/*! 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 54 A long, straight wire carries a ... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 28: Problem 54

A long, straight wire carries a current of 8.60 A. An electron is traveling in the vicinity of the wire. At the instant when the electron is \(4.50 \mathrm{~cm}\) from the wire and traveling at a speed of \(6.00 \times 10^{4} \mathrm{~m} / \mathrm{s}\) directly toward the wire, what are the magnitude and direction (relative to the direction of the current) of the force that the magnetic field of the current exerts on the electron?

Short Answer

Expert verified
The magnitude of the force is equal to \(2.06 \times 10^{-16} N\). The direction of the force on the electron is to the left relative to the direction of the current in the wire.

Step by step solution

01

Calculate Magnetic Field

Use Ampere's Law in the form \(B = \frac{{\mu I}}{{2 \pi r}}\) to calculate the magnitude of the magnetic field at the location of the electron. Here, \(\mu\) is the permeability of free space equal to \(4\pi \times 10^{-7} T m/A\), \(I = 8.60 A\) is the current in the wire, and \(r = 0.045 m\) is the distance of the electron from the wire.
02

Calculate Force

Use the Lorentz force formula \(F = |e|vBsin\theta\) to calculate the magnitude of the force on the electron due to the magnetic field. \(e =1.60219 脳 10^{-19} C\) is the charge of the electron, \(v = 6.00 脳 10^4 m/ s\) is the speed of the electron and \(\theta =90掳\) is the angle between the velocity vector of the electron and the direction of the magnetic field, which is perpendicular in this case due to the direction of current in wire and the path of electron.
03

Determine Direction of Force

Use the right-hand rule. Point the thumb in the direction of the negatively charged particle's velocity (opposite direction for an electron), and curl fingers in the direction of the magnetic field, the push of the palm gives the direction of the force. In this case, if the current in wire is upward and the electron is moving towards the wire, then the force on the electron will be to the left.

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.

Ampere's Law
Ampere's Law helps us determine the magnetic field generated by an electric current. In a simple form for a long, straight wire, the formula is given by:
  • \(B = \frac{{\mu I}}{{2 \pi r}}\)
Here, \(B\) represents the magnetic field strength, \(\mu\) is the permeability of free space which is approximately \(4\pi \times 10^{-7} \, T \, m/A\), \(I\) is the current flowing through the wire, and \(r\) is the radial distance from the wire to the point you are interested in.

In our problem, a current of \(8.60 \, A\) flows through the wire and the electron is \(4.50 \, cm\) away, which is \(0.045 \, m\) when converted to meters. Plugging these values in, we get the magnetic field around the wire at the electron鈥檚 position.

This formula shows how the magnetic field intensity reduces with distance, spreading outward along concentric circles around the wire. Knowing the magnetic field helps us move onto calculating the force exerted on a charge in motion, like an electron.
Lorentz Force
The Lorentz force is what we need to find how much force a magnetic field exerts on a moving charge. The formula is:
  • \(F = |e|vB\sin\theta\)
Here, \(F\) is the force magnitude, \(|e|\) is the absolute charge of an electron \(\approx 1.60219 \times 10^{-19} \, C\), \(v\) the velocity of the electron, \(B\) the magnetic field strength, and \(\theta\) the angle between the electron鈥檚 velocity and the magnetic field.

In this situation, \(\theta = 90^\circ\) since the electron is moving directly towards the wire. This means they are perpendicular, making \(\sin\theta = 1\). Because of this, the force formula simplifies to \(F = |e|vB\).

As the electron moves closer or faster, the force increases if the magnetic field remains constant. This is critical in designing applications where precise control over charged particles is needed, like in particle accelerators or certain types of sensors.
Right-Hand Rule
While calculating the force's magnitude is essential, understanding its direction makes it whole. The right-hand rule is your friend here when navigating through magnetic and electric forces visually.

This rule involves using your right hand to determine the direction of force on a moving charge in a magnetic field:
  • Point the thumb in the direction of the velocity of the positive charge (reverse this for a negative charge like an electron).
  • Fingers point in the direction of the magnetic field.
  • The palm then indicates the direction of the force.
For an electron traveling towards the wire with an upward current, the force points to the left. This happens because we swap the thumb direction due to the electron's negative charge. Thus, understanding this helps predict how a charged particle will move, facilitating control over systems involving magnetic fields and currents.

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

We can estimate the strength of the magnetic field of a refrigerator magnet in the following way: Imagine the magnet as a collection of current-loop magnetic dipoles. (a) Derive the force between two current loops with radius \(R\) and current \(I\) separated by distance \(d \ll R\). Very close to the wire its magnetic field is about the same as for an infinitely long wire, and Eq.( 28.11 ) can be used. (b) Using Eq. ( 28.17 ), express the current \(I\) in terms of the magnetic field at the middle of the loop, and express the radius \(R\) in terms of the area of the loop. In this way, derive an expression for the force \(F\) between two identical current loops separated by a small distance \(d\) in terms of their mutual area \(A\) and center magnetic field \(B\). (c) Rearrange your result to obtain an expression for the magnetic field of a dipole with area \(A\) in terms of the force \(F\) from an identical dipole separated by a small distance \(d\). (d) Now notice that the force it takes to separate one magnet from your refrigerator is nearly the same as the force it takes to separate two magnets stuck together. Estimate that force \(F\). (e) Estimate the area of a refrigerator magnet. (f) Assume that when these magnets are stuck together or to the refrigerator, they are separated by an effective distance \(d=25 \mu \mathrm{m}\). Use the formula derived above to estimate the magnetic field strength of the magnet.

A long, straight wire carries a 13.0 A current. An electron is fired parallel to this wire with a velocity of \(250 \mathrm{~km} / \mathrm{s}\) in the same direction as the current, \(2.00 \mathrm{~cm}\) from the wire. (a) Find the magnitude and direction of the electron's initial acceleration. (b) What should be the magnitude and direction of a uniform electric field that will allow the electron to continue to travel parallel to the wire? (c) Is it necessary to include the effects of gravity? Justify your answer.

A \(+6.00 \mu\) C point charge is moving at a constant \(8.00 \times 10^{6} \mathrm{~m} / \mathrm{s}\) in the \(+y\) -direction, relative to a reference frame. At the instant when the point charge is at the origin of this reference frame, what is the magnetic-field vector \(\overrightarrow{\boldsymbol{B}}\) it produces at the following points: (a) \(x=0.500 \mathrm{~m}, y=0, z=0\) (b) \(x=0, y=-0.500 \mathrm{~m}, z=0\) (c) \(x=0, \quad y=0, \quad z=+0.500 \mathrm{~m}\) (d) \(x=0, \quad y=-0.500 \mathrm{~m}\) \(z=+0.500 \mathrm{~m} ?\)

A solid conductor with radius \(a\) is supported by insulating disks on the axis of a conducting tube with inner radius \(b\) and outer radius \(c\) (Fig. E28.39). The central conductor and tube carry equal currents \(I\) in opposite directions. The currents are distributed uniformly over the cross sections of each conductor. Derive an expression for the magnitude of the magnetic field (a) at points outside the central, solid conductor but inside the tube \((ac)\)

Two concentric circular loops of wire lie on a tabletop, one inside the other. The inner wire has a diameter of \(20.0 \mathrm{~cm}\) and carries a clockwise current of \(12.0 \mathrm{~A}\), as viewed from above, and the outer wire has a diameter of \(30.0 \mathrm{~cm} .\) What must be the magnitude and direction (as viewed from above) of the current in the outer wire so that the net magnetic field due to this combination of wires is zero at the common center of the wires?
See all solutions

Recommended explanations on Physics Textbooks

Classical Mechanics

Read Explanation

Turning Points in Physics

Read Explanation

Fluids

Read Explanation

Medical Physics

Read Explanation

Geometrical and Physical Optics

Read Explanation

Linear Momentum

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.