/*! 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 34 A straight, vertical wire carrie... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 27: Problem 34

A straight, vertical wire carries a current of 2.60 A down- ward in a region between the poles of a large superconducting electromagnet, where the magnetic field has magnitude \(B=0.588 \mathrm{~T}\) and is horizontal. What are the magnitude and direction of the magnetic force on a \(1.00 \mathrm{~cm}\) section of the wire that is in this uniform magnetic field, if the magnetic field direction is (a) east; (b) south; (c) \(30.0^{\circ}\) south of west?

Short Answer

Expert verified
The magnitude of the force on the wire when the magnetic field is east is 0.0153 N towards the north, 0 when it is towards the south, and 0.0132 N north of east when it is \(30.0^{\circ}\) south of west.

Step by step solution

01

Identify given quantities

The current \(I\) is 2.60 A, the magnetic field strength \(B\) is 0.588 T, and the length of the wire \(L\) is 1.00 cm or 0.01 m (remember we should always convert cm to m).
02

Calculate the force for each given direction

We know from the formula \(F=BIL\sin(\theta)\) that the force is directly proportional to the sine of the angle between the direction of the current and the magnetic field direction. Since the current is downward, for (a) \(F=BIL\sin(90)\) because the magnetic field direction is east (perpendicular to the current), (b) \(F=BIL\sin(0)\) because the magnetic field is south (same direction as current). And for (c) since the current is downwards (south) and the magnetic field is \(30.0^{\circ}\) south of west, therefore the angle between them would be \(90+30=120^{\circ}\), so \(F=BIL\sin(120)\).
03

Compute the magnetic forces

Substitute the given values into the formula to compute the force in each case. For (a), the magnitude of the force \(F\) would be equal to (0.588 T) * (2.60 A) * (0.01 m) * \(\sin(90^{\circ}) = 0.0153 N\). Also, the direction of force would be to the north due to the right-hand rule. For (b), since the angle is 0^{\circ}, the force is 0. And for (c), the force will be (0.588 T) * (2.60 A) * (0.01 m) * \(\sin(120^{\circ}) = 0.0132 N\), and the direction will be north of east due to the right-hand rule.

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.

Current-Carrying Wire
A current-carrying wire is one through which electric current flows. In this context, the wire carries a current of 2.60 A. The wire's orientation is vertical, meaning the current flows downward through it. Understanding current-carrying wires is crucial because they generate magnetic fields around them. This happens as electric charges move through the wire.
This scenario is a practical example of how this interaction plays out: a wire in the presence of a magnetic field experiences a force. This force's strength and direction depend on factors like:
  • The amount of current flowing through the wire
  • The length of the wire
  • The angle between the magnetic field lines and the wire
To solve problems regarding forces on a wire, we often use the formula for magnetic force: \[ F = BIL \sin(\theta) \] where \( F \) is the magnetic force, \( B \) the magnetic field strength, \( I \) the current, \( L \) the wire length, and \( \theta \) the angle between the current direction and the magnetic field lines.
Magnetic Field
Magnetic fields are invisible fields surrounding magnets or electric currents. In the exercise, the wire is placed in a magnetic field of 0.588 T (teslas), which is quite strong. A magnetic field exerts forces on moving charged particles, including those in current-carrying wires.
The magnetic field affects how the wire experiences forces when carrying an electric current. The force鈥檚 direction and magnitude can change based on the angle of the field relative to the wire. Causes of magnetic fields involve:
  • Permanent magnets that have a constant magnetic field
  • Electric currents creating a magnetic field around the conductor
  • Changing electric fields generating magnetic fields
Understanding magnetic fields involves recognizing the directions in which they act, which is often mapped using field lines. These lines indicate the path a north pole would take if placed in the field, flowing from north to south poles.
Right-Hand Rule
The right-hand rule is a mnemonic for determining the direction of the magnetic force on a current-carrying conductor in a magnetic field. It's a simple yet powerful tool used in electromagnetism. When faced with problems involving forces on a wire, the right-hand rule aids in visualizing the force direction.
To apply the right-hand rule:
  • Point your thumb in the direction of the current.
  • Position your fingers such that they point in the direction of the magnetic field.
  • Your palm or the force's resultant direction points perpendicular to both.
In the exercise, when calculating forces, this rule helps confirm the force directions: for instance, northward or northeastward, depending on the specific arrangement of current and magnetic field directions. It becomes crucial when figuring out whether the force is towards or away from specific points of the compass.
Electromagnetism
Electromagnetism is the branch of physics that studies electromagnetic forces. It's one of the fundamental forces of nature and describes interactions between electric charges and magnetic fields. When a wire carries a current, it interacts with nearby magnetic fields, demonstrating electromagnetic principles.
The scenarios described in the problem highlight concepts of electromagnetism such as:
  • The production of magnetic fields by electric currents
  • The interaction between currents and external magnetic fields
  • How this interaction results in forces acting on the current-carrying wire
Electromagnetic forces are calculated using the formula \( F = BIL\sin(\theta) \), which combines concepts of electric currents (from moving charges), magnetic fields (produced by these currents), and resulting forces. Such interactions form the basis for numerous technologies, including electric motors and transformers, where controlling electromagnetic forces is vital.

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

You measure the charge-to-mass ratio \(q / m\) for a particle with positive charge in the following way: The particle starts from rest, is accelerated through a potential difference \(\Delta V,\) and attains a velocity with magnitude \(v .\) It then enters a region of uniform magnetic field \(B=0.200 \mathrm{~T}\) that is directed perpendicular to the velocity; the particle moves in a path that is an arc of a circle of radius \(R .\) You measure \(R\) as a function of \(\Delta V\). You plot your data as \(R^{2}\) (in units of \(\mathrm{m}^{2}\) ) versus \(\Delta V\) (in \(\mathrm{V}\) ) and find that the values lie close to a straight line that has slope \(1.04 \times 10^{-6} \mathrm{~m}^{2} / \mathrm{V} .\) What is the value of \(q / m\) for this particle?

A uniform bar of length \(L\) carries a current \(I\) in the direction from point \(a\) to point \(b\) (Fig. \(\mathbf{P 2 7 . 6 6}\) ). The bar is in a uniform magnetic field that is directed into the page. Consider the torque about an axis perpendicular to the bar at point \(a\) that is due to the force that the magnetic field exerts on the bar. (a) Suppose that an infinitesimal section of the bar has length \(d x\) and is located a distance \(x\) from point \(a\). Calculate the torque \(d \tau\) about point \(a\) due to the magnetic force on this infinitesimal section. (b) Use \(\tau=\int_{a}^{b} d \tau\) to calculate the total torque \(\tau\) on the bar. (c) Show that \(\tau\) is the same as though all of the magnetic force acted at the midpoint of the bar.

Magnetic Moment of the Hydrogen Atom. In the Bohr model of the hydrogen atom (see Section 39.3 ), in the lowest energy state the electron orbits the proton at a speed of \(2.2 \times 10^{6} \mathrm{~m} / \mathrm{s}\) in a circular orbit of radius \(5.3 \times 10^{-11} \mathrm{~m}\). (a) What is the orbital period of the electron? (b) If the orbiting electron is considered to be a current loop, what is the current \(I ?\) (c) What is the magnetic moment of the atom due to the motion of the electron?

A horizontal rectangular surface has dimensions \(2.80 \mathrm{~cm}\) by \(3.20 \mathrm{~cm}\) and is in a uniform magnetic field that is directed at an angle of \(30.0^{\circ}\) above the horizontal. What must the magnitude of the magnetic field be to produce a flux of \(3.10 \times 10^{-4} \mathrm{~Wb}\) through the surface?

Fusion Reactor. If two deuterium nuclei (charge \(+e,\) mass \(3.34 \times 10^{-27} \mathrm{~kg}\) ) get close enough together, the attraction of the strong nuclear force will fuse them to make an isotope of helium, releasing vast amounts of energy. The range of this force is about \(10^{-15} \mathrm{~m} .\) This is the principle behind the fusion reactor. The deuterium nuclei are moving much too fast to be contained by physical walls, so they are confined magnetically. (a) How fast would two nuclei have to move so that in a head-on collision they would get close enough to fuse? (Assume their speeds are equal. Treat the nuclei as point charges, and assume that a separation of \(1.0 \times 10^{-15} \mathrm{~m}\) is required for fusion. \()\) (b) What strength magnetic field is needed to make deuterium nuclei with this speed travel in a circle of diameter \(2.50 \mathrm{~m} ?\)
See all solutions

Recommended explanations on Physics Textbooks

Famous Physicists

Read Explanation

Further Mechanics and Thermal Physics

Read Explanation

Fields in Physics

Read Explanation

Fluids

Read Explanation

Astrophysics

Read Explanation

Quantum Physics

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.