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Chapter 22: Problem 69

Parts \(a\) and \(b\) of the drawing show the same uniform and constant (in time) magnetic field \(\overrightarrow{\mathbf{B}}\) directed perpendicularly into the paper over a rectangular region. Outside this region, there is no field. Also shown is a rectangular coil (one turn), which lies in the plane of the paper. In part \(a\) the long side of the coil (length \(=L)\) is just at the edge of the field region, while in part \(b\) the short side (width \(=W\) is just at the edge. It is known that \(L / W=3.0 .\) In both parts of the drawing the coil is pushed into the field with the same velocity \(\overrightarrow{\mathbf{v}}\) until it is completely within the field region. The magnitude of the average emf induced in the coil in part \(a\) is \(0.15 \mathrm{~V}\). What is its magnitude in part \(b\) ?

Short Answer

Expert verified
The magnitude of the EMF in part (b) is 0.45 V.

Step by step solution

01

Understand Faraday's Law of Induction

According to Faraday's law, the electromotive force (EMF) induced in a circuit is equal to the rate of change of magnetic flux through the circuit. Mathematically, this is given by \( \text{EMF} = - \frac{d\Phi}{dt} \), where \( \Phi \) is the magnetic flux.
02

Calculate Magnetic Flux in Part (a)

In part (a), the coil enters the magnetic field with its longer side first. The magnetic flux \( \Phi \) through the coil is given by \( \Phi = B \times A \), where \( A = L \times W \) is the area of the coil. The rate of change of flux is related to the motion of the coil entering the magnetic field.
03

Apply Faraday's Law to Part (a)

The rate of change of magnetic flux as the coil moves into the field is \( \frac{d\Phi}{dt} = B \times v \times W \) (since the width \( W \) is the dimension along which the rate of entry changes). The given average EMF is 0.15 V, so we have \( 0.15 = B \times v \times W \).
04

Calculate Magnetic Flux in Part (b)

In part (b), the coil enters the magnetic field with its shorter side first. Thus, the change in flux depends on the movement of the length \( L \) of the coil into the field: \( \frac{d\Phi}{dt} = B \times v \times L \).
05

Apply Ratio of Length to Width

Given that \( L/W = 3.0 \), it follows from substitution into the equations from steps 3 and 4 that \( L = 3W \). Substituting, we find that \( \frac{\text{EMF}_{b}}{\text{EMF}_{a}} = \frac{L}{W} = 3.0 \), since \( L = 3W \) allows increasing in the rate of change of flux in part (b) relative to part (a).
06

Solve for EMF in Part (b)

From Step 5, we know \( \text{EMF}_{b} = 3 \times 0.15 \). Calculate \( \text{EMF}_{b} = 0.45 \). Thus the magnitude of EMF in part (b) is 0.45 V.

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Key Concepts

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

Faraday's Law
Faraday's Law is a fundamental concept in electromagnetism, stating that the electromotive force (EMF) is induced in a closed circuit proportional to the rate of change of magnetic flux through the circuit. This concept explains how a change in the magnetic environment around a coil of wire will cause a current to be induced. In mathematical terms, Faraday's Law is expressed as: \( \text{EMF} = - \frac{d\Phi}{dt} \), where \( \text{EMF} \) is the induced voltage, and \( \Phi \) represents the magnetic flux. The negative sign in the formula is indicative of Lenz's Law, which states that the induced EMF will produce a current whose magnetic field opposes the change in the original magnetic field. This is why your fridge magnet doesn't fly off - the induced currents create resistance to movement.
Faraday's Law is pivotal in understanding how devices like generators and transformers operate, converting mechanical energy into electrical energy and vice versa.
Magnetic Flux
Magnetic flux, denoted as \( \Phi \), gives us a measure of the number of magnetic field lines passing perpendicularly through a given area. To calculate magnetic flux, you use the formula \( \Phi = B \times A \times \cos\theta \), where \( B \) stands for the magnetic field strength, \( A \) is the area, and \( \theta \) is the angle between the magnetic field lines and the perpendicular to the area. For most simple cases, \( \theta \) is 0 degrees and \( \cos\theta \) simplifies to 1, so \( \Phi = B \times A \).
  • Imagine a sheet of paper, with a precisely uniform magnetic field hitting it perpendicularly. Here, the magnetic flux through the paper will be simply the product of the field and the area of the sheet.
  • The greater the magnetic flux, the stronger the induced current when changes occur, as in our case with the coil moving into the magnetic field.
Understanding magnetic flux is crucial for calculating how much EMF will be generated as a coil interacts with the magnetic field.
Electromotive Force (EMF)
Electromotive Force (EMF) is not an actual force, but rather a term that describes the potential difference in a circuit, driving the flow of electrons. Expressed in volts, it proves essential in understanding electrical circuits and devices like batteries.
While Faraday's Law tells us how EMF is induced, the actual concept breaks down to a few key points:
  • EMF is directly related to the change in magnetic flux. A greater rate of change results in higher EMF. This is fundamental in the scenario where our coil moves from no magnetic field into an area with a magnetic field.
  • In the original exercise, knowing the dimensions and orientation of the coil enabled calculation of how the magnetic flux changed and hence directly figured out the induced EMF in different scenarios.
By mastering how EMF works, you're better equipped to solve problems involving the conversion of mechanical movements into electrical signals.

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Most popular questions from this chapter

Interactive Solution \(\underline{22.55}\) at offers one approach to problems such as this one. The secondary coil of a step-up transformer provides the voltage that operates an electrostatic air filter. The turns ratio of the transformer is 50: 1 . The primary coil is plugged into a standard \(120-\mathrm{V}\) outlet. The current in the secondary coil is \(1.7 \times 10^{-3} \mathrm{~A} .\) Find the power consumed by the air filter.

Mutual induction can be used as the basis for a metal detector. A typical setup uses two large coils that are parallel to each other and have a common axis. Because of mutual induction, the ac generator connected to the primary coil causes an emf of \(0.46 \mathrm{~V}\) to be induced in the secondary coil. When someone without metal objects walks through the coils, the mutual inductance and, thus, the induced emf do not change much. But when a person carrying a hand gun walks through, the mutual inductance increases. The change in emf can be used to trigger an alarm. If the mutual inductance increases by a factor of three, find the new value of the induced emf.

Concept Questions The rechargeable batteries for a laptop computer need a much smaller voltage than what a wall socket provides. Therefore, a transformer is plugged into the wall socket and produces the necessary voltage for charging the batteries. (a) Is the transformer a step-up or a step-down transformer? (b) Is the current that goes through the batteries greater than, equal to, or smaller than the current coming from the wall socket? (c) If the transformer has a negligible resistance, is the electric power delivered to the batteries greater than, equal to, or less than the power coming from the wall socket? In all cases, provide a reason for your answer. Problem The batteries of a laptop computer are rated at \(9.0 \mathrm{~V},\) and a current of \(225 \mathrm{~mA}\) is used to charge them. The wall socket provides a voltage of \(120 \mathrm{~V}\). (a) Determine the turns ratio of the transformer, (b) What is the current coming from the wall socket? (c) Find the power delivered by the wall socket and the power sent to the batteries. Be sure your answers are consistent with your answers to the Concept Questions.

Interactive LearningWare 22.2 at reviews the fundamental approach in problems such as this. A constant magnetic field passes through a single rectangular loop whose dimensions are \(0.35 \mathrm{~m} \times 0.55 \mathrm{~m}\). The magnetic field has a magnitude of \(2.1 \mathrm{~T}\) and is inclined at an angle of \(65^{\circ}\) with respect to the normal to the plane of the loop. (a) If the magnetic field decreases to zero in a time of \(0.45 \mathrm{~s}\), what is the magnitude of the average emf induced in the loop? (b) If the magnetic field remains constant at its initial value of \(2.1 \mathrm{~T},\) what is the magnitude of the rate \(\Delta \mathrm{A} / \Delta t\) at which the area should change so that the average emf has the same magnitude as in part (a)?

Concept Questions The drawing shows a straight wire carrying a current \(I\). Above the wire is a rectangular loop that contains a resistor \(R\). (a) Does the magnetic field produced by the current \(I\) penetrate the loop and generate a magnetic flux? (b) When is there an induced current in the loop, if the current \(I\) is constant or if it is decreasing in time? (c) When there is an induced magnetic field produced by the loop, does it always have a direction that is opposite to the direction of the magnetic field produced by the current \(I\) ? Provide a reason for each answer. Problem If the current \(I\) is decreasing in time, what is the direction of the induced current through the resistor \(R\) - left to right or right to left? Give your reasoning.
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