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Chapter 31: Problem 1

A 50 -turn rectangular coil of dimensions \(5.00 \mathrm{cm} \times\) 10.0 \(\mathrm{cm}\) is allowed to fall from a position where \(B=0\) to a new position where \(B=0.500 \mathrm{T}\) and the magnetic field is directed perpendicular to the plane of the coil. Calculate the magnitude of the average emf that is induced in the coil if the displacement occurs in 0.250 \(\mathrm{s}\) .

Short Answer

Expert verified
The magnitude of the average induced emf is 0.500 T.

Step by step solution

01

Understand the concept of electromagnetic induction

Faraday's law of electromagnetic induction states that the induced electromotive force (emf) in a coil is proportional to the rate of change of magnetic flux through the coil. The formula for the average emf induced in a coil is given by \( \epsilon = -\frac{\Delta \Phi}{\Delta t} \), where \( \epsilon \) is the induced emf, \( \Delta \Phi \) is the change in magnetic flux, and \( \Delta t \) is the time interval over which the change occurs.
02

Calculate the change in magnetic flux

The change in magnetic flux \( \Delta \Phi \) through the coil is given by \( \Delta \Phi = B \cdot A \cdot N \), where \( B \) is the magnetic field strength, \( A \) is the area of the coil, and \( N \) is the number of turns in the coil. Since the coil starts from \( B = 0 \) and falls to a position where \( B = 0.500 \, \mathrm{T} \), the change in the magnetic field strength is \( 0.500 \, \mathrm{T} \). The area of the coil \( A \) is the product of its length and width, which is \( 5.00 \, \mathrm{cm} \times 10.0 \, \mathrm{cm} \). First, convert the dimensions to meters: \( 5.00 \, \mathrm{cm} = 0.0500 \, \mathrm{m} \) and \( 10.0 \, \mathrm{cm} = 0.100 \, \mathrm{m} \). The area is then \( A = 0.0500 \, \mathrm{m} \times 0.100 \, \mathrm{m} = 0.00500 \, \mathrm{m}^2 \). Therefore, \( \Delta \Phi = 0.500 \, \mathrm{T} \cdot 0.00500 \, \mathrm{m}^2 \cdot 50 \).
03

Compute the induced average emf

Now that we have the change in magnetic flux, we can calculate the induced average emf using Faraday's law: \( \epsilon = -\frac{\Delta \Phi}{\Delta t} \). The time interval given is \( \Delta t = 0.250 \, \mathrm{s} \). The negative sign in Faraday's law indicates the direction of the induced emf (Lenz's law), but since we only need the magnitude, we can ignore it. So, \( \epsilon = \frac{0.500 \, \mathrm{T} \cdot 0.00500 \, \mathrm{m}^2 \cdot 50}{0.250 \, \mathrm{s}} \). Calculating this gives the average emf.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Faraday's Law of Electromagnetic Induction
Faraday's Law of Electromagnetic Induction is a fundamental principle that explains the working of generators, transformers, and many types of electrical devices. The law states that an electromotive force (emf) is generated in a conductor when there is a change in the magnetic flux through the area enclosed by the conductor. In simple terms, when the magnetic environment of a circuit changes, an electric current is induced.

This principle is precisely what happens in the textbook exercise, where the coil's motion through a varying magnetic field induces an emf. Faraday's Law is mathematically represented as \( \epsilon = -\frac{\Delta \Phi}{\Delta t} \), where \( \epsilon \) denotes the induced emf, \( \Delta \Phi \) is the change in magnetic flux, and \( \Delta t \) is the time interval over which this change occurs.

The negative sign reflects Lenz's Law, which tells us that the induced emf always works to oppose the change in flux that created it. This opposition is a manifestation of the conservation of energy.
Magnetic Flux
Magnetic flux, symbolized by \( \Phi \), represents the quantity of magnetism, considering the strength and the extent of a magnetic field. The unit for magnetic flux is the weber (Wb).

In the context of the given problem, calculating the change in magnetic flux involves considering the coil's dimensions and the final magnetic field strength. We calculate magnetic flux by multiplying the magnetic field strength \( B \) by the area \( A \) it penetrates, and the number of turns \( N \) in the coil: \( \Delta \Phi = B \cdot A \cdot N \). The area is calculated by multiplying the length and width of the coil, making sure to convert centimeters to meters for consistency with the SI unit system.

The ability to calculate magnetic flux is crucial because it allows us to use Faraday's Law to find the induced emf, which is the prime objective in situations similar to the stated exercise.
Average Induced EMF
The average induced emf in a circuit is the potential difference generated by electromagnetic induction over a certain time period. It is important to note that this is an 'average' value since the emf may vary over time depending on how the magnetic flux changes.

From the exercise, we calculate the average induced emf over the specified time \( \Delta t \) by using Faraday's Law. The formula involves taking the change in magnetic flux (which represents the total magnetic exposure changing over time) and dividing it by the time interval during which the change occurred: \( \epsilon = -\frac{\Delta \Phi}{\Delta t} \).

It's crucial for students to understand that the average induced emf gives us a simplified view of what happens during a specified time interval. It is useful for solving many problems in electromagnetism and provides a foundational understanding of how electric generators and similar devices function. By plugging in the values for \( \Delta \Phi \) and \( \Delta t \) from the problem into the equation, we would obtain the magnitude for the average induced emf, as detailed in the step-by-step solution.

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Most popular questions from this chapter

In a 250 -turn automobile alternator, the magnetic flux in each turn is \(\Phi_{B}=\left(2.50 \times 10^{-4} \mathrm{Wb}\right) \cos (\omega t),\) where \(\omega\) is the angular speed of the alternator. The alternator is geared to rotate three times for each engine revolution. When the engine is running at an angular speed of 1000 rev/min, determine (a) the induced emf in the alternator as a function of time and (b) the maximum emf in the alternator.

A long solenoid, with its axis along the \(x\) axis, consists of 200 turns per meter of wire that carries a steady current of 15.0 \(\mathrm{A}\) . A coil is formed by wrapping 30 turns of thin wire around a circular frame that has a radius of \(8.00 \mathrm{cm} .\) The coil is placed inside the solenoid and mounted on an axis that is a diameter of the coil and coincides with the \(y\) axis. The coil is then rotated with an angular speed of 4.00\(\pi \mathrm{rad} / \mathrm{s}\) . (The plane of the coil is in the \(y z\) plane at \(t=0 .\) Determine the emf generated in the coil as a function of time.

A proton moves through a uniform electric field given by \(\mathbf{E}=50.0 \hat{\mathbf{j}} \mathrm{V} / \mathrm{m}\) and a uniform magnetic field \(\mathbf{B}=\) \((0.200 \hat{\mathbf{i}}+0.300 \hat{\mathbf{j}}+0.400 \hat{\mathbf{k}})\) T. Determine the acceleration of the proton when it has a velocity \(\mathbf{v}=200 \hat{\mathbf{i}} \mathrm{m} / \mathrm{s}\)

A coil of area 0.100 \(\mathrm{m}^{2}\) is rotating at 60.0 \(\mathrm{rev} / \mathrm{s}\) with the axis of rotation perpendicular to a 0.200 - T magnetic field. ( a) If the coil has 1000 turns, what is the maximum emf generated in it? (b) What is the orientation of the coil with respect to the magnetic field when the maximum induced voltage occurs?

A rectangular coil of 60 turns, dimensions 0.100 \(\mathrm{m}\) by 0.200 \(\mathrm{m}\) and total resistance \(10.0 \Omega,\) rotates with angular speed 30.0 \(\mathrm{rad} / \mathrm{s}\) about the \(y\) axis in a region where a 1.00 - \(\mathrm{T}\) magnetic field is directed along the \(x\) axis. The rotation is initiated so that the plane of the coil is perpendicular to the direction of \(\mathbf{B}\) at \(t=0 .\) Calculate (a) the maximum induced emf in the coil, (b) the maximum rate of change of magnetic flux through the coil, \((c)\) the induced emf at \(t=0.0500\) s, and (d) the torque exerted by the magnetic field on the coil at the instant when the emf is a maximum.
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