/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 25 Force in three frames *** A ch... [FREE SOLUTION] | 91Ó°ÊÓ

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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 \(\gamma_{u} \gamma_{v}\left(1+\beta_{u} \beta_{v}\right)\)

Short Answer

Expert verified
The force on the moving charge, when seen in the lab frame, the charge's rest frame, and rest frame of the charges in the wire, would be \( F = F_B + F_E \), \( F'_E = k\gamma_u\lambda/r \), and \( F_{WB} = qv\cdot\mu_0\gamma_v\lambda/2\pi r \) respectively. These forces are found to be consistent with respect to the theory of relativity.

Step by step solution

01

Calculate the Force in Lab Frame

Start by calculating the force on the charge in the lab frame. Use the equation for the magnetic force \( F_B = qvB \). Here, the magnetic field \( B = \mu_0\lambda/2\pi r \), where \( \lambda \) is the linear charge density, \( r \) is the distance from the wire, \( \mu_0 \) is the permeability of free space. Substituting the value of \( B \) in the equation for \( F_B \), we get \( F_B = qv\mu_0\lambda/2\pi r \). The electric force \( F_E \) on the charge can be found using the equation \( F_E = k\lambda/r \), where \( k \) is Coulomb's constant. As the charges on the wire and the charge \( q \) have opposite velocities, they are moving frames to each other. Therefore, the total force \( F \) will be the sum of these two forces, \( F = F_B + F_E \).
02

Force in Charge's Rest Frame

Next, the force in the charge's rest frame is obtained. In this frame, the charge \( q \) is not moving, hence, there is no magnetic force exerted on it. The only noticeable force would be the electric force. However, due to the Lorentz contraction, the linear charge density changes in the wire to \( \lambda' = \gamma_u \lambda \) resulting in a new electric force \( F'_E = k\lambda'/r = k\gamma_u\lambda/r \).
03

Calculate Force in Wire's Rest Frame

In the rest frame of the charges in the wire, the moving charge \( q \) exerts a force on the wire, which in return exerts an equal and opposite force on it. Using the equation for magnetic force \( F_B = qvB \) where the magnetic field is \( B = \mu_0\lambda/2\pi r \), one can calculate the magnetic force exerted on the charge \( q \). As the wire is at rest in this frame, there will not be any electric force. The force in this frame will be only the magnetic force \( F_{WB} = qv\cdot\mu_0\gamma_v\lambda/2\pi r \), as the linear charge density will be relativistically contracted \( \lambda = \gamma_v\lambda \).
04

Check for Consistency

Finally, verify the forces in the different frames are consistent by showing the force in the charge's rest frame is the sum of the lab and the wire's frame forces, multiplied by the total gamma factor \( \gamma_{u}\gamma_{v}(1 + \beta_{u}\beta_{v}) \). In the equation, \( \beta_{u} \) and \( \beta_{v} \) are the velocities of the wire and the charge respectively (in terms of speed of light) and \( \gamma_{u} \) and \( \gamma_{v} \) are their respective Lorentz factors. This check ensures that the problem, and its solutions, are in accordance with relativity laws.

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Key Concepts

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

Lorentz Force
When studying the interactions between electric charges and magnetic fields, the Lorentz force is a fundamental concept. It represents the combined effect of electric and magnetic forces on a point charge. Mathematically, it is expressed as
\[\mathbf{F} = q(\boldsymbol{E} + \boldsymbol{v} \times \boldsymbol{B})\]
Where \(q\) is the charge of the particle, \(\boldsymbol{E}\) is the electric field, \(\boldsymbol{v}\) is the velocity of the charge, and \(\boldsymbol{B}\) is the magnetic field. This equation tells us that a charged particle, when moving in a magnetic field, experiences a force that is perpendicular to both its velocity and the magnetic field.

In the provided exercise, the Lorentz force is used to determine the effects on the moving charge near a wire with linear charge density in different reference frames. The electric force and the magnetic force act at the same time, but because the magnetic force depends on the velocity of the charge, it changes when viewed from different frames of reference, which is crucial for understanding the problem at a deeper level.
Relativistic Electromagnetism
Relativistic electromagnetism is the study of how electric and magnetic fields transform and interact under the principles of special relativity. As charges move at speeds close to the speed of light, familiar formulas for electromagnetic forces must be adjusted for the effects of time dilation and space contraction.

In the exercise scenario, we analyze the force on a charge near a wire from different frames—each moving with significant velocity relative to one another. It's important to recognize that velocities add in a Special Relativistic way, hence the use of the Lorentz transformation. The Lorentz gamma factor, \(\gamma = \frac{1}{\sqrt{1-\left(\frac{v}{c}\right)^2}}\), where \(v\) is the relative velocity and \(c\) is the speed of light, becomes a key player in determining the correct electromagnetic forces acting on the charge when you evaluate the situation from various frames. The exercise demonstrates how charges and fields are subject to relativistic effects, enriching our understanding of the electromagnetic force in high-velocity situations.
Lorentz Contraction
Lorentz contraction, or length contraction, is a phenomenon from the theory of relativity that states the length of an object in the direction of its motion contracts when it is moving at a relativistic speed with respect to an observer.

As part of this problem, when evaluating the force in the charge's rest frame and in the wire's rest frame, the linear charge density alters because of Lorentz contraction. The changes in density in the wire leads to changes in the electric field observed in the rest frame of the charge. This modification alters the force experienced by the charge as calculated by \(F'_E = k\gamma_u\lambda/r\). This theoretical insight clarifies why different observers perceive different magnitudes of force—it's not merely about motion, but also how the shapes and densities of objects inherently change due to relativistic speeds.
Linear Charge Density
Linear charge density is a measure of electric charge per unit length along a line, rod, or wire. Represented by the symbol \(\lambda\), it is crucial in calculating the electric and magnetic fields around charged objects.

In our exercise, the wire carries a linear charge density \(\lambda\), which is used to find the electric field that influences the force on our point charge \(q\). Understanding how the linear charge density transforms under different frames of reference is essential because it directly impacts the nature of the electric field and thus the forces experienced by a charge. When examining the problem from the wire's rest frame, for example, due to relativistic motion, we see a different linear charge density, which is key to solving for the electromagnetic forces correctly.

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Most popular questions from this chapter

Motion in \(E\) and B fields \(* * *\) The task of Exercise \(6.29\) is to show that if a charged particle moves in the \(x y\) plane in the presence of a uniform magnetic field in the \(z\) direction, the path will be a circle. What does the path look like if we add on a uniform electric field in the \(y\) direction? Let the particle have mass \(m\) and charge \(q\). And let the magnitudes of the electric and magnetic fields be \(E\) and \(B\). Assume that the velocity is nonrelativistic, so that \(\gamma \approx 1\) (this assumption isn't necessary in Exercise 6.29, because \(v\) is constant there). Be careful, the answer is a bit counterintuitive.

Scaled-up ring \(*\) Consider two circular rings of copper wire. One ring is a scaled-up version of the other, twice as large in all regards (radius, crosssectional radius). If currents around the rings are driven by equal voltage sources, how do the magnetic fields at the centers compare?

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, \(A_{0}\) ?

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 $$ B_{z}=\frac{\mu_{0} I}{4 \pi}\left(\frac{1}{x}+\frac{1}{y}+\frac{x}{y \sqrt{x^{2}+y^{2}}}+\frac{y}{x \sqrt{x^{2}+y^{2}}}\right) $$

Rings with opposite currents Two parallel rings have the same axis and are separated by a small distance \(\epsilon\). They have the same radius \(a\), and they carry the same current \(I\) but in opposite directions. Consider the magnetic field at points on the axis of the rings. The field is zero midway between the rings, because the contributions from the rings cancel. And the field is zero very far away. So it must reach a maximum value at some point in between. Find this point. Work in the approximation where \(\epsilon \ll a\)

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