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Zero force in any frame \(* *\) A neutral wire carries current \(I\). A stationary charge is nearby. There is no electric field from the neutral wire, so the electric force on the charge is zero. And although there is a magnetic field, the charge isn't moving, so the magnetic force is also zero. The total force on the charge is therefore zero. Hence it must be zero in every other frame. Verify this, in a particular case, by using the Lorentz transformations to find the \(\mathbf{E}\) and \(\mathbf{B}\) fields in a frame moving parallel to the wire with velocity \(\mathbf{v}\).

Short Answer

Expert verified
By applying the Lorentz transformations to the electric and magnetic fields, it is verified that the total force experienced by the stationary charge in proximity to the current-carrying neutral wire remains zero, regardless of the frame of observation.

Step by step solution

01

Start with equations for electric and magnetic fields

The electric and magnetic fields at a point in space due to a current carrying wire can be given by Coulomb's law and Ampere’s law, respectively.
02

Identify the state in initial frame

In the rest frame, the electric field \( \mathbf{E_0} \) is zero (because the wire is neutral) and magnetic field \( \mathbf{B_0} \) can be calculated from Ampere's law. Since the charge is stationary, there is no magnetic force. Therefore, the resultant force, \( \mathbf{F_0} \), is zero.
03

Apply Lorentz transformations

Using Lorentz transformations, we can calculate the electric and magnetic fields in the new frame.In the frame moving parallel to the wire, the new electric field \( \mathbf{E'} \) and magnetic field \( \mathbf{B'} \) can be related to the fields in the rest frame \( \mathbf{E_0} \) and \( \mathbf{B_0} \) as follows: \[ \mathbf{E'} = \mathbf{E_0} + \mathbf{v} \times \mathbf{B_0} \] \[ \mathbf{B'} = \mathbf{B_0} - \frac{1}{c^{2}}( \mathbf{v} \times \mathbf{E_0} ) \]
04

Calculate the new fields and force

Substitute \( \mathbf{E_0} \) and \( \mathbf{B_0} \) to find \( \mathbf{E'} \) and \( \mathbf{B'} \). From these, calculate the force \( \mathbf{F'} \) acting on the charge in the moving frame using the equation \( \mathbf{F'} = q(\mathbf{E'} + \mathbf{v} \times \mathbf{B'}) \).
05

Confirmation of result

it can be checked whether the total force \( \mathbf{F'} \) indeed remains zero in the new frame, thereby verifying the initial statement.

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Key Concepts

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

Understanding the Electric Field
An electric field is a region around a charged particle or object within which a force would be exerted on other charges. In the context of a neutral wire, such as the one given in the exercise, there is no net electric field surrounding the wire because the positive and negative charges are evenly distributed, effectively canceling each other out.
Therefore, a stationary charge near the neutral wire experiences zero electric force. It’s important to remember that electric fields originate from charged particles. When charges are in balance, as in a neutral object, their electric fields negate each other, resulting in no net electric field.
The Role of the Magnetic Field
Magnetic fields are created by moving charges or currents, such as the current flowing through the wire in the exercise. According to Ampère’s law, the magnetic field around a current carrying wire loops in a circular path around the wire.
This means that in the initial rest frame of the wire, there is a magnetic field \( \mathbf{B_0} \) present. However, a stationary charge does not experience a magnetic force because the magnetic force depends on the motion of the charge. The magnetic force formula \( \mathbf{F_m} = q(\mathbf{v} \times \mathbf{B}) \) shows that when velocity \( \mathbf{v} \) is zero, the force is consequently zero as well, ensuring no magnetic force acts on a stationary charge.
Exploring the Current Carrying Wire
A current carrying wire is critical in studying electromagnetic fields. As the wire carries current, it creates a surrounding magnetic field. This is explained by the right-hand rule, where if you point your thumb in the direction of the current, your fingers curl in the direction of the magnetic loops surrounding the wire. This wire does not produce an electric field because it is neutral.
In the exercise, a neutral wire ensures that any effects due to electric fields are absent because of the inherent charge balance in the wire. The speed of moving charges in the wire generates the magnetic field, which we need to consider when applying transformations to see effects in different frames.
Relativistic Effects and Lorentz Transformations
Relativistic effects become significant when charges are considered in different frames of motion, often involving large velocities. Lorentz transformations help us understand how measurements of electric and magnetic fields change between these frames. When a frame moves with velocity \( \mathbf{v} \) parallel to the wire, Lorentz transformations modify the fields as follows:
  • The electric field in the moving frame \( \mathbf{E'} \) is given by \( \mathbf{E_0} + \mathbf{v} \times \mathbf{B_0} \).
  • The magnetic field is given by \( \mathbf{B'} = \mathbf{B_0} - \frac{1}{c^{2}}( \mathbf{v} \times \mathbf{E_0} ) \).
This procedure reveals that even when transforming frames, the analysis guarantees that the results agree with the principle that physical laws --- including forces --- are consistent across all frames. This means the total zero force observed in one frame remains zero in all frames, confirming the principle of relativity.

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Most popular questions from this chapter

Scaled-down solenoid ** Consider two solenoids, one of which is a tenth-scale model of the other. The larger solenoid is 2 meters long, 1 meter in diameter, and is wound with \(1 \mathrm{~cm}\) diameter copper wire. When the coil is connected to a \(120 \mathrm{~V}\) direct-current generator, the magnetic field at its center is 1000 gauss. The scaled-down model is exactly onetenth the size in every linear dimension, including the diameter of the wire. The number of turns is the same, and it is designed to provide the same central field. (a) Show that the voltage required is the same, namely \(120 \mathrm{~V}\). (b) Compare the coils with respect to the power dissipated and the difficulty of removing this heat by some cooling means.

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?

E and B for a point charge ** (a) Use the Lorentz transformations to show that the \(\mathbf{E}\) and \(\mathbf{B}\) fields due to a point charge moving with constant velocity \(\mathbf{v}\) are related by \(\mathbf{B}=\left(\mathbf{v} / c^{2}\right) \times \mathbf{E}\) (b) If \(v \ll c\), then \(\mathbf{E}\) is essentially obtained from Coulomb's law, and \(\mathbf{B}\) can be calculated from the Biot-Savart law. Calculate \(\mathbf{B}\) this way, and then verify that it satisfies \(\mathbf{B}=\left(\mathbf{v} / c^{2}\right) \times\) E. (It may be helpful to think of the point charge as a tiny rod of charge, in order to get a handle on the \(d l\) in the BiotSavart law.)

Helmholtz coils One way to produce a very uniform magnetic field is to use a very long solenoid and work only in the middle section of its interior. This is often inconvenient, wasteful of space and power. Can you suggest ways in which two short coils or current rings might be arranged to achieve good uniformity over a limited region? Hint: Consider two coaxial current rings of radius \(a\), separated axially by a distance \(b\). Investigate the uniformity of the field in the vicinity of the point on the axis midway between the two coils. Determine the magnitude of the coil separation \(b\) that for given coil radius \(a\) will make the field in this region as nearly uniform as possible.

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.

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