/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 13 A self-induced emf in a solenoid... [FREE SOLUTION] | 91Ó°ÊÓ

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A self-induced emf in a solenoid of inductance \(L\) changes in time as \(\boldsymbol{\varepsilon}=\boldsymbol{\varepsilon}_{0} e^{-k t} .\) Find the total charge that passes through the solenoid, assuming the charge is finite.

Short Answer

Expert verified
The total charge that passes through the solenoid for finite time is \(0\)

Step by step solution

01

Understanding the concept

The induced emf within a solenoid is given by Faraday's law as the negative rate of change of magnetic flux through the coil. It's given by the equation \( \varepsilon = - L \frac{dI}{dt}\), where \(\varepsilon\) is the induced emf, \(L\) is the inductance of the solenoid and \(I\) is the current. \(I = \frac{dQ}{dt}\) where \(Q\) is the total charge that flows through the solenoid.
02

Substituting given values

We are given \(\varepsilon = \varepsilon0e^{-kt}\) and \(I = \frac{dQ}{dt}\). Substitute these values in Step 1 equation \( \varepsilon = - L \frac{dI}{dt}\) we get, \( \varepsilon0e^{-kt} = - L \frac{d(\frac{Q}{dt})}{dt}\).
03

Solving the equation

We can now integrate both sides of the equation with respect to time to get the total charge \(Q\), \[Q = - \int \frac{\varepsilon0e^{-k t}}{L} dt\]. The integral resolves to \(\frac{\varepsilon0}{kL} [ - e^{-k t}] \). Here, \(Q\) is the total charge passed through the solenoid.
04

Evaluating the result

When the total charge is finite, as time approaches infinity, \( e^{-kt}\) approaches zero. So, evaluating the expression at infinity gives us, \[Q = - \frac{\varepsilon0}{kL} [ - e^{-k \infty}] = - \frac{\varepsilon0}{kL} \times 0 = 0.\] As a result, no charge passes through the solenoid when charge is finite.

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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 principle in electromagnetism that describes the process of electromagnetic induction. This law states that an electromotive force (emf) is induced in a circuit when the magnetic flux passing through it changes over time. The magnetic flux is the product of the magnetic field, the area of the loop, and the cosine of the angle between the field and the normal to the loop's area.

Mathematically, Faraday's Law is expressed as \(\varepsilon = -\frac{d\Phi_B}{dt}\), where \(\varepsilon\) is the induced emf and \(\Phi_B\) is the magnetic flux through the circuit. The negative sign in the equation, known as Lenz's Law, indicates that the induced emf will create a current whose magnetic field opposes the change in the original magnetic flux, adhering to the principle of conservation of energy.
Induced EMF
Induced emf refers to the voltage generated within a conductor or coil when it is exposed to a changing magnetic field. This is the basic operating principle behind many electrical generators and transformers. The induced emf prompts the flow of electric charge, or current, when the circuit is closed.

In the context of the solenoid in the provided exercise, the induced emf, denoted by \(\varepsilon\), is described by a time-dependent function \(\varepsilon = \varepsilon_0e^{-kt}\), which indicates that the emf decreases exponentially over time. The factor \(\varepsilon_0\) represents the initial emf, and \(k\) is a constant that determines the rate of decay of the emf with respect to time.
Magnetic Flux
Magnetic flux, symbolized by \(\Phi_B\), measures the quantity of the magnetic field (B) that passes through a given area (A), considering the orientation of the field relative to the area. It is calculated by the formula \(\Phi_B = B \cdot A \cdot \cos(\theta)\), where \(\theta\) is the angle between the magnetic field lines and the perpendicular to the area's surface.

The relevance of magnetic flux to the inductance of a solenoid is that changes in flux through the solenoid's coils induce an emf. In the exercise, the changing magnetic flux underlies the time-varying induced emf within the solenoid, which in turn affects the charge flow through the solenoid.
Total Charge Through Solenoid
The total charge \(Q\) that moves through the solenoid over a period can be determined by integrating the current that the solenoid carries over time. Current, \(I\), is defined as the rate of flow of charge, \(I = \frac{dQ}{dt}\). The relationship between the current and induced emf in an inductance is given by \(\varepsilon = -L\frac{dI}{dt}\), which can be re-written in terms of charge as \(\varepsilon = -L\frac{d^2Q}{dt^2}\).

From the step-by-step solution, by performing the integration to find the total charge, it turns out that for a finite charge, \(Q = 0\). This implies that eventually, no net charge moves through the solenoid, a key concept that might not be immediately obvious without understanding the underlying principles of Faraday's Law and the behavior of inductance and emf over time.
Time-Dependent EMF
The time-dependent emf in the exercise's solenoid demonstrates how the induced emf changes as a function of time. In many practical applications, the form of the time dependence can take various shapes – linear, sinusoidal, exponential, etc. The exponential decay of emf over time in this particular exercise, expressed as \(\varepsilon = \varepsilon_0e^{-kt}\), suggests that the emf strength decreases rapidly initially and then tapers off more slowly.

This time-dependent behavior plays a critical role in determining the total charge flow through the solenoid, as it impacts the rate at which the charge is transferred. The decreasing emf over time means that the force driving the charge through the coil diminishes, leading to a decrease in the rate at which the charge flows, and ultimately, under the specified conditions, leading to no net transfer of charge.

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Most popular questions from this chapter

A coiled telephone cord forms a spiral with 70 turns, a diameter of \(1.30 \mathrm{cm},\) and an unstretched length of \(60.0 \mathrm{cm} .\) Determine the self-inductance of one conductor in the unstretched cord.

A series \(R L\) circuit with \(L=3.00 \mathrm{H}\) and a series \(R C\) circuit with \(C=3.00 \mu \mathrm{F}\) have equal time constants. If the two circuits contain the same resistance \(R,\) (a) what is the value of \(R\) and \((\mathrm{b})\) what is the time constant?

A toroid has a major radius \(R\) and a minor radius \(r,\) and is tightly wound with \(N\) turns of wire, as shown in Figure P32.12. If \(R \gg r,\) the magnetic field in the region enclosed by the wire of the torus, of cross-sectional area \(A=\pi r^{2},\) is essentially the same as the magnetic field of a solenoid that has been bent into a large circle of radius \(R\) Modeling the field as the uniform field of a long solenoid, show that the self-inductance of such a toroid is approximately $$L \approx \frac{\mu_{0} N^{2} A}{2 \pi R}$$ (An exact expression of the inductance of a toroid with a rectangular cross section is derived in Problem \(64 .\) )

An inductor that has an inductance of \(15.0 \mathrm{H}\) and a resistance of \(30.0 \Omega\) is connected across a \(100-\mathrm{V}\) battery. What is the rate of increase of the current (a) at \(t=0\) and (b) at \(t=1.50 \mathrm{s} ?\)

This problem extends the reasoning of Section \(26.4,\) Problem \(26.37,\) Example \(30.6,\) and Section 32.3. (a) Consider a capacitor with vacuum between its large, closely spaced, oppositely charged parallel plates. Show that the force on one plate can be accounted for by thinking of the electric field between the plates as exerting a "negative pressure" equal to the energy density of the electric field. (b) Consider two infinite plane sheets carrying electric currents in opposite directions with equal linear current densities \(J_{s} .\) Calculate the force per area acting on one sheet due to the magnetic field created by the other sheet. (c) Calculate the net magnetic field between the sheets and the field outside of the volume between them. (d) Calculate the energy density in the magnetic field between the sheets. (e) Show that the force on one sheet can be accounted for by thinking of the magnetic field between the sheets as exerting a positive pressure equal to its energy density. This result for magnetic pressure applies to all current configurations, not just to sheets of current.

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