especially at relativistic speeds where kinetic energy significantly alters the mass-energy equivalence of an object—as given by Einstein's famous equation, E=mc².
Electric Field Strength
Electromagnetism is a branch of physics that teaches us about electric fields, which are areas around charged particles or objects where forces are exerted on other charges within the field. The strength of this electric field is a measure of the force a charge would experience per unit charge. In simpler terms, it’s like the gravitational pull you feel on Earth, but for electric charges.

In the context of our exercise, we're looking at the electric field strength caused by a moving beam of protons. The electric field at a certain distance from the center of the beam can be thought of as being similar to the field around a charged wire, which is defined by the linear charge density and the distance from it. Here, the electric field decreases as the distance from the beam increases, which is also inversely proportional to the linear charge density. The linear charge density, in turn, depends on the current and velocity of the protons—this directly ties back to the kinetic energy of the protons we discussed earlier. The higher the kinetic energy, the higher the velocity and the lower the linear charge density for a given current, which then affects the electric field strength around the proton beam.
Magnetic Field Strength
Magnetic fields are another fundamental aspect of electromagnetism and are intimately connected to electric fields. Magnetic fields are produced by moving electric charges—such as the current of protons in our beam—and these fields exert forces on other moving charges and magnetic materials. The magnetic field strength, often symbolized as B, represents the magnitude of this field and determines the force exerted on a charge moving within it.

For the proton beam in our exercise, Ampere's law is used to relate the current flowing through the beam to the magnetic field it produces. In the steps provided, Ampere's law leads us to the expression for the magnetic field strength in terms of the current and distance from the beam, but it's also dependent on the permeability of free space, \( \mu_0 \). This value represents how much resistance the vacuum of space provides to the magnetic field. Knowing the magnetic field strength is crucial for applications such as particle beam steering and focusing in accelerators, MRI machines in medicine, and numerous other technologies.

The exercise and solution guide students through the calculation based on the principles of electromagnetism, which include understanding the relationship between electric current and magnetic fields as well as the underlying physics of charged particle motion.

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Most popular questions from this chapter

Hall voltage \(* *\) A Hall probe for measuring magnetic fields is made from arsenic-doped silicon, which has \(2 \cdot 10^{21}\) conduction electrons per \(\mathrm{m}^{3}\) and a resistivity of \(0.016 \mathrm{ohm}-\mathrm{m}\). The Hall voltage is measured across a ribbon of this \(n\)-type silicon that is \(0.2 \mathrm{~cm}\) wide, \(0.005\) \(\mathrm{cm}\) thick, and \(0.5 \mathrm{~cm}\) long between thicker ends at which it is connected into a \(1 \mathrm{~V}\) battery circuit. What voltage will be measured across the \(0.2 \mathrm{~cm}\) dimension of the ribbon when the probe is inserted into a field of 1 kilogauss?

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.

Right-angled wire ** A wire carrying current \(I\) runs down the \(y\) axis to the origin, thence out to infinity along the positive \(x\) axis. Show that the magnetic field at any point in the \(x y\) plane (except right on one of the axes) is given by $$ B_{z}=\frac{\mu_{0} I}{4 \pi}\left(\frac{1}{x}+\frac{1}{y}+\frac{x}{y \sqrt{x^{2}+y^{2}}}+\frac{y}{x \sqrt{x^{2}+y^{2}}}\right) $$

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, \(A_{0}\) ?

Field at the center of a disk * A disk with radius \(R\) and surface charge density \(\sigma\) spins with angular frequency \(\omega\). What is the magnetic field at the center?

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