Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 19: Problem 86

An electromagnet is made by inserting a soft iron core into a solenoid. The solenoid has 1800 turns, radius \(2.0 \mathrm{cm},\) and length $15 \mathrm{cm} .\( When \)2.0 \mathrm{A}$ of current flows through the solenoid, the magnetic field inside the iron core has magnitude 0.42 T. What is the relative permeability \(\overline{K_{\mathrm{B}}}\) of the iron core? (See Section 19.10 for the definition of \(\boldsymbol{K}_{\mathbf{B}} .\) )

Short Answer

Expert verified
Answer: The relative permeability of the iron core is 139.

Step by step solution

01

Write down the formula for the magnetic field inside a solenoid

The formula for the magnetic field (\(B\)) inside a solenoid with \(n\) turns per unit length carrying a current \(I\) is: \(B = μ_0nI\) where \(μ_0\) is the permeability of free space, which is \(4\pi × 10^{-7}\) Tm/A.
02

Identify the information given in the problem and compute the turns per unit length of the solenoid

The information given in the problem is: - Solenoid turns: 1800 - Solenoid radius: 2.0 cm - Solenoid length: 15 cm - Current through the solenoid: 2.0 A - Magnetic field inside the iron core: 0.42 T First, we need to find the number of turns per unit length (\(n\)). We do this by dividing the total number of turns (1800) by the length of the solenoid (15 cm or 0.15 m): \(n = \frac{1800}{0.15 \mathrm{m}} = 12000 \mathrm{m}^{-1}\)
03

Rearrange the formula and use it to find the permeability of the iron core

Now, we need to find the permeability of the iron core (\(μ\)), which is related to the product of the permeability of free space (\(μ_0\)) and the relative permeability of the iron core (\(\overline{K_B}\)): \(μ = μ_0\overline{K_B}\) Using the formula for the magnetic field inside a solenoid, we can get the permeability of the iron core by rearranging the formula as follows: \(μ = \frac{B}{nI}\) Then, we can plug in the values given in the problem: \(μ = \frac{0.42 \mathrm{T}}{12000 \mathrm{m}^{-1} \times 2.0 \mathrm{A}} = 1.75 × 10^{-5} \mathrm{Tm/A}\)
04

Calculate the relative permeability of the iron core

Now that we have the permeability of the iron core (\(μ\)), we can find the relative permeability \(𝑘_B\) by using the relation between \(μ\), \(μ_0\), and \(𝑘_B\): \(\overline{K_B} = \frac{μ}{μ_0}\) Plugging in the values of \(μ\) and \(μ_0\): \(\overline{K_B} = \frac{1.75 × 10^{-5} \mathrm{Tm/A}}{4\pi × 10^{-7} \mathrm{Tm/A}} = 139\)
05

Final Answer

The relative permeability of the iron core is 139.

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.

Start your free trial

Over 30 million students worldwide already upgrade their learning with Vaia!

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

Electrons in a television's CRT are accelerated from rest by an electric field through a potential difference of \(2.5 \mathrm{kV} .\) In contrast to an oscilloscope, where the electron beam is deflected by an electric ficld, the beam is deflected by a magnetic field. (a) What is the specd of the electrons? (b) The beam is deflected by a perpendicular magnetic field of magnitude $0.80 \mathrm{T}$. What is the magnitude of the acceleration of the electrons while in the field? (c) What is the speed of the electrons after they travel 4.0 mm through the magnetic field? (d) What strength electric field would give the electrons the same magnitude acceleration as in (b)? (c) Why do we have to use an clectric ficld in the first place to get the electrons up to speed? Why not use the large acceleration due to a magnetic field for that purpose?
A uniform magnetic field of \(0.50 \mathrm{T}\) is directed to the north. At some instant, a particle with charge \(+0.020 \mu \mathrm{C}\) is moving with velocity \(2.0 \mathrm{m} / \mathrm{s}\) in a direction \(30^{\circ}\) north of east. (a) What is the magnitude of the magnetic force on the charged particle? (b) What is the direction of the magnetic force?
A solenoid has 4850 turns per meter and radius \(3.3 \mathrm{cm}\) The magnetic field inside has magnitude 0.24 T. What is the current in the solenoid?
In a certain region of space, there is a uniform electric field \(\overrightarrow{\mathbf{E}}=3.0 \times 10^{4} \mathrm{V} / \mathrm{m}\) directed due east and a uniform magnetic field \(\mathbf{B}=0.080\) T also directed due east. What is the electromagnetic force on an electron moving due south at \(5.0 \times 10^{6} \mathrm{m} / \mathrm{s} ?\)
An electromagnetic flowmeter is to be used to measure blood speed. A magnetic field of \(0.115 \mathrm{T}\) is applied across an artery of inner diameter \(3.80 \mathrm{mm}\) The Hall voltage is measured to be \(88.0 \mu \mathrm{V} .\) What is the average speed of the blood flowing in the artery?
See all solutions

Recommended explanations on Physics Textbooks

Atoms and Radioactivity

Read Explanation

Mechanics and Materials

Read Explanation

Translational Dynamics

Read Explanation

Modern Physics

Read Explanation

Waves Physics

Read Explanation

Electromagnetism

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.

Sign-up for free