/*! 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 65 A long straight wire carries a c... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 19: Problem 65

A long straight wire carries a current of 3.2 \(A\) in the positive \(x\)-direction. An electron, traveling at \(6.8 脳 10 6\) m/s in the positive \(x\) -direction, is 4.6 cm from the wire. What force acts on the electron?

Short Answer

Expert verified
Question: Calculate the magnetic force acting on an electron moving parallel to a wire carrying a current of 3.2 A. The electron is moving at a speed of 6.8 x 10^6 m/s and is located at a distance of 4.6 cm from the wire. Answer: The magnetic force acting on the electron is approximately -3.02 x 10^(-17) N.

Step by step solution

01

Write down the known values and formula.

First, let's collect all the information given in the exercise. We have the current in the wire (\(I = 3.2 A\)), the electron's speed (\(v = 6.8 \times 10^6 m/s\)), and its distance from the wire (\(r = 4.6 cm = 0.046 m\)). The charge of the electron is \(q = -1.6 \times 10^{-19} C\). The formula to calculate the force on a moving charged particle in a magnetic field is \(F = qvB \sin{\theta}\).
02

Calculate the magnetic field strength due to the wire's current.

To find the magnetic field strength at the location of the electron, we can apply Ampere's law. The magnetic field strength (\(B\)) at a distance \(r\) from a straight wire carrying current \(I\) is described by the following formula: \(B = \frac{\mu_0 I}{2 \pi r}\), where \(\mu_0\) is the permeability of free space and has a value of \(4\pi \times 10^{-7} Tm/A\). Now, we can plug in the values we know: \(B = \frac{4\pi \times 10^{-7} \times 3.2}{2 \pi \times 0.046}\)
03

Simplify the expression for the magnetic field strength.

Now, we simplify the expression to find the magnetic field strength: \(B = \frac{4 \times 10^{-7} \times 3.2}{0.046} = 2.78 \times 10^{-5} T\)
04

Determine the angle between the velocity vector and the magnetic field.

Since the electron is moving parallel to the wire in the positive \(x\)-direction, the direction of its velocity vector is perpendicular to the magnetic field lines around the wire, which are concentric circles centered on the wire. So, the angle between the velocity vector and the magnetic field (\(\theta\)) is 90 degrees or \(\frac{\pi}{2}\) radians.
05

Calculate the magnetic force on the electron.

Now, we can use the formula for the magnetic force on a moving charged particle to find the force experienced by the electron: \(F = qvB \sin{\theta} = (-1.6 \times 10^{-19}) \times (6.8 \times 10^6) \times (2.78 \times 10^{-5}) \times \sin{\frac{\pi}{2}}\)
06

Simplify the expression for the magnetic force.

Finally, we simplify the expression to find the magnetic force on the electron: \(F = -1.6 \times 10^{-19} \times 6.8 \times 10^6 \times 2.78 \times 10^{-5} = -3.02 \times 10^{-17} N\) The force that acts on the electron is approximately \(-3.02 \times 10^{-17} N\). The negative sign indicates that the force is acting in the direction opposite to the electron's motion, which is expected due to the repulsion between the negatively charged electron and the straight wire's magnetic field lines.

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!

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

A square loop of wire of side \(3.0 \mathrm{cm}\) carries \(3.0 \mathrm{A}\) of current. A uniform magnetic field of magnitude \(0.67 \mathrm{T}\) makes an angle of \(37^{\circ}\) with the plane of the loop. (a) What is the magnitude of the torque on the loop? (b) What is the net magnetic force on the loop?
The magnetic field in a cyclotron is \(0.50 \mathrm{T}\). What must be the minimum radius of the dees if the maximum proton speed desired is $1.0 \times 10^{7} \mathrm{m} / \mathrm{s} ?$
Electrons in a television's CRT are accelerated from rest by an electric field through a potential difference of \(2.5 \mathrm{kV} .\) In contrast to an oscilloscope, where the electron beam is deflected by an electric ficld, the beam is deflected by a magnetic field. (a) What is the specd of the electrons? (b) The beam is deflected by a perpendicular magnetic field of magnitude $0.80 \mathrm{T}$. What is the magnitude of the acceleration of the electrons while in the field? (c) What is the speed of the electrons after they travel 4.0 mm through the magnetic field? (d) What strength electric field would give the electrons the same magnitude acceleration as in (b)? (c) Why do we have to use an clectric ficld in the first place to get the electrons up to speed? Why not use the large acceleration due to a magnetic field for that purpose?
Prove that the time for one revolution of a charged particle moving perpendicular to a uniform magnetic field is independent of its speed. (This is the principle on which the cyclotron operates.) In doing so, write an expression that gives the period \(T\) (the time for one revolution) in terms of the mass of the particle, the charge of the particle, and the magnetic field strength.
A proton moves in a helical path at speed \(v=4.0\) $\times 10^{7} \mathrm{m} / \mathrm{s}$ high above the atmosphere, where Earth's magnetic field has magnitude \(B=1.0 \times 10^{-6} \mathrm{T}\). The proton's velocity makes an angle of \(25^{\circ}\) with the magnetic field. (a) Find the radius of the helix. [Hint: Use the perpendicular component of the velocity.] (b) Find the pitch of the helix-the distance between adjacent "coils." [Hint: Find the time for one revolution; then find how far the proton moves along a field line during that time interval.]
See all solutions

Recommended explanations on Physics Textbooks

Oscillations

Read Explanation

Kinematics Physics

Read Explanation

Electric Charge Field and Potential

Read Explanation

Electromagnetism

Read Explanation

Physical Quantities And Units

Read Explanation

Electricity and Magnetism

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.