/*! 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 48 The current in the windings of a... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 28: Problem 48

The current in the windings of a toroidal solenoid is \(2.400 \mathrm{~A}\). 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
The relative permeability is the value obtained from step 2 and the magnetic susceptibility is the value obtained from step 3.

Step by step solution

01

Calculating The Magnetic Field Inside a Toroidal Solenoid

The magnetic field inside a toroidal solenoid can be given by the formula, \(B=\frac{{\mu_0 \cdot I \cdot N}}{2\pi r}\), where \(B\) is the magnetic field, \(\mu_0\) is the permeability of free space, \(I\) is the current, \(N\) is the number of turns, and \(r\) is the radius. With given values, \(B = 1.940 T,\) \(I = 2.4 A,\) \(N = 500,\) and \(r = 25cm = 0.25m\). We can substitute these values into the formula.
02

Finding the Relative Permeability

After substituting the values in the formula, we get \(1.940=\frac{{\mu_0 \cdot 2.400 \cdot 500}}{2\pi \cdot 0.25}\), where \(\mu_0 = 4\pi \times 10^{-7}\). Solving this equation will give us the value of the relative permeability, \(\mu_r\). This result will be the value of \(\mu=\mu_0\mu_r\), from which we can get the relative permeability, \(\mu_r=\frac{\mu}{\mu_0}\).
03

Calculating Magnetic Susceptibility

Once we have the relative permeability, we can find the magnetic susceptibility by using the formula \(\chi = \mu_r -1\). By substituting the value of \(\mu_r\) from step 2 into this formula, we can find the magnetic susceptibility.

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.

Relative Permeability
Relative permeability is a measure of how easily a material can support the formation of a magnetic field within itself. It is a dimensionless quantity that represents the ratio of the permeability of a material (\(\mu\)) to the permeability of free space (\(\mu_0\)). In simpler terms, it tells us how much "better" or "worse" a medium is compared to a vacuum in terms of conducting magnetic flux. The permeability of free space (\(\mu_0\)) is a universal constant and is approximately equal to \(4\pi \times 10^{-7} \text{ T}\cdot\enspace\text{m/A}\).
To find relative permeability (\(\mu_r\)), we often use practical formulas based on experimental measurements. By using the provided formula for the magnetic field in a toroidal solenoid, students can substitute known values such as the magnetic field strength (\(1.940\\text{ T}\)) and rearrange the equations to solve for the material's contribution, arriving at a clear understanding of the material鈥檚 relative permeability.
Magnetic Susceptibility
Magnetic susceptibility (\(\chi\)) is a property that indicates the degree to which a material can be magnetized in an applied magnetic field. It is related closely to relative permeability. Whereas relative permeability is expressed as a ratio, magnetic susceptibility directly measures the degree of magnetization. Calculated using the formula \(\chi = \mu_r - 1\), magnetic susceptibility quantifies the magnetizing potential of a material after accounting for the influence of free space.
If the susceptibility is positive, the material can become magnetized when placed in an external magnetic field and is classified as paramagnetic. In contrast, if it is negative, the material is diamagnetic and will repel the magnetic field. Through calculation of susceptibility, students gain insight into the intrinsic properties of the magnetic material involved in the solenoid setup.
Toroidal Solenoid
A toroidal solenoid, or simply a toroid, is a solenoid bent into a circular shape to form a closed-loop. This spatial arrangement helps to confine the magnetic field within the loop, vastly reducing magnetic leakage outside the coil. Let鈥檚 break it down further.
The structure of a toroid consists of a core wrapped by several turns of wire. Its geometric setup and the number of turns determine the strength and directionality of the magnetic field within it. Given these attributes, toroids are highly efficient at maintaining strong magnetic fields. Students often encounter equations relating to toroids, such as the calculation of magnetic field strength using the formula \(B = \frac{{\mu I N}}{2\pi r}\), where the variables denote standard magnetic elements 鈥� including current (\(I\)), number of turns (\(N\)), and radius (\(r\)).
Understanding how a toroidal solenoid operates is key in mastering topics of electromagnetism, considering their previous widespread application in transformers, inductors, and magnetic storage devices.
Magnetic Material
Magnetic materials are those that respond noticeably to a magnetic field. They can be broadly classified into diamagnetic, paramagnetic, and ferromagnetic materials based on their magnetic susceptibility. The behavior of these materials in response to an external magnetic field is pivotal in creating and utilizing magnetic fields for technological applications.
In a toroidal solenoid, the inclusion of a magnetic material impacts the overall magnetic field strength, as these materials can enhance the field considerably if possessing high permeability. Students learn to observe this interplay through equations that simulate real-world conditions. For instance, in this exercise, the magnetic material within the toroid increases the magnetic field substantially compared to if the toroid were filled with air alone. - **Diamagnetic materials** slightly oppose the external field, with small, negative susceptibility. - **Paramagnetic materials** align with the field proportionally to their positive susceptibility. - **Ferromagnetic materials** strongly align with a field, exhibiting large, positive susceptibility. Understanding these aspects helps students grasp why certain materials are used in conjunction with solenoids to modify and control magnetic behavior.

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 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 \((ac)\)

The law of Biot and Savart in Eq. ( 28.7 ) generalizes to the case of surface currents as $$ \overrightarrow{\boldsymbol{B}}=\frac{\mu_{0}}{4 \pi} \int \frac{\sigma \overrightarrow{\boldsymbol{v}} \times \hat{\boldsymbol{r}}}{r^{2}} d a $$ where \(\sigma\) is the local charge density, \(\overrightarrow{\boldsymbol{v}}\) is the local velocity, and \(d a\) is a differential area element. Re-visit Challenge Problem 28.76 and use the above equation as an alternative means to derive the magnetic field at the center of the cylinder. Use the following steps: (a) Write the charge density \(\sigma\). (b) The origin is at the center of the cylinder. What is the vector \(\vec{v}\) that points from the element with coordinates \((x, y, z)=(x, R \cos \phi, R \sin \phi)\) to the origin? (c) What is the velocity \(\overrightarrow{\boldsymbol{v}}\) of the element? (d) What is the vector product \(\overrightarrow{\boldsymbol{v}} \times \hat{\boldsymbol{r}} ?\) (e) An area element on the cylinder may be written as \(d a=R d x d \phi .\) Use this and the previously established information to write the generalized law of Biot and Savart as a double integral. Evaluate the integral to determine the magnetic field \(\vec{B}\) at the center of the cylinder. (f) Is your result consistent with your result in Challenge Problem \(28.76 ?\)

A long, straight wire with a circular cross section of radius \(R\) carries a current \(I\). Assume that the current density is not constant across the cross section of the wire, but rather varies as \(J=\alpha r,\) where \(\alpha\) is a constant. (a) \(\mathrm{By}\) the requirement that \(J\) integrated over the cross section of the wire gives the total current \(I,\) calculate the constant \(\alpha\) in terms of \(I\) and \(R .\) (b) Use Ampere's law to calculate the magnetic field \(B(r)\) for (i) \(r \leq R\) and (ii) \(r \geq R .\) Express your answers in terms of \(I\).

The solenoid is removed from the enclosure and then used in a location where the earth's magnetic field is \(50 \mu \mathrm{T}\) and points horizontally. A sample of bacteria is placed in the center of the solenoid, and the same current is applied that produced a magnetic field of \(150 \mu \mathrm{T}\) in the lab. Describe the field experienced by the bacteria: The field (a) is still \(150 \mu \mathrm{T} ;\) (b) is now \(200 \mu \mathrm{T} ;\) (c) is between 100 and \(200 \mu \mathrm{T},\) depending on how the solenoid is oriented; (d) is between 50 and \(150 \mu \mathrm{T}\), depending on how the solenoid is oriented.

A very long, straight horizontal wire carries a current such that \(8.20 \times 10^{18}\) electrons per second pass any given point going from west to east. What are the magnitude and direction of the magnetic field this wire produces at a point \(4.00 \mathrm{~cm}\) directly above it?
See all solutions

Recommended explanations on Physics Textbooks

Space Physics

Read Explanation

Oscillations

Read Explanation

Solid State Physics

Read Explanation

Measurements

Read Explanation

Physical Quantities And Units

Read Explanation

Further Mechanics and Thermal 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.