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Consider a positively charged particle moving at constant speed parallel to a current-carrying wire, in the direction of the current. As you know (after studying Chapters 27 and 28 ), the particle is attracted to the wire by the magnetic force due to the current. Now suppose another observer moves along with the particle, so according to him the particle is at rest. Of course, a particle at rest feels no magnetic force. Does that observer see the particle attracted to the wire or not? How can that be? (Either answer seems to lead to a contradiction: If the particle is attracted, it must be by an electric force because there is no magnetic force, but there is no electric field from a neutral wire; if the particle is not attracted, the observer sees that the particle is, in fact, moving toward the wire.)

Short Answer

Expert verified
Answer: In the stationary observer's frame, the charged particle experiences a magnetic force that attracts it towards the wire. In contrast, when the observer moves along with the charged particle, the particle appears to be at rest and does not experience any magnetic force. However, due to special relativity and length contraction, the moving observer perceives the wire as negatively charged, causing the positively charged particle to be attracted to the wire by an electric force. In both reference frames, the charged particle is attracted to the wire by the combination of electric and magnetic forces, but the contribution of each force depends on the observer's reference frame.

Step by step solution

01

Analyze stationary observer's perspective

In the stationary observer's frame, the positively charged particle is moving parallel to the current-carrying wire. According to the magnetic force experienced by a moving charged particle, the charged particle will feel an attraction towards the wire due to the magnetic field generated by the current-carrying wire. This force can be calculated using the equation F = q*v*B (where q is the charge of the particle, v is its velocity, and B is the magnetic field).
02

Analyze moving observer's perspective

When the observer is moving along with the charged particle, the particle appears to be at rest in this reference frame. Hence, from this perspective, the charged particle should not experience any magnetic force, as F = q*(0)*B = 0. However, the particle should still appear to be attracted to the wire, which seems contradictory.
03

Resolve the contradiction using special relativity

The resolution to this apparent contradiction can be found in special relativity. When the observer moves along with the charged particle, the moving observer is now in a different reference frame. In this new reference frame, the electrons in the current-carrying wire are moving at a different velocity (since the observer is now moving too). This change in the reference frame leads to length contraction, which causes the electrons in the wire to appear denser to the moving observer.
04

Determine the force acting on the charged particle

Due to the length contraction, the moving observer perceives the wire to be negatively charged. As a result, the positively charged particle is now attracted to the wire by the electric force between them. This electric force, acting on the charged particle, keeps it attracted to the wire in the moving observer's frame while satisfying the absence of magnetic force. Thus, there is no contradiction, as the stationary observer sees the magnetic force acting on the particle, and the moving observer sees the electric force acting on the particle. In both reference frames, the charged particle is attracted to the wire by the combination of electric and magnetic forces, but the contribution of each force depends on the reference frame of the observer.

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Most popular questions from this chapter

Consider motion in one spatial dimension. For any velocity \(v_{n}\) define the parameter \(\theta\) via the relation \(v=c \tanh \theta,\) where \(c\) is the speed of light in vacuum. This quantity is called the velocity parameter or the rapidity corresponding to velocity \(v\). a) Prove that for two velocities, which add via a Lorentz transformation, the corresponding velocity parameters simply add algebraically, that is, like Galilean velocities. b) Consider two reference frames in motion at speed \(v\) in the \(x\) -direction relative to one another, with axes parallel and origins coinciding when clocks at the origin in both frames read zero. Write the Lorentz transformation between the two coordinate systems entirely in terms of the velocity parameter corresponding to \(v\) and the coordinates.

Find the value of \(g\), the gravitational acceleration at Earth's surface, in light-years per year per year, to three significant figures.

In the twin paradox example, Alice boards a spaceship that flies to a space station 3.25 light-years away and then returns with a speed of \(0.650 c\). View the trip in terms of Alice's reference frame. a) Show that Alice must travel with a speed of \(0.914 c\) to establish a relative speed of \(0.650 c\) with respect to Earth when she is returning to Earth. b) Calculate the time duration for Alice's return flight to Earth at the speed of \(0.914 c\)

In the twin paradox example (in Section 35.2 ), Alice boards a spaceship that flies to space station 3.25 light-years away and then returns with a speed of \(0.65 c .\) a) Calculate the total distance Alice traveled during the trip, as measured by Alice. b) Using the total distance from part (a), calculate the total time duration for the trip, as measured by Alice.

\- 35.87 Although it deals with inertial reference frames, the special theory of relativity describes accelerating objects without difficulty. Of course, uniform acceleration no longer means \(d v / d t=g,\) where \(g\) is a constant, since that would have \(v\) exceeding \(c\) in a finite time. Rather, it means that the acceleration experienced by the moving body is constant: In each increment of the body's own proper time, \(d \tau,\) the body experiences a velocity increment \(d v=g d \tau\) as measured in the inertial frame in which the body is momentarily at rest. (As it accelerates, the body encounters a sequence of such frames, each moving with respect to the others.) Given this interpretation: a) Write a differential equation for the velocity \(v\) of the body, moving in one spatial dimension, as measured in the inertial frame in which the body was initially at rest (the "ground frame"). You can simplify your equation by remembering that squares and higher powers of differentials can be neglected. b) Solve the equation from part (a) for \(v(t),\) where both \(v\) and \(t\) are measured in the ground frame. c) Verify that \(v(t)\) behaves appropriately for small and large values of \(t\). d) Calculate the position of the body, \(x(t),\) as measured in the ground frame. For convenience, assume that the body is at rest at ground-frame time \(t=0\) and at ground-frame position \(x=c^{2} / g\). e) Identify the trajectory of the body on a space-time diagram (a Minkowski diagram, for Hermann Minkowski) with coordinates \(x\) and \(c t,\) as measured in the ground frame. f) For \(g=9.81 \mathrm{~m} / \mathrm{s}^{2},\) calculate how much time it takes the body to accelerate from rest to \(70.7 \%\) of \(c\), measured in the ground frame, and how much ground-frame distance the body covers in this time.

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