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Chapter 27: Problem 69

A long copper wire carries current in the east direction. The electrons are moving with a drift velocity \(\overrightarrow{\mathbf{u}}\). An. observer now moves with the velocity \(\overrightarrow{\mathbf{u}}\). In the frame of this observer: (a) electric field is present (b) magnetic field due to wire is zero (c) only magnetic field is present (d) none of the above

Short Answer

Expert verified
In the observer's frame, both electric and magnetic fields are zero; answer is (d) none of the above.

Step by step solution

01

Understand the Motion of Electrons

The copper wire carries a current, denoting that electrons within the wire are moving with a drift velocity \( \overrightarrow{\mathbf{u}} \) opposite to the current (west direction). Current direction is east since electrons move in opposite direction of conventional current.
02

Observer's Frame of Reference

The observer is moving with the same velocity \( \overrightarrow{\mathbf{u}} \) as the electrons (west). Relative to the observer, electrons appear stationary because both move at the same velocity.
03

Magnetic Field in Observer's Frame

Since the relative motion between the electrons and the observer is zero (as electrons appear stationary to the observer), there should be no change in magnetic field observed. In the reference frame of the observer, magnetic fields due to moving charges would be zero since there's no motion of charge relative to the observer.
04

Electric Field in Observer's Frame

The observer sees electrons as stationary (no drift velocity component), which implies there's no electric field created by their motion. Thus, the electric field would also appear to be zero from their perspective.
05

Conclusion Based on Observations

As both the electric and magnetic fields are zero according to the observer's frame where electrons are stationary, the correct answer is (d) none of the above.

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Key Concepts

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

Drift Velocity
Drift velocity is the average velocity that charge carriers, such as electrons, attain due to an electric field in a material like copper. When a current flows through a conductor, electrons in the material do not travel in a straight line. Instead, they move randomly due to thermal energy but gradually drift in a specific direction. This drift direction is opposite to that of the current due to electrons being negatively charged.
  • In a copper wire carrying an electric current, electrons move with this drift velocity.
  • The electrons' drift poses the current, often opposite to the assumed positive flow of conventional current.
When an observer moves with this drift velocity, as in the original problem, the electrons appear stationary relative to the observer. This observation directly influences conclusions about the electric and magnetic fields encountered.鈥�
Frame of Reference
The frame of reference is a perspective or "point of view" that affects how we measure and perceive physical quantities like velocity and fields. In the context of the original exercise, the observer's frame of reference is crucial.
  • An observer moving with the same velocity as the drifting electrons shares their frame of reference.
  • In this scenario, electrons appear stationary, causing certain dynamic effects, like magnetic fields due to their motion, to be non-apparent.
Consequently, the relative velocity between the observer and the electrons becomes zero. This means from the observer's perspective, the whole situation simplifies: without relative motion, effects reliant on motion like magnetic fields are neutralized.
Electric Fields
Electric fields represent regions where a charge experiences a force; they are integral to understanding how charges like electrons interact within materials.
  • In a practical scenario, moving electrons in a wire create a net electric field that drives the current.
  • However, when examined from a differing frame of reference, such as that of an observer moving alongside the electrons, these fields behave differently.
To the moving observer, these electric fields diminish or disappear. That's because the potential driving the drift velocity is aligned with the observer's velocity, rendering the field effect unobservable, a concept that aligns with the conclusion of the zero electric field in the observer's frame in the given exercise.

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Most popular questions from this chapter

In a certain region uniform electric field \(E\) and magnetic field \(B\) are present in the opposite direction. At the instant \(t=0\), a particle of mass \(m\) carrying a charge \(q\) is given velocity \(v_{0}\) at an angle \(\theta\), with the \(y\) -axis, in the \(y-z\) plane. The time after which the speed of the particle would be minimum is equal to: (a) \(\frac{m v_{0}}{q E}\) (b) \(\frac{m v_{0} \sin \theta}{q E}\) (c) \(\frac{m v_{0} \cos \theta}{q E}\) (d) \(\frac{2 \pi m}{q B}\)

A charged particle of mass \(m\) and charge \(q\) is accelerated through a potential difference of \(V\) volt. It enters a region of uniform magnetic field which is directed perpendicular to the direction of motion of the particie. The particle will move on a circular path of radius given by: (a) \(\sqrt{\left(\frac{V m}{q B^{2}}\right)}\) (b) \(\frac{2 V m}{q B^{2}}\) (c) \(\sqrt{\left(\frac{2 V_{m}}{q}\right)}\left(\frac{1}{B}\right)\) (d) \(\sqrt{\left(\frac{V m}{q}\right)}\left(\frac{1}{B}\right)\)

The restoring couple in the moving coil galvanometer is because of : (a) magnetic field (b) material of the coil (c) twist produced in the suspension (d) current in the coil

A beam of protons with a velocity of \(4 \times 10^{5} \mathrm{~ms}^{-1}\) enters a uniform magnetic field of \(0.3 \mathrm{~T}\). The velocity makes an angle of \(60^{\circ}\) with the magnetic field. The radius of the helical path taken by the proton is: (a) \(4.4 \mathrm{~cm}\) (b) \(1.2 \mathrm{~cm}\) (c) \(4.4 \mathrm{~cm}\) (d) \(2.2 \mathrm{~cm}\)

The ratio of the energy required to set-up in a cube of side \(10 \mathrm{~cm}\) in a uniform magnetic field of \(4 \mathrm{~Wb} / \mathrm{m}^{2}\) and a uniform electric field of of \(10^{6} \mathrm{~V} / \mathrm{m}\), is: (a) \(1.44 \times 10^{7}\) (b) \(1.44 \times 10^{-5}\) (c) \(1.44 \times 10^{6}\) (d) \(1.44 \times 10^{3}\)
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