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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 sclf-induced \(\mathrm{cmf}\) 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
The inductance of the inductor is \(0.25 H\) and the average magnetic flux through each turn when the current is \(0.720 A\) is \(0.00045 Wb\).

Step by step solution

01

Calculate the Inductance

We use the definition of inductance to find \(L = ε / di/dt\). Here, \(ε = 0.0160 V\) and \(di/dt = 0.0640 A/s\). Substituting these values into the formula, we get \(L = 0.0160 V / 0.0640 A/s = 0.25 H\).
02

Calculate the Magnetic Flux

Next, we calculate the magnetic flux through each turn of the coil. The formula for the magnetic flux is \(φ = Li/N\). Using the values \(L = 0.25 H\), \(i = 0.720 A\) and \(N = 400\), we substitute into the formula to get \(φ = 0.25 H * 0.720 A / 400 = 0.00045 Wb\). This is the magnetic flux through each turn of the coil at the moment when the current is \(0.720 A\).

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Key Concepts

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

Self-induced EMF
Electromagnetic Induction is a fundamental concept in physics where a change in magnetic field induces an electromotive force (EMF). A self-induced EMF occurs when the change in current through an inductor itself generates this EMF. The magnitude of the self-induced EMF, denoted as \( \varepsilon \), is directly proportional to the rate of change of current \( \frac{di}{dt} \). This relationship is expressed through the formula: \( \varepsilon = -L \frac{di}{dt} \). Here, \( L \) represents the inductance of the coil, which is a measure of how effectively a coil can induce an EMF in itself due to a change in current.

This induced EMF always acts to oppose the change in current due to Lenz's Law, which is why a negative sign appears in the equation. Understanding self-induced EMF is crucial for grasping how inductors function in circuits, helping to smooth out fluctuations in current and maintain stability within electronic devices.
Magnetic Flux
Magnetic Flux is a measure of the quantity of magnetism, taking account of the strength and the extent of a magnetic field. It is represented with the symbol \( \Phi \) and is measured in Webers (Wb). Magnetic flux through a single loop is calculated with the formula: \( \Phi = L \cdot i \). Here, \( L \) represents the inductance, and \( i \) denotes the current flowing through the coil.

In the context of a solenoid or coil, magnetic flux is key to understanding how a magnetic field interacts with the material within and around the coil. Higher magnetic flux implies a stronger magnetic field, essential for the functioning of devices like transformers, inductors, and coils used in various electronic applications.
  • A change in magnetic flux can induce an EMF, leading to current flow in the circuit.
  • Consistent magnetic flux means stable electromotive properties within the coil.
Understanding magnetic flux is integral to exploring how electrical energy functions in many practical applications.
Solenoid
A solenoid is essentially a long coil of wire, and it functions as an electromagnet when electric current passes through it. Solenoids are characterized by their ability to produce a uniform magnetic field in a controlled manner.

The magnetic field generated by a solenoid is strong and concentrated in the center of the coil, making it a versatile component in many electromagnetic devices. Solenoids are fundamental to designing motors, inductors, and electromagnets. They rely heavily on the principles of electromagnetic induction, where:
  • Tighter coil turns produce a more concentrated magnetic field.
  • The strength of the magnetic field is adjustable by controlling the current flowing through the solenoid.
Solenoids are practical for various engineering applications, offering a straightforward method to convert electrical energy into mechanical motion.
Coil Turns
Coil Turns refer to the number of loops or turns within a coil structure, directly affecting the electrical characteristics of a device that employs electromagnetic induction. The number of coil turns, denoted as \( N \), significantly influences both the inductance and the magnetic flux thereby enhancing or limiting the performance of the coil.

More turns in a coil can lead to greater inductance, where \( L \propto N^2 \), meaning that as the number of turns increases, the inductance increases exponentially. Adequate coil turns ensure that coils have sufficient magnetic flux to do work effectively. Consider these points:
  • Larger numbers of coil turns contribute to stronger magnetic fields per unit current.
  • Coil turns enable tuning of the inductance to meet specific circuit requirements.
Thus, the thoughtful design and arrangement of coil turns are crucial to crafting efficient and effective electromagnetic systems.

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Most popular questions from this chapter

A solcnoid \(25.0 \mathrm{~cm}\) long and with a cross-sectional area of \(0.500 \mathrm{~cm}^{2}\) contains 400 tums of wirc and carrics a currcnt of \(80.0 \mathrm{~A}\). Calculate: (a) the magnetic field in the solenoid; (b) the energy density in the magnetic field if the solenoid is filled with air; (c) the total energy contained in the coil's magnetic field (assume the field is uniform); (d) the inductance of the solenoid.

An inductor with an inductance of \(2.50 \mathrm{II}\) and a resistance of \(8.00 \Omega\) is connected to the terminals of a battery with an emf of \(6.00 \mathrm{~V}\) and negligible internal resistance. Find (a) the initial rate of increase of current in the circuit; (b) the rate of increase of current at the instant when the current is \(0.500 \mathrm{~A}\) : (c) the current \(0.250 \mathrm{~s}\) after the circuit is closed; (d) the final steady-state current.

An air-filled toroidal solenoid has a mean radius of \(15.0 \mathrm{~cm}\) and a cross-sectional area of \(5.00 \mathrm{~cm}^{2}\). When the current is \(12.0 \mathrm{~A}\), the encrgy stored is \(0.390 \mathrm{~J}\). How many turns does the winding have?

I-C Oscillations. 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 os cillations and the period of these ascillations (the time for one oscillation)? (b) What is the initial charge on the capacitor? (c) Ilow much energy is initially stored in the capacitor? (d) What is the charge on the capacitor \(0.0230 \mathrm{~s}\) after the connection to the inductor is made? Interpret the sign of your answer. (c) 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 part (d), how much electrical energy is stored in the capacitor and how much is stored in the inductor?

Solar Magnetic Energy. Magnetic fields within a sunspot can be as strong as 0.4 T. (By comparison, the earth's magnetic field is about \(1 / 10,000\) as strong. ) Sunspots can be as large as \(25,000 \mathrm{~km}\) in radius. The material in a sunspot has a density of about \(3 \times 10^{-4} \mathrm{~kg} / \mathrm{m}^{3}\). Assume \(\mu\) for the sunspot material is \(\mu_{0}\). If \(100 \%\) of the magneticfield energy stored in a sunspot could be used to eject the sunspot's material away from the sun's surface, at what speed would that material be cjected? Compare to the sun's escape speed, which is about \(6 \times 10^{5} \mathrm{~m} / \mathrm{s} .\) (Hint: Calculate the kinetic energy the magnetic field could supply to \(1 \mathrm{~m}^{3}\) of sunspot material.)

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