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Chapter 29: Problem 2

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) \(0.24 \mathrm{~V}\).

Short Answer

Expert verified
Answer: The magnitude of the induced potential difference in the solenoid is 0.24 V.

Step by step solution

01

Write down Faraday's law of electromagnetic induction

Faraday's law states that the induced electromotive force (EMF) in any closed circuit is equal to the rate of change of the magnetic flux through the circuit. The formula for Faraday's law is: $$ \varepsilon = -N \frac{\Delta \Phi}{\Delta t} $$ Where \(\varepsilon\) is the induced EMF, \(N\) is the number of turns, \(\Delta \Phi\) is the change in magnetic flux, and \(\Delta t\) is the time in which the change occurs.
02

Calculate the change in magnetic flux

The magnetic flux, \(\Phi\), through the solenoid is given by the product of the magnetic field, \(B\), and the cross-sectional area, \(A\), of the solenoid. So, the change in magnetic flux can be written as: $$ \Delta \Phi = A \Delta B $$ We are given that \(\Delta B = 0.20 \mathrm{~T} / \mathrm{s}\), and the cross-sectional area \(A = 60 \mathrm{~cm}^{2}\). We need to convert the area to \(\mathrm{m}^{2}\) (1 cm = 0.01 m): $$ A = 60 \mathrm{~cm}^{2} \times (0.01)^2 = 0.006 \mathrm{~m}^{2} $$ Now, we can calculate the change in magnetic flux: $$ \Delta \Phi = 0.006 \mathrm{~m}^{2} \cdot 0.20 \mathrm{~T} = 0.0012 \mathrm{~T} \cdot \mathrm{m}^{2} $$
03

Calculate the induced EMF

Now, we can plug the values into Faraday's law and solve for the induced EMF: $$ \varepsilon = -N \frac{\Delta \Phi}{\Delta t} = -200 \cdot \frac{0.0012 \mathrm{~T} \cdot \mathrm{m}^{2}}{1 \mathrm{s}} $$ Keep in mind that the minus sign in Faraday's law indicates the direction (polarity) of the induced EMF, but we only need its magnitude. So, we can take the absolute value of the expression: $$ |\varepsilon| = 200 \cdot 0.0012 \mathrm{~T} \cdot \mathrm{m}^{2} \cdot \mathrm{s}^{-1} = 0.24 \mathrm{~V} $$
04

Choose the correct answer

The magnitude of the induced potential difference in the solenoid is \(0.24 \mathrm{~V}\), which matches answer choice (d). So, the correct answer is: (d) \(0.24 \mathrm{~V}\).

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Most popular questions from this chapter

A \(100 .-V\) battery is connected in series with a \(500 .-\Omega\) resistor. According to Faraday's Law of Induction, current can never change instantaneously, so there is always some "stray" inductance. Suppose the stray inductance is \(0.200 \mu \mathrm{H}\). How long will it take the current to build up to within \(0.500 \%\) of its final value of \(0.200 \mathrm{~A}\) after the resistor is con- nected to the battery?

An electromagnetic wave propagating in vacuum has electric and magnetic fields given by \(\vec{E}(\vec{x}, t)=\vec{E}_{0} \cos (\vec{k} \cdot \vec{x}-\omega t)\) and \(\vec{B}(\vec{x}, t)=\vec{B}_{0} \cos (\vec{k} \cdot \vec{x}-\omega t)\) where \(\vec{B}_{0}\) is given by \(\vec{B}_{0}=\vec{k} \times \vec{E}_{0} / \omega\) and the wave vector \(\vec{k}\) is perpendicular to both \(\vec{E}_{0}\) and \(\vec{B}_{0} .\) The magnitude of \(\vec{k}\) and the angular frequency \(\omega\) satisfy the dispersion relation, \(\omega /|\vec{k}|=\left(\mu_{0} \epsilon_{0}\right)^{-1 / 2},\) where \(\mu_{0}\) and \(\epsilon_{0}\) are the permeability and permittivity of free space, respectively. Such a wave transports energy in both its electric and magnetic fields. Calculate the ratio of the energy densities of the magnetic and electric fields, \(u_{B} / u_{E}\), in this wave. Simplify your final answer as much as possible.

A metal loop has an area of \(0.100 \mathrm{~m}^{2}\) and is placed flat on the ground. There is a uniform magnetic field pointing due west, as shown in the figure. This magnetic field initially has a magnitude of \(0.123 \mathrm{~T}\), which decreases steadily to \(0.075 \mathrm{~T}\) during a period of \(0.579 \mathrm{~s}\). Find the potential difference induced in the loop during this time.

An emf of \(20.0 \mathrm{~V}\) is applied to a coil with an inductance of \(40.0 \mathrm{mH}\) and a resistance of \(0.500 \Omega\). a) Determine the energy stored in the magnetic field when the current reaches one fourth of its maximum value. b) How long does it take for the current to reach this value?

A wedding ring (of diameter \(2.0 \mathrm{~cm}\) ) is tossed into the air and given a spin, resulting in an angular velocity of 17 rotations per second. The rotation axis is a diameter of the ring. Taking the magnitude of the Earth's magnetic field to be \(4.0 \cdot 10^{-5} \mathrm{~T}\), what is the maximum induced potential difference in the ring?
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