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Two coils are wound around the same cylindrical form, like the coils in Example 30.1. When the current in the first coil is decreasing at a rate of -0.242 A/s, the induced emf in the second coil has magnitude \(1.65 \times 10^{-3}\) V. (a) What is the mutual inductance of the pair of coils? (b) If the second coil has 25 turns, what is the flux through each turn when the current in the first coil equals 1.20 A? (c) If the current in the second coil increases at a rate of 0.360 A/s, what is the magnitude of the induced emf in the first coil?

Short Answer

Expert verified
(a) \( M = 6.82 \times 10^{-3} \) H; (b) \( \Phi = 3.27 \times 10^{-4} \) Wb; (c) \( \varepsilon = 2.46 \times 10^{-3} \) V.

Step by step solution

01

Understanding Mutual Inductance

Mutual inductance occurs when a change in current in one coil induces an emf in the other. The formula for induced electromotive force (emf) is given by Faraday's law of induction: \( \varepsilon = M \frac{dI}{dt} \), where \( \varepsilon \) is the induced emf, \( M \) is the mutual inductance, and \( \frac{dI}{dt} \) is the rate of change of current. Here, \( \varepsilon = 1.65 \times 10^{-3} \) V and \( \frac{dI}{dt} = -0.242 \) A/s.
02

Calculate Mutual Inductance, M

To find the mutual inductance \( M \), rearrange the formula to solve for \( M \): \( M = \frac{\varepsilon}{\frac{dI}{dt}} \). Substituting the given values, \( M = \frac{1.65 \times 10^{-3}}{-0.242} \). Calculate \( M \).
03

Solve for Mutual Inductance

Perform the calculation: \( M = \frac{1.65 \times 10^{-3}}{0.242} \approx 6.82 \times 10^{-3} \) H. The mutual inductance is \( 6.82 \times 10^{-3} \) Henries.
04

Understand Magnetic Flux

Magnetic flux \( \Phi \) through each turn of the second coil when there’s current in the first coil is given by \( \Phi = \frac{M I_1}{N_2} \), where \( M \) is mutual inductance, \( I_1 \) is the current in the first coil (1.20 A), and \( N_2 \) is the number of turns (25 turns).
05

Calculate Magnetic Flux, \( \Phi \)

Substitute the values into the formula: \( \Phi = \frac{6.82 \times 10^{-3} \times 1.20}{25} \). Calculate \( \Phi \).
06

Solve for Magnetic Flux

Perform the calculation: \( \Phi = \frac{6.82 \times 10^{-3} \times 1.20}{25} \approx 3.27 \times 10^{-4} \) Wb (Webers) per turn.
07

Understand Induced EMF in Coil 1

For part (c), consider the induced emf in the first coil when the current in the second coil changes. Use the same formula: \( \varepsilon = M \frac{dI}{dt} \), but now \( \frac{dI}{dt} = 0.360 \) A/s.
08

Calculate Induced EMF in Coil 1

Substitute the values into the formula: \( \varepsilon = 6.82 \times 10^{-3} \times 0.360 \). Calculate \( \varepsilon \).
09

Solve for Induced EMF in Coil 1

Perform the calculation: \( \varepsilon = 6.82 \times 10^{-3} \times 0.360 \approx 2.46 \times 10^{-3} \) V. The induced emf in the first coil is \( 2.46 \times 10^{-3} \) Volts.

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Key Concepts

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

Faraday's law of induction
Faraday's law of induction is a fundamental principle in electromagnetism, describing how a changing magnetic field within a circuit induces an electromotive force (emf). In simpler terms, it explains how electricity can be generated by moving a coil within a magnetic field. This law is expressed mathematically as \( \varepsilon = - \frac{d\Phi}{dt} \), where \( \varepsilon \) is the induced emf and \( \frac{d\Phi}{dt} \) is the rate of change of magnetic flux. The negative sign indicates the direction of the induced emf opposes the change in magnetic flux, a concept known as Lenz's Law.
When applied to mutual inductance, Faraday's law takes the form \( \varepsilon = M \frac{dI}{dt} \). Here, \( M \) is the mutual inductance between two coils, and \( \frac{dI}{dt} \) is the rate at which current changes in one coil. Therefore, as the current in the first coil changes, it creates a changing magnetic field that induces an emf in the second coil. This principle is crucial for understanding transformers and many electrical devices.
Magnetic flux
Magnetic flux, denoted by \( \Phi \), represents the quantity of magnetic field passing through a given area. Imagine it as the number of magnetic field lines penetrating through a surface. The greater the number of lines crossing the surface, the higher the magnetic flux.
The formula for magnetic flux is \( \Phi = B \cdot A \cdot \cos(\theta) \), where \( B \) is the magnetic field's strength, \( A \) is the area through which the field lines pass, and \( \theta \) is the angle between the magnetic field and the perpendicular to the surface.
  • Magnetic flux is measured in Webers (Wb).
  • When computing flux through a coil, the number of turns affects the calculation, since more turns means more field lines are captured.
In the context of mutual inductance, flux through each turn of the coil can be found using \( \Phi = \frac{M I_1}{N_2} \). Here, \( M \) refers to mutual inductance, \( I_1 \) is the current in the first coil, and \( N_2 \) is the number of turns in the second coil. This calculation helps in determining the magnetic interaction between the coils.
Induced electromotive force
Induced electromotive force, or induced emf, is the potential difference generated in a circuit due to a change in magnetic flux. This induction is central to devices such as generators, where it converts mechanical energy to electrical energy. The formula linking induced emf to mutual inductance is \( \varepsilon = M \frac{dI}{dt} \), with \( M \) as the mutual inductance and \( \frac{dI}{dt} \) the change rate of current.
In practical terms:
  • Induced emf allows power to be transferred across components without direct connection.
  • The magnitude of induced emf depends on how quickly the magnetic environment changes.
  • Devices utilizing induced emf are often highly efficient, making them essential in modern technology.
The reciprocal nature of mutual inductance—where changing the current in one coil induces emf in another, and vice versa—is a practical manifestation of this principle. Using the problem's data, when the current rate of change in the second coil is 0.360 A/s, the first coil experiences an induced emf calculated as \( 2.46 \times 10^{-3} \) V, demonstrating the power of mutual induction.

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Most popular questions from this chapter

Your latest invention is a car alarm that produces sound at a particularly annoying frequency of 3500 Hz. To do this, the car-alarm circuitry must produce an alternating electric current of the same frequency. That's why your design includes an inductor and a capacitor in series. The maximum voltage across the capacitor is to be 12.0 V. To produce a sufficiently loud sound, the capacitor must store 0.0160 J of energy. What values of capacitance and inductance should you choose for your car-alarm circuit?

A toroidal solenoid with mean radius \(r\) and cross-sectional area \(A\) is wound uniformly with \(N_1\) turns. A second toroidal solenoid with \(N_2\) turns is wound uniformly on top of the first, so that the two solenoids have the same cross-sectional area and mean radius. (a) What is the mutual inductance of the two solenoids? Assume that the magnetic field of the first solenoid is uniform across the cross section of the two solenoids. (b) If \(N_1 = 500\) turns, \(N_2 = 300\) turns, \(r = 10.0\) cm, and \(A = 0.800 \, \mathrm{cm}^2\), what is the value of the mutual inductance?

(a) A long, straight solenoid has \(N\) turns, uniform cross sectional area \(A\), and length \(l\). Show that the inductance of this solenoid is given by the equation \(L = \mu_0 AN^2/l\). Assume that the magnetic field is uniform inside the solenoid and zero outside. (Your answer is approximate because \(B\) is actually smaller at the ends than at the center. For this reason, your answer is actually an upper limit on the inductance.) (b) A metallic laboratory spring is typically 5.00 cm long and 0.150 cm in diameter and has 50 coils. If you connect such a spring in an electric circuit, how much self-inductance must you include for it if you model it as an ideal solenoid?

It is proposed to store \(1.00 \, \mathrm{kW} \cdot \mathrm{h} = 3.60 \times 10^6\) J of electrical energy in a uniform magnetic field with magnitude 0.600 T. (a) What volume (in vacuum) must the magnetic field occupy to store this amount of energy? (b) If instead this amount of energy is to be stored in a volume (in vacuum) equivalent to a cube 40.0 cm on a side, what magnetic field is required?

In an \(L\)-\(C\) circuit, \(L = 85.0\) mH and \(C = 3.20 \, \mu \mathrm{F}\). During the oscillations the maximum current in the inductor is 0.850 mA. (a) What is the maximum charge on the capacitor? (b) What is the magnitude of the charge on the capacitor at an instant when the current in the inductor has magnitude 0.500 mA?

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