/*! 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 100 Two very long straight parallel ... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 9: Problem 100

Two very long straight parallel wires carry steady currents \(i\) and \(2 i\) in opposite directions. The distence between the wires is \(d\). At a certain instant of time, a point charge \(q\) is at a polnt equidistant from the two wires in the plane of the wires. Its instantaneous velocity \(\vec{v}\) is perpendicular to this plane. The magnitude of the force due to the magnetic fleld acting on the charge at this instant is a \(\frac{\mu_{0} i q v}{2 \pi d}\) h \(\frac{\mu_{0} i q v}{\pi d}\) c. \(\frac{3 \mu_{0} l q v}{2 \pi d}\) d zero

Short Answer

Expert verified
Option b, \(\frac{\mu_{0} i q v}{\pi d}\), is correct.

Step by step solution

01

Define the situation

We have two parallel wires separated by a distance \(d\). They carry currents \(i\) and \(2i\) in opposite directions. A point charge \(q\) with velocity \(\vec{v}\) perpendicular to the plane of these wires is equidistant from them.
02

Use Amp猫re's Law to find magnetic fields

According to Amp猫re's Law, the magnetic field \(B\) at a distance \(r\) from a very long straight wire carrying a current \(I\) is given by \(B = \frac{\mu_0 I}{2\pi r}\). For each wire, calculate the magnetic field at the location of the charge.
03

Calculate magnetic field from each wire

Since the charge is equidistant from both wires, the distance from each to the charge is \( \frac{d}{2} \). - For the wire with current \(i\): \[ B_1 = \frac{\mu_0 i}{2\pi (d/2)} = \frac{\mu_0 i}{\pi d} \]- For the wire with current \(2i\): \[ B_2 = \frac{\mu_0 (2i)}{2\pi (d/2)} = \frac{2 \mu_0 i}{\pi d} \]
04

Determine the net magnetic field

The magnetic fields created by the currents in opposite directions will add up at the location of the charge. Since they are in opposite directions, the net magnetic field \(B_{net}\) is:\[ B_{net} = B_2 - B_1 = \frac{2\mu_0 i}{\pi d} - \frac{\mu_0 i}{\pi d} = \frac{\mu_0 i}{\pi d} \] upward.
05

Calculate the force on the charge

The force \(F\) on a charge \(q\) moving in a magnetic field \(B\) with velocity \(\vec{v}\) is given by \(F = qvB\). Substitute \(B = \frac{\mu_0 i}{\pi d}\) to find the force:\[ F = qv \left( \frac{\mu_0 i}{\pi d} \right) = \frac{\mu_0 i q v}{\pi d} \]
06

Compare with given options

The calculated force \( \frac{\mu_0 i q v}{\pi d} \) matches option b. Therefore, this is the correct magnitude of the magnetic force acting on the charge at this instant.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

Key Concepts

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

Amp猫re's Law
Amp猫re's Law is a fundamental principle in electromagnetism that relates magnetic fields to the electric currents that produce them. It's an integral part of Maxwell's equations and helps us calculate the magnetic field generated by a current distribution. The law is expressed mathematically as: \[ B = \frac{\mu_0 I}{2\pi r} \]where:
  • \( B \) is the magnetic field.
  • \( \mu_0 \) is the permeability of free space.
  • \( I \) is the current through the wire.
  • \( r \) is the radial distance from the wire.
Using Amp猫re's Law, we can easily determine the magnetic field at any point as long as we know the current and distance from the source. This concept is crucial when dealing with long, straight conductors like the wires in our problem setup.
Magnetic Fields of Parallel Currents
The interaction between magnetic fields created by parallel currents is quite fascinating. When two straight, parallel wires carry currents, they generate magnetic fields that influence one another. If the currents flow in the same direction, the magnetic attraction brings the wires closer. Conversely, if the currents flow in opposite directions, they repel each other.

In the scenario from the original problem, the wires are carrying currents in opposite directions, leading to repulsion. For a point equidistant from both wires, like the location of the charge in our instance, each wire contributes its field:
  • Wire with current \( i \) creates a field \( B_1 = \frac{\mu_0 i}{\pi d} \).
  • Wire with current \( 2i \) creates a field \( B_2 = \frac{2 \mu_0 i}{\pi d} \).
These fields add up linearly, resulting in a net field pointing upward due to the direction of current flow in the respective wires.
Force on a Moving Charge in a Magnetic Field
The force experienced by a moving charge in a magnetic field is governed by the principles of the Lorentz force law. This force can significantly affect the path of the charged particle. The formula for the force \( F \) is given by:\[ F = qvB \sin \theta \]where:
  • \( q \) is the charge.
  • \( v \) is the velocity of the charge.
  • \( B \) is the magnetic field.
  • \( \theta \) is the angle between the velocity and magnetic field (90 degrees in this problem, so \( \sin \theta = 1 \)).
For our point charge in the problem, the magnetic force is simplified to:\[ F = qv \left( \frac{\mu_0 i}{\pi d} \right) = \frac{\mu_0 i q v}{\pi d} \]

This result is found by understanding how these components interact. It matches perfectly with option (b) given in the original exercise, confirming the calculated magnitude of the force.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

An electron is moving along positive \(x\) -axis. A uniform electric field exists toward negative \(y\) -axis. What should be the direction of magnetic field of suitable magnitude so that net force on the electron is zero? a. Positive \(z\) -axis b Negative \(z\) -axis c. Positive \(y\) -axis d. Negative \(y\) -axis

A current carrying loop lies on a smooth horizontal plane. Then, a it is possible to establish a uniform magnetic field in the region so that the loop starts rotating about its own axis b. it is possible to establish a uniform magnetic field in the region so that the loop will tip over about any of the point c. it is not possible that loop will tip over about any of the point whatever be the direction of established magnetic field (uniform) d both (a) and (b) are correct.

Four parallel conductors, carrying equal currents, pass. vertically through the four comers of a square WXYZ. In two conductors, the current is flowing into the page, and in the other two out of the page. In what directions must the currents flow to produce a resultant magnetic field in the direction shown at \(O\), the center of the square? \(9.278\) Into the page \(\quad\) Out of the page A \(W\) and \(Y\) and \(Z\) \(\mathbf{b} X\) and \(Z\) and \(Y\) c. \(W\) and \(Z\) \(X\) and \(Y\) d. \(W\) and \(X\) \(\boldsymbol{Y}\) and \(\mathrm{Z}\)

A particle of charge per unit mass \(\alpha\) is released from origin with velocity \(\bar{V}=-V_{0} \hat{j}\) in a magnetic field $$ \vec{B}=-B_{0} \hat{k} \text { for } x \leq \frac{\sqrt{3}}{2} \frac{v_{0}}{B_{0} \alpha} $$ and \(\bar{B}=0\) for \(x>\frac{\sqrt{3}}{2} \frac{v_{0}}{B_{0} \alpha}\) The \(x\) -coordinate of the particle at time \(t\left(>\frac{\pi}{3 B_{0} \alpha}\right)\) would be a. \(\frac{\sqrt{3}}{2} \frac{v_{0}}{B_{0} \alpha}+\frac{\sqrt{3}}{2} v_{0}\left(t-\frac{\pi}{B_{0} \alpha}\right)\). b. \(\frac{\sqrt{3}}{2} \frac{v_{0}}{B_{0} \alpha}+v_{0}\left(t-\frac{\pi}{3 B_{0} \alpha}\right)\) c. \(\frac{\sqrt{3}}{2} \frac{v_{0}}{B_{0} \alpha}+\frac{v_{0}}{2}\left(t-\frac{\pi}{3 B_{0} \alpha}\right)\) d. \(\frac{\sqrt{3}}{2} \frac{v_{0}}{B_{0} \alpha}+\frac{v_{0} t}{2}\)

A square loop of side a carries a current \(I\). The magnetic induction \(B\) at point \(P\), lying on the axis of the loop and at a distance \(x\) from the center of loop is a. \(\frac{4 \mu_{0} i a^{2}}{3 \pi\left(4 x^{2}+a^{2}\right) \sqrt{4 x^{2}+2 a^{2}}}\) b. \(\frac{\sqrt{3} \mu_{0} i a^{2}}{\pi\left(4 x^{2}+a^{2}\right) \sqrt{4 x^{2}+2 a^{2}}}\) c. \(\frac{\mu_{0} i a^{2}}{\pi\left(4 x^{2}+a^{2}\right) \sqrt{4 x^{2}+2 a^{2}}}\) d. \(\frac{4 \mu_{0} i a^{2}}{\pi\left(4 x^{2}+a^{2}\right) \sqrt{4 x^{2}+2 a^{2}}}\)
See all solutions

Recommended explanations on Physics Textbooks

Magnetism and Electromagnetic Induction

Read Explanation

Geometrical and Physical Optics

Read Explanation

Electromagnetism

Read Explanation

Wave Optics

Read Explanation

Fundamentals of Physics

Read Explanation

Physics of Motion

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.