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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: \(B_2 = \frac{\mu_0^2 n^2 A_2 i_0 \omega \sin (\omega t)}{R}\)

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Most popular questions from this chapter

Which of the following statements regarding self induction is correct? a) Self-induction occurs only when a direct current is flowing through a circuit. b) Self-induction occurs only when an alternating current is flowing through a circuit. c) Self-induction occurs when either a direct current or an alternating current is flowing through a circuit. d) Self-induction occurs when either a direct current or an alternating current is flowing through a circuit as long as the current is varying.

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 solenoid with 200 turns and a cross-sectional area of \(60 \mathrm{~cm}^{2}\) has a magnetic field of \(0.60 \mathrm{~T}\) along its axis. If the field is confined within the solenoid and changes at a rate of \(0.20 \mathrm{~T} / \mathrm{s}\), the magnitude of the induced potential difference in the solenoid will be a) \(0.0020 \mathrm{~V}\). b) \(0.02 \mathrm{~V}\). c) \(0.001 \mathrm{~V}\). d) \(0.24 \mathrm{~V}\).

A solid metal disk of radius \(R\) is rotating around its center axis at a constant angular speed of \(\omega .\) The disk is in a uniform magnetic field of magnitude \(B\) that is oriented normal to the surface of the disk. Calculate the magnitude of the potential difference between the center of the disk and the outside edge.

People with pacemakers or other mechanical devices as implants are often warned to stay away from large machinery or motors. Why?
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