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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

A short coil with radius \(R=10.0 \mathrm{~cm}\) contains \(N=30.0\) turns and surrounds a long solenoid with radius \(r=8.00 \mathrm{~cm}\) containing \(n=60\) turns per centimeter. The current in the short coil is increased at a constant rate from zero to \(i=2.00 \mathrm{~A}\) in a time of \(t=12.0 \mathrm{~s}\). Calculate the induced potential difference in the long solenoid while the current is increasing in the short coil.

What is the resistance in an RL circuit with \(L=36.94 \mathrm{mH}\) if the time taken to reach \(75 \%\) of its maximum current value is \(2.56 \mathrm{~ms} ?\)

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 V. b) \(0.02 \mathrm{~V}\). c) 0.001 V. d) \(0.24 \mathrm{~V}\).

A circular coil of wire with 20 turns and a radius of \(40.0 \mathrm{~cm}\) is laying flat on a horizontal table as shown in the figure. There is a uniform magnetic field extending over the entire table with a magnitude of \(5.00 \mathrm{~T}\) and directed to the north and downward, making an angle of \(25.8^{\circ}\) with the horizontal. What is the magnitude of the magnetic flux through the coil?

Your friend decides to produce electrical power by turning a coil of \(1.00 \cdot 10^{5}\) circular loops of wire around an axis parallel to a diameter in the Earth's magnetic field, which has a local magnitude of \(0.300 \mathrm{G}\). The loops have a radius of \(25.0 \mathrm{~cm} .\) a) If your friend turns the coil at a frequency of \(150.0 \mathrm{~Hz}\) what peak current will flow in a resistor, \(R=1500 . \Omega\) connected to the coil? b) The average current flowing in the coil will be 0.7071 times the peak current. What will be the average power obtained from this device?
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