/*! 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 28 A long solenoid with cross-secti... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 29: Problem 28

A long solenoid with cross-sectional area \(A_{1}\) surrounds another long solenoid with cross-sectional area \(A_{2}

Short Answer

Expert verified
Answer: The expression for the magnetic field in the inner solenoid due to the induced current from the outer solenoid is \(B_2 = \frac{\mu_0^2 n^2 A_2 i_0 \omega \sin (\omega t)}{R}\), where \(B_2\) is the magnetic field in the inner solenoid, \(\mu_0\) is the permeability of free space, \(n\) is the number of turns per unit length, \(A_2\) is the cross-sectional area of the inner solenoid, \(i_0\) is the amplitude of the current in the outer solenoid, \(\omega\) is the angular frequency, \(t\) is time, and \(R\) is the resistance of the inner solenoid.

Step by step solution

01

Find the magnetic field produced by the outer solenoid

The current in the outer solenoid is given by \(i=i_0 \cos{\omega t}\). Using Ampere's Law, the magnetic field produced by the outer solenoid can be calculated as follows: \(B_1 = \mu_0 n i_0 \cos (\omega t)\) Here, \(B_1\) represents the magnetic field produced by the outer solenoid, \(\mu_0\) is the permeability of free space, \(n\) is the number of turns per unit length, and \(i_0\) is the amplitude of the current.
02

Find the induced EMF in the inner solenoid

To find the induced EMF in the inner solenoid, we will use Faraday's Law, which states that the induced EMF is equal to the negative rate of change of the magnetic flux through the solenoid: \(EMF = -\frac{d\Phi}{dt}\) Since the inner solenoid is surrounded by the outer solenoid, the magnetic flux through the inner solenoid is equal to the product of the magnetic field produced by the outer solenoid and the cross-sectional area of the inner solenoid: \(\Phi = B_1 A_2\) Now, taking the derivative with respect to time: \(EMF = -\frac{d}{dt} (B_1 A_2) = - A_2 \frac{d}{dt} (\mu_0 n i_0 \cos (\omega t))\) \(EMF = A_2 \mu_0 n i_0 \omega \sin (\omega t)\)
03

Find the induced current in the inner solenoid

The induced current in the inner solenoid is related to the induced EMF and the resistance of the inner solenoid using Ohm's Law: \(I_2 = \frac{EMF}{R}\) Substituting the previously calculated EMF value: \(I_2 = \frac{A_2 \mu_0 n i_0 \omega \sin (\omega t)}{R}\)
04

Find the magnetic field in the inner solenoid due to the induced current

Now that we have found the induced current in the inner solenoid, we can find the magnetic field in the inner solenoid due to the induced current using Ampere's Law: \(B_2 = \mu_0 n I_2\) Substituting the value for \(I_2\): \(B_2 = \mu_0 n \frac{A_2 \mu_0 n i_0 \omega \sin (\omega t)}{R}\) Simplifying, we get the final expression for the magnetic field in the inner solenoid due to the induced current: \(B_2 = \frac{\mu_0^2 n^2 A_2 i_0 \omega \sin (\omega t)}{R}\)

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!

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

Faraday's Law of Induction states a) that a potential difference is induced in a loop when there is a change in the magnetic flux through the loop. b) that the current induced in a loop by a changing magnetic field produces a magnetic field that opposes this change in magnetic field. c) that a changing magnetic field induces an electric field. d) that the inductance of a device is a measure of its opposition to changes in current flowing through it. e) that magnetic flux is the product of the average magnetic field and the area perpendicular to it that it penetrates.

A 100 -turn solenoid of length \(8 \mathrm{~cm}\) and radius \(6 \mathrm{~mm}\) carries a current of 0.4 A from right to left. The current is then reversed so that it flows from left to right. By how much does the energy stored in the magnetic field inside the solenoid change?

A rectangular wire loop (dimensions of \(h=15.0 \mathrm{~cm}\) and \(w=8.00 \mathrm{~cm}\) ) with resistance \(R=5.00 \Omega\) is mounted on a door. The Earth's magnetic field, \(B_{\mathrm{E}}=2.6 \cdot 10^{-5} \mathrm{~T}\), is uniform and perpendicular to the surface of the closed door (the surface is in the \(x z\) -plane). At time \(t=0,\) the door is opened (right edge moves toward the \(y\) -axis) at a constant rate, with an opening angle of \(\theta(t)=\omega t,\) where \(\omega=3.5 \mathrm{rad} / \mathrm{s}\) Calculate the direction and the magnitude of the current induced in the loop, \(i(t=0.200 \mathrm{~s})\).

A rectangular conducting loop with dimensions \(a\) and \(b\) and resistance \(R\), is placed in the \(x y\) -plane. A magnetic field of magnitude \(B\) passes through the loop. The magnetic field is in the positive \(z\) -direction and varies in time according to \(B=B_{0}\left(1+c_{1} t^{3}\right),\) where \(c_{1}\) is a constant with units of \(1 / \mathrm{s}^{3}\) What is the direction of the current induced in the loop, and what is its value at \(t=1 \mathrm{~s}\) (in terms of \(a, b, R, B_{0},\) and \(\left.c_{1}\right) ?\)

A circuit contains a 12.0 -V battery, a switch, and a light bulb connected in series. When the light bulb has a current of 0.100 A flowing in it, it just starts to glow. This bulb draws \(2.00 \mathrm{~W}\) when the switch has been closed for a long time. The switch is opened, and an inductor is added to the circuit, in series with the bulb. If the light bulb begins to glow \(3.50 \mathrm{~ms}\) after the switch is closed again, what is the magnitude of the inductance? Ignore any time to heat the filament, and assume that you are able to observe a glow as soon as the current in the filament reaches the 0.100 - A threshold.
See all solutions

Recommended explanations on Physics Textbooks

Magnetism

Read Explanation

Solid State Physics

Read Explanation

Further Mechanics and Thermal Physics

Read Explanation

Translational Dynamics

Read Explanation

Electricity and Magnetism

Read Explanation

Fields in 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.