/*! 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 toroidal solenoid with 400 tur... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

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?

Short Answer

Expert verified
(a) 0.167 T; (b) 98.2% is due to atomic currents.

Step by step solution

01

Understand the Toroidal Solenoid Setup

A toroidal solenoid is a coil of wire in the shape of a donut. It has 400 turns, a mean radius of 6.0 cm, carries a current of 0.25 A, and a core with a relative permeability of 80. The goal is first to find the magnetic field inside the core.
02

Use the Formula for the Magnetic Field in a Toroid

The formula for the magnetic field inside a toroid is \( B = \frac{\mu N I}{2\pi r} \), where \( \mu \) is the permeability of the core, \( N \) is the number of turns, \( I \) is the current, and \( r \) is the mean radius of the toroid.
03

Calculate Permeability of the Core

The permeability of the core \( \mu \) is given by \( \mu = \mu_0 \times \mu_r \), where \( \mu_0 \) is the permeability of free space (approximately \( 4\pi \times 10^{-7} \) Tm/A), and \( \mu_r \) is the relative permeability (80).
04

Substitute Values into the Magnetic Field Formula

Substitute \( \mu_0 = 4\pi \times 10^{-7} \) Tm/A, \( \mu_r = 80 \), \( N = 400 \), \( I = 0.25 \) A, and \( r = 0.06 \) m into the formula:\[ B = \frac{(4\pi \times 10^{-7}) \times 80 \times 400 \times 0.25}{2\pi \times 0.06} \]
05

Simplify and Calculate the Magnetic Field

Simplify and calculate the expression:\[ B = \frac{(4 \times 80 \times 400 \times 0.25)}{2 \times 0.06} \times 10^{-7} \approx 0.167 \text{ T} \]Thus, the magnetic field in the core is approximately 0.167 T.
06

Determine the Contribution from Atomic Currents

The total magnetic field is due to both the current flowing in the wire and atomic currents in the core. The contribution from atomic currents (magnetization) can be found by:\[ B_m = (\mu - \mu_0) \frac{N I}{2\pi r} \]Substitute the values:\[ B_m = ((4\pi \times 10^{-7} \times 80) - (4\pi \times 10^{-7})) \frac{400 \times 0.25}{2\pi \times 0.06} \]Solve to find \( B_m \approx 0.164 \text{ T} \).
07

Calculate the Fraction Due to Atomic Currents

The fraction of the magnetic field due to atomic currents is the ratio \( \frac{B_m}{B} \). Compute:\[ \frac{0.164}{0.167} \approx 0.982 \]Thus, about 98.2% of the magnetic field is due to atomic currents.

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 91Ó°ÊÓ!

Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Magnetic Field Calculation
Calculating the magnetic field inside a toroidal solenoid involves understanding its unique characteristics. A toroidal solenoid is shaped like a donut, providing a closed-loop path for the magnetic flux. This configuration ensures that the magnetic field inside is uniform and effectively contained within the core.

The formula used to find the magnetic field (B) in a toroid is \( B = \frac{\mu N I}{2\pi r} \), where:
  • \( \mu \) = permeability of the core
  • \( N \) = number of turns in the coil
  • \( I \) = the current flowing through the coil
  • \( r \) = mean radius of the toroid
To solve for \( B \), you substitute these values into the formula, ensuring the units are consistent. This formula links the physical makeup of the solenoid to the magnetic field it creates, allowing precise calculations.
Relative Permeability
Relative permeability \( \mu_r \) is a measure of how much more or less a material responds to magnetization compared to a vacuum. It describes the material's ability to permit the passage of a magnetic field relative to free space. In simple terms, whether a material enhances or weakens an external magnetic field.

For magnetic field calculations in core materials, it's essential to consider this factor. In a toroidal solenoid with a core, such as our exercise, the relative permeability greatly influences the magnetic field strength. The higher the relative permeability, the more the core will contribute to the magnetic field.

The formula for the core permeability is: \( \mu = \mu_0 \times \mu_r \), where \( \mu_0 \) is the permeability of free space, and \( \mu_r \) is given in the exercise, in this case, 80. This depiction allows calculation of the effective permeability, taking into account the material's enhancement over free space.
Atomic Currents Contribution
Atomic currents arise from the alignment of atoms in a material with an external magnetic field. These are minuscule loops of current created by electron spins and orbital motions in atoms.

In a toroidal solenoid, part of the magnetic field is contributed by these atomic currents. They result from the relative permeability of the core material, adding to the field generated by the actual electric current in the coils. This boost from atomic currents is expressed mathematically as:
  • Field due to atomic currents, \( B_m = (\mu - \mu_0) \frac{N I}{2\pi r} \)
By this formula, you can calculate how much of the magnetic field is enhanced by these atomic movements, leading to a significant fraction in total field strength. Understanding this helps explain why the total magnetic field is much larger than it would be in the absence of a core material.
Permeability of Free Space
The permeability of free space, denoted as \( \mu_0 \), is a fundamental physical constant expressing how magnetic fields interact in a vacuum. This constant is approximately \( 4\pi \times 10^{-7} \; \text{Tm/A} \).

It serves as a baseline to measure how much different materials alter magnetic fields compared to emptiness. In calculations involving magnetic fields, \( \mu_0 \) forms the foundation upon which the permeability of actual materials (with their respective relative permeabilities) are built and normalized.

For instance, in our toroidal solenoid exercise, \( \mu_0 \) is used to calculate the effective permeability of the core material with \( \mu = \mu_0 \times \mu_r \). This application reveals the magnetic enhancement due to the core material, highlighting its role as a foundational aspect in the study of electromagnetism.

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 \(-4.80-\mu \mathrm{C}\) charge is moving at a constant speed of \(6.80 \times 10^{5} \mathrm{m} / \mathrm{s}\) in the \(+x\) -direction relative to a reference frame. At the instant when the point charge is at the origin, what is the magnetic-field vector it produces at the following points: (a) \(x=0.500 \mathrm{m}, y=0, z=0 ;\) (b) \(x=0\) \(y=0.500 \mathrm{m}, \quad z=0 ; \quad\) (c) \(x=0.500 \mathrm{m}, \quad y=0.500 \mathrm{m}, \quad z=0\) (d) \(x=0, y=0, z=0.500 \mathrm{m} ?\)

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.

Two long, parallel wires are separated by a distance of 2.50 \(\mathrm{cm} .\) The force per unit length that each wire exerts on the other is \(4.00 \times 10^{-5} \mathrm{N} / \mathrm{m},\) and the wires repel each other. The current in one wire is 0.600 A. (a) What is the current in the second wire? (b) Are the two currents in the same direction or in opposite directions?

BIO EMF. Currents in de transmission lines can be 100 A or more. Some people have expressed concern that the electromagnetic fields (EMFs) from such lines near their homes could cause health dangers. For a line with current 150 \(\mathrm{A}\) and at a height of 8.0 \(\mathrm{m}\) above the ground, what magnetic field does the line produce at ground level? Express your answer in teslas and as a percent of the earth's magnetic field, which is 0.50 gauss. Does this seem to be cause for worry?

A toroidal solenoid with 500 turns is wound on a ring with a mean radius of 2.90 \(\mathrm{cm} .\) Find the current in the winding that is required to set up a magnetic field of 0.350 T in the ring (a) if the ring is made of annealed iron \(\left(K_{\mathrm{m}}=1400\right)\) and \((\mathrm{b})\) if the ring is made of silicon steel \(\left(K_{\mathrm{m}}=5200\right)\)

See all solutions

Recommended explanations on Physics Textbooks

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.