/*! 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 56 The current in the windings of a... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 28: Problem 56

The current in the windings of a toroidal solenoid is 2.400 A. There are 500 turns, and the mean radius is 25.00 \(\mathrm{cm} .\) The toroidal solenoid is filled with a magnetic material. The magnetic field inside the windings is found to be 1.940 T. Calculate (a) the relative permeability and (b) the magnetic susceptibility of the material that fills the toroid.

Short Answer

Expert verified
Relative permeability is approximately 2023.92, and magnetic susceptibility is 2022.92.

Step by step solution

01

Understanding the Relationship

For a toroidal solenoid, the magnetic field inside the windings filled with magnetic material is given by the equation: \( B = \mu H \), where \( B \) is the magnetic field, \( \mu \) is the permeability of the material, and \( H \) is the magnetic field intensity.
02

Calculating the Magnetic Field Intensity (H)

The magnetic field intensity \( H \) in a toroid is given by the formula: \(H = \frac{NI}{2\pi r}\), where \( N \) is the number of turns, \( I \) is the current, and \( r \) is the mean radius. Convert the mean radius to meters: \( r = 0.25 \) m. Then, substitute into the formula: \( H = \frac{500 \times 2.400}{2\pi \times 0.25} \approx 763.94 \text{ A/m} \).
03

Relating to Permeability of Free Space

The permeability of the material \( \mu \) can be expressed in terms of the permeability of free space \( \mu_0 \) and the relative permeability \( \mu_r \): \( \mu = \mu_0 \mu_r \). The permeability of free space \( \mu_0 \) is \( 4\pi \times 10^{-7} \text{ Tm/A} \).
04

Calculating Relative Permeability (mu_r)

Using the equation \( B = \mu H = \mu_0 \mu_r H \), solve for \( \mu_r \): \( \mu_r = \frac{B}{\mu_0 H} \). Substituting the given values, \( B = 1.940 \) T, and \( H = 763.94 \text{ A/m} \), we find \( \mu_r = \frac{1.940}{(4\pi \times 10^{-7}) \times 763.94} \approx 2023.92 \).
05

Calculating the Magnetic Susceptibility (chi_m)

The magnetic susceptibility \( \chi_m \) is related to relative permeability by the equation: \( \mu_r = 1 + \chi_m \). Therefore, \( \chi_m = \mu_r - 1 \). Substituting the calculated \( \mu_r = 2023.92 \), we find \( \chi_m = 2023.92 - 1 = 2022.92 \).

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.

Magnetic Susceptibility
Magnetic susceptibility is a key concept in electromagnetism that measures how much a material will become magnetized in an applied magnetic field.
This property is represented by the symbol \(\chi_m\) and is a dimensionless quantity.
The higher the magnetic susceptibility, the more a material can be magnetized.It directly relates to how materials respond to magnetic fields. Understanding susceptibility allows us to determine the degree of magnetization a substance can achieve under an external magnetic force.
Supposing a material has high magnetic susceptibility, it means it can easily be magnetized.To calculate magnetic susceptibility, we use the equation:
  • \(\chi_m = \mu_r - 1\)
where:
  • \(\mu_r\) is the relative permeability of the material.
If you find that a material has a magnetic susceptibility of \( 2022.92 \), as calculated, this indicates a significant ability to be magnetized, reflecting the material's effectiveness in enhancing magnetic field lines.
Toroidal Solenoid
A toroidal solenoid is a coil that is shaped into a torus, or a doughnut-like shape.
This design helps contain the magnetic field within the coil, making it extremely efficient in creating a uniform magnetic field inside the core.The magnetic field in a toroidal solenoid primarily depends on the following factors:
  • The number of turns \( N \) of the coil.
  • The current \( I \) flowing through the coil.
  • The mean radius \( r \) of the torus.
The calculation of the magnetic field intensity, \( H \), is essential for understanding the behavior inside a toroidal solenoid and is given by the formula:
  • \(H = \frac{NI}{2\pi r}\)
This formula helps to determine the strength of the magnetic field created within the toroid.With 500 turns, a current of 2.400 A, and a mean radius of 0.25 m, you can calculate that the magnetic field intensity \( H \) is approximately 763.94 A/m.
This high intensity field is indicative of the strong magnetic environment created within the toroid's core.
Magnetic Field Intensity
Magnetic field intensity, often denoted as \( H \), is a measure of the magnetizing force in a material. It represents the level of external magnetic field applied to a material in order to establish a magnetic field within it.In calculations, the magnetic field intensity is expressed as:
  • \(H = \frac{NI}{2\pi r}\)
where:
  • \(N\) is the number of coil turns.
  • \(I\) is the current in amperes.
  • \(r\) is the mean radius of the coil.
This quantification helps us understand how effectively the coil can concentrate the magnetic field within its structure.Magnetic field intensity is crucial in applications like transformers and inductors, where controlling magnetic fields is vital.
In a toroidal solenoid, it plays a central role in strengthening the core's magnetic properties, enabling efficient energy transfer and storage.

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 closed curve encircles several conductors. The line integral \(\oint \vec{B} \cdot d \vec{l}\) around this curve is \(3.83 \times 10^{-4} \mathrm{T} \cdot \mathrm{m}\) . (a) What is the net current in the conductors? (b) If you were to integrate around the curve in the opposite direction, what would be the value of the line integral? Explain.

A conductor is made in the form of a hollow cylinder with inner and outer radii \(a\) and \(b\) respectively. It carries a current \(I\) uniformly distributed over its cross section. Derive expressions for the magnitude of the magnetic field in the regions (a) \(r < a ;\) (b) \(a < r < b ;(c) r > b\) .

The Magnetic Field from a Lightning Bolt. Lightning bolts can carry currents up to approximately 20 kA. We can model such a current as the equivalent of a very long, straight wire. (a) If you were unfortunate enough to be 5.0 m away from such a lightning bolt, how large a magnetic field would you experience? (b) How does this field compare to one you would experience by being 5.0 \(\mathrm{cm}\) from a long, straight household current of 10 A ?

A toroidal solenoid with 400 turns of wire and a mean radius of 6.0 \(\mathrm{cm}\) carries a current of 0.25 A. The relative permeability of the core is \(80 .\) (a) What is the magnetic field in the core? (b) What part of the magnetic field is due to atomic currents?

Two long, straight conducting wires with linear mass density \(\lambda\) are suspended from cords so that they are each horizontal, parallel to each other, and a distance \(d\) apart. The back ends of the wires are connected to each other by a slack, low-resistance connecting wire. A charged capacitor (capacitance \(C )\) is now added to the system; the positive plate of the capacitor (initial charge \(+Q_{0}\) ) is connected to the front end of one of the wires, and the negative plate of the capacitor (initial charge \(-Q_{0} )\) is connected to the front end of the other wire (Fig. P28.87). Both of these connections are also made by slack, low-resistance wires. When the connection is made, the wires are pushed aside by the repulsive force between the wires, and each wire has an initial horizontal velocity of magnitude \(v_{0} .\) Assume that the time constant for the capacitor to discharge is negligible compared to the time it takes for any appreciable displacement in the position of the wires to occur. (a) Show that the initial speed \(v_{0}\) of either wire is given by $$ v_{0}=\frac{\mu_{0} Q_{0}^{2}}{4 \pi \lambda R C d} $$ where \(R\) is the total resistance of the circuit. (b) To what height \(h\) will each wire rise as a result of the circuit connection?
See all solutions

Recommended explanations on Physics Textbooks

Scientific Method Physics

Read Explanation

Turning Points in Physics

Read Explanation

Electromagnetism

Read Explanation

Linear Momentum

Read Explanation

Medical Physics

Read Explanation

Electric Charge Field and Potential

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.