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At the instant when the current in an inductor is increasing at a rate of \(0.0640 \mathrm{A} / \mathrm{s},\) the magnitude of the self-induced emf is 0.0160 \(\mathrm{V}\) (a) What is the inductance of the inductor? (b) If the inductor is a solenoid with 400 turns, what is the average magnetic flux through each turn when the current is 0.720 \(\mathrm{A} ?\)

Short Answer

Expert verified
(a) The inductance is 0.25 H. (b) The magnetic flux through each turn is approximately 4.5 × 10^{-4} Wb.

Step by step solution

01

Understanding the emf Formula

The formula for induced emf \( \varepsilon \) in a coil is given by \( \varepsilon = -L \frac{di}{dt} \), where \( L \) is the inductance and \( \frac{di}{dt} \) is the rate of change of current. Here, \( \varepsilon = 0.0160 \) V and \( \frac{di}{dt} = 0.0640 \) A/s. Our task is to find \( L \).
02

Solving for Inductance

Rearranging the formula \( \varepsilon = -L \frac{di}{dt} \) gives us \( L = -\frac{\varepsilon}{\frac{di}{dt}} \). Substituting the values, we have \( L = -\frac{0.0160}{0.0640} \). Calculating this gives \( L = 0.25 \) H (Henrys).
03

Understanding Magnetic Flux in a Solenoid

The magnetic flux through a single loop is given by \( \Phi = \frac{BA}{N} \), but for a solenoid, the relevant expression involves the current and inductance, \( L = N \frac{\Phi}{I} \). We have \( N = 400 \) turns, and the current \( I = 0.720 \) A, with the previously found \( L = 0.25 \) H.
04

Solving for Magnetic Flux

Rearrange the formula \( L = N \frac{\Phi}{I} \) to find \( \Phi = \frac{L \cdot I}{N} \). Substituting the known values, \( \Phi = \frac{0.25 \times 0.720}{400} \), we find \( \Phi \approx 4.5 \times 10^{-4} \, \text{Wb} \) (Weber) per turn.

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Key Concepts

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

Inductance
Inductance is a fundamental concept in electromagnetism, describing a coil's ability to resist changes in electric current. It acts similarly to inertia in mechanical systems. The higher the inductance, the more a coil opposes changes in current. It is measured in Henry (H). When you have an inductor, like a solenoid, and the current changes, the inductor will generate an electromotive force (emf) opposing that change.

This phenomenon leads to the creation of a magnetic field around the coil. The inductance, symbolized by \( L \), is determined by factors such as the number of turns in the coil, the coil's cross-sectional area, and the material inside the coil. For example, in our problem, using the formula \( \varepsilon = -L \frac{di}{dt} \), we found the inductance as \( L = 0.25 \) H by using the given change rate in current and the induced emf.

Inductance is crucial in designing circuits, particularly those used in transformers and inductors.
Self-induced emf
Self-induced emf is a property of a coil where a change in the electric current flowing through it induces an emf in the same coil. This self-induction is described by Lenz's law, which states that the induced emf always works against the change in current that created it.
  • The self-induced emf, \( \varepsilon \), can be calculated with the formula \( \varepsilon = -L \frac{di}{dt} \), where \( L \) is the inductance and \( \frac{di}{dt} \) is the rate at which the current changes.
  • The negative sign indicates that the induced emf opposes the change in current, aligning with Lenz's Law.
Understanding self-induced emf helps in analyzing how inductors behave in circuits and calculating the energy stored within a magnetic field. As in our exercise, the self-induced emf aligns directly with how quickly the current is changing, and the inductor's ability to resist that change.
Magnetic Flux
Magnetic flux represents the total magnetic field passing through a specific area, conceptualized as magnetic lines making their way through a loop or coil. It is measured in Weber (Wb). High magnetic flux means a strong magnetic field presence through the coil's surface.

For solenoids, magnetic flux is interconnected with current and number of turns via the inductor's formula. For instance, the inductance \( L = 0.25 \) H, number of turns \( N = 400 \), and current \( I = 0.720 \) A guide us in finding the flux through each loop:
  • The formula \( L = N \frac{\Phi}{I} \) helps in determining magnetic flux.
  • Rearranging to \( \Phi = \frac{L \cdot I}{N} \), gives us an average flux of approximately \( 4.5 \times 10^{-4} \) Wb per turn.
Thus, magnetic flux connects electric currents, magnetic fields, and coil geometry, integral to device functionality from transformers to sensors.

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Most popular questions from this chapter

Two coils have mutual inductance \(M=3.25 \times 10^{-4} \mathrm{H}\) The current \(i_{1}\) in the first coil increases at a uniform rate of 830 \(\mathrm{A} / \mathrm{s} .\) (a) What is the magnitude of the induced emf in the second coil? Is it constant? (b) Suppose that the current described is in the second coil rather than the first. What is the magnitude of the induced emf in the first coil?

Two toroidal solenoids are wound around the same form so that the magnetic field of one passes through the turns of the other. Solenoid 1 has 700 turns, and solenoid 2 has 400 turns. When the current in solenoid 1 is 6.52 \(\mathrm{A}\) , the average flux through each turn of solenoid 2 is 0.0320 \(\mathrm{Wb}\) . (a) What is the mutual inductance of the pair of solenoids? (b) When the current in solenoid 2 is 2.54 \(\mathrm{A}\) , what is the average flux through each turn of solenoid 1 1 ?

A capacitor with capacitance \(6.00 \times\) \(10^{-5} \mathrm{F}\) is charged by connecting it to a \(12.0-\mathrm{V}\) battery. The capacitor is disconnected from the battery and connected across an inductor with \(L=1.50 \mathrm{H}\) (a) What are the angular frequency \(\omega\) of the electrical oscillations and the period of these oscillations (the time for one oscillation)? (b) What is the initial charge on the capacitor? (c) How much energy is initially stored in the capacitor? (d) What is the charge on the capacitor 0.0230 s after the connection to the inductor is made? Interpret the sign of your answer.(e) At the time given in part (d), what is the current in the inductor' Interpret the sign of your answer. (f) At the time given in par (d), how much electrical energy is stored in the capacitor and how much is stored in the inductor?

An air-filled toroidal solenoid has 300 turns of wire, a mean radius of \(12.0 \mathrm{cm},\) and a cross-sectional area of 4.00 \(\mathrm{cm}^{2} .\) If the current is \(5.00 \mathrm{A},\) calculate: (a) the magnetic field in the solenoid; (b) the self-inductance of the solenoid; (c) the energy stored mean radius of \(12.0 \mathrm{cm},\) and a cross-sectional area of 4.00 \(\mathrm{cm}^{2} .\) If the current is \(5.00 \mathrm{A},\) calculate: (a) the magnetic field in the solenoid; (b) the self-inductance of the solenoid; (c) the energy stored

An inductor is connected to the terminals of a battery that has an emf of 12.0 \(\mathrm{V}\) and negligible internal resistance. The current is 4.86 \(\mathrm{mA}\) at 0.940 \(\mathrm{ms}\) after the connection is completed. After a long time the current is 6.45 \(\mathrm{mA} .\) What are (a) the resistance \(R\) of the inductor and (b) the inductance \(L\) of the inductor?

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