/*! 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 40 A magnetic field of \(1.30 \math... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

A magnetic field of \(1.30 \mathrm{T}\) is to be set up in an iron-core toroid. The toroid has a mean radius of \(10.0 \mathrm{cm},\) and magnetic permeability of \(5000 \mu_{0}\). What current is required if the winding has 470 turns of wire? The thickness of the iron ring is small compared to \(10 \mathrm{cm},\) so the field in the material is nearly uniform.

Short Answer

Expert verified
After carrying out the calculations in Step 4, the required current for the winding with 470 turns of wire should be approximately \(1.10 \mathrm{A}\).

Step by step solution

01

Identify the known values

From the problem, we know the magnetic field \(B = 1.30 \mathrm{T}\), the mean radius \(r = 10.0 \mathrm{cm} = 0.10 \mathrm{m}\), the number of turns \(N = 470\) and the magnetic permeability \( \mu = 5000 \mu_{0}\), where \( \mu_{0} = 4\pi \times 10^{-7} \mathrm{T m/A}\).
02

Use Ampere's Law

The magnetic field in a toroid of mean radius r, with N turns of wire carrying a current I is given by Ampere's Law. The formula is \(B = \frac{\mu_0 \mu_r N I }{2 \pi r}\), where \(B\) is the magnetic field, \(\mu_0\) is the permeability of free space, \(\mu_r\) is the relative permeability, \(N\) is the number of turns, \(I\) is the current, and \(r\) is the radius.
03

Rearrange the formula to solve for the required current I

Rearranging the formula from step 2, the current \(I\) can be given by \(I = \frac{B \times 2\pi r}{\mu_0 \mu_r N}\)
04

Substitute the values into the formula and compute

Substitute the known values from Step 1 into the rearranged formula from Step 3 and solve for current: \(I = \frac{1.30 \mathrm{T} \times 2\pi \times 0.10 \mathrm{m}}{4\pi \times 10^{-7} \mathrm{T m/A} \times 5000 \times 470}\)

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.

Ampere's Law
Ampere's Law is a fundamental principle that relates the magnetic field in space to the current that produces it. It is especially useful in scenarios involving symmetrical configurations, such as toroids. In a toroid, the magnetic field forms a loop inside the donut-shaped coil. This field is uniform inside the toroid and zero outside due to its symmetrical nature.

According to Ampere's Law, the magnetic field \(B\) within a toroid can be calculated through the equation:
  • \( B = \frac{\mu_0 \mu_r N I}{2 \pi r} \)
Here, \(\mu_0\) is the permeability of free space, \(\mu_r\) is the magnetic permeability, \(N\) is the number of turns, \(I\) is the current, and \(r\) is the mean radius of the toroid.
This formula highlights how the magnetic field depends on various factors. Changing the current or number of turns will affect the field strength proportionately. Understanding Ampere's Law is key to calculating magnetic fields in practical devices, including toroids.
Magnetic Permeability
Magnetic permeability is a property that characterizes how a material responds to a magnetic field. It is often symbolized as \(\mu\), with \(\mu_0\) representing the permeability of free space and \(\mu_r\) representing the relative permeability of the material. Together, they define the total permeability: \(\mu = \mu_0 \mu_r\).

In the context of an iron-core toroid, high magnetic permeability, such as \(5000 \mu_0\), means the material is highly effective in concentrating magnetic lines of force. This amplification allows stronger magnetic fields with less current.
  • Greater \(\mu_r\) increases the magnetic field's strength within the toroid.
  • The use of iron significantly raises \(\mu_r\) compared to air or vacuum (where \(\mu_r = 1\)).
By selecting materials with high \(\mu_r\), engineers can design efficient magnetic circuits, which is critical for many electromagnetic devices.
Current Calculation
Calculating the current required to generate a specified magnetic field in a toroid involves rearranging the formula derived from Ampere's Law. The goal is to solve for the current \(I\).

Re-arrangement of Ampere’s equation for current, \(I\):
  • \( I = \frac{B \times 2 \pi r}{\mu_0 \mu_r N} \)
Here’s how to compute it step-by-step:1. **Substitute the given values** into the formula:
  • \( B = 1.30 \) Tesla
  • \( r = 0.10 \) meters
  • \( N = 470 \) turns
  • \( \mu_0 = 4 \pi \times 10^{-7} \) T m/A
  • \( \mu_r = 5000 \)
2. **Calculate the terms** and rearrange: \( I = \frac{1.30 \times 2 \pi \times 0.10}{4 \pi \times 10^{-7} \times 5000 \times 470} \)By computing these values, you can find the current necessary to produce the desired magnetic field in the toroid. This step ensures you understand how variations in any factor (like coil turns \(N\) or radius \(r\)) can change the required current.

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

What current is required in the windings of a long solenoid that has 1000 turns uniformly distributed over a length of \(0.400 \mathrm{m},\) to produce at the center of the solenoid a magnetic field of magnitude \(1.00 \times 10^{-4} \mathrm{T} ?\)

A very large parallel-plate capacitor carries charge with uniform charge per unit area \(+\sigma\) on the upper plate and \(-\sigma\) on the lower plate. The plates are horizontal and both move horizontally with speed \(v\) to the right. (a) What is the magnetic field between the plates? (b) What is the magnetic field close to the plates but outside of the capacitor? (c) What is the magnitude and direction of the magnetic force per unit area on the upper plate? (d) At what extrapolated speed \(v\) will the magnetic force on a plate balance the electric force on the plate? Calculate this speed numerically.

The magnetic field \(40.0 \mathrm{cm}\) away from a long straight wire carrying current \(2.00 \mathrm{A}\) is \(1.00 \mu \mathrm{T}\). (a) At what distance is it \(0.100 \mu \mathrm{T} ?\) (b) What If? At one instant, the two conductors in a long household extension cord carry equal 2.00 -A currents in opposite directions. The two wires are \(3.00 \mathrm{mm}\) apart. Find the magnetic field \(40.0 \mathrm{cm}\) away from the middle of the straight cord, in the plane of the two wires. (c) At what distance is it one tenth as large? (d) The center wire in a coaxial cable carries current \(2.00 \mathrm{A}\) in one direction and the sheath around it carries current \(2.00 \mathrm{A}\) in the opposite direction. What magnetic field does the cable create at points outside?

A nonconducting ring of radius \(R\) is uniformly charged with a total positive charge \(q\). The ring rotates at a constant angular speed \(\omega\) about an axis through its center, perpendicular to the plane of the ring. What is the magnitude of the magnetic field on the axis of the ring a distance \(R / 2\) from its center?

A single-turn square loop of wire, \(2.00 \mathrm{cm}\) on each edge, carries a clockwise current of 0.200 A. The loop is inside a solenoid, with the plane of the loop perpendicular to the magnetic field of the solenoid. The solenoid has 30 turns/cm and carries a clockwise current of \(15.0 \mathrm{A}\) Find the force on each side of the loop and the torque acting on the loop.

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.