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When the current in a toroidal solenoid is changing at a rate of 0.0260 A/s, the magnitude of the induced emf is 12.6 mV. When the current equals 1.40 A, the average flux through each turn of the solenoid is 0.00285 Wb. How many turns does the solenoid have?

Short Answer

Expert verified
The solenoid has approximately 170 turns.

Step by step solution

01

Understand the Problem

We need to find the number of turns in a toroidal solenoid. We know the rate of change of the current, the induced emf, and the average magnetic flux through each turn when a certain current is flowing.
02

Identify the Relevant Equation for Induced EMF

The electromotive force (emf) induced in the solenoid can be expressed using Faraday's Law: \[ |\varepsilon| = N \left| \frac{d\Phi}{dt} \right| \] where \( |\varepsilon| \) is the magnitude of the induced emf, \( N \) is the number of turns, and \( \frac{d\Phi}{dt} \) is the rate of change of magnetic flux.
03

Express the Rate of Change of Magnetic Flux

The rate of change of magnetic flux through each turn can also be expressed as \[ \frac{d\Phi}{dt} = \frac{dI}{dt} \cdot \Phi_{\text{avg}} \]where \( \frac{dI}{dt} = 0.0260 \text{ A/s} \) and \( \Phi_{\text{avg}} = 0.00285 \text{ Wb} \).
04

Calculate the Rate of Change of Flux

Substitute the given values into the expression for the rate of change of magnetic flux: \[ \frac{d\Phi}{dt} = 0.0260 \cdot 0.00285 = 0.0000741 \text{ Wb/s} \].
05

Relate EMF with Number of Turns

Substitute the values of induced emf and the calculated \( \frac{d\Phi}{dt} \) into Faraday's Law to solve for \( N \):\[ |\varepsilon| = N \cdot \frac{d\Phi}{dt} \].Given \( |\varepsilon| = 12.6 \text{ mV} = 0.0126 \text{ V} \), we have: \[ 0.0126 = N \times 0.0000741 \].
06

Solve for Number of Turns

Rearrange the equation to solve for \( N \):\[ N = \frac{0.0126}{0.0000741} \approx 170 \].

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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 that explains how a change in magnetic field can induce an electromotive force (EMF) in a circuit. This law is crucial in understanding how transformers and electrical generators work. According to Faraday's Law, the induced EMF in any closed circuit is equal to the negative rate of change of the magnetic flux through the circuit. Mathematically, it can be expressed as: \( \varepsilon = - \frac{d\Phi}{dt} \). The negative sign comes from Lenz's Law, which states that the induced EMF always acts in such a way as to oppose the change in flux that produces it.

In simpler terms, if the magnetic environment of a coil is changing, an EMF is induced in the coil trying to counteract the change. Hence, a rapid change in magnetic flux will result in a stronger induced EMF, while a slower change will result in a weaker EMF.
Magnetic Flux
Magnetic Flux refers to the measure of the amount of magnetic field passing through a given area. It is often described through lines of magnetic field, which when dense, indicate a stronger magnetic field. The magnetic flux (\( \Phi \)) can be calculated using the formula: \( \Phi = B \cdot A \cdot \cos(\theta) \), where \( B \) is the magnetic field 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.

In cases like our solenoid problem, the average magnetic flux through each turn is already given as \( 0.00285 \text{ Wb} \). When the current varies, its change directly affects how much flux penetrates the coil, which is essential for analyzing induced electric fields in coils according to Faraday's Law.
Toroidal Solenoid
A toroidal solenoid is a type of coil shaped like a doughnut, made by wrapping a wire many times over a circular core. This configuration is advantageous for its ability to confine the magnetic field within its windings, creating a highly efficient magnetic circuit. In our exercise, the toroidal solenoid's role is to allow the magnetic environment to influence each turn of the wire, which in turn induces an EMF when the current in the solenoid changes.

Because the magnetic field inside a toroidal solenoid is largely uniform and concentrated within the coil, this allows us to easily calculate the average magnetic flux through each turn, a critical factor when using Faraday’s Law to determine the number of turns.
Number of Turns in Coil
The number of turns in a coil, represented by \( N \), is the count of how many loops the wire makes around the core of the solenoid. This parameter is central to determining the strength of electromagnetic effects in devices. In our exercise, it's crucial because Faraday’s Law expression for induced EMF directly involves \( N \) as: \( \varepsilon = N \left| \frac{d\Phi}{dt} \right| \).

As the number of turns increases, the induced EMF for a given rate of change in magnetic flux will be stronger, since the coil has more loops to intercept the magnetic field. Our task in the original exercise involved calculating \( N \) by using known values of EMF, the rate of change of current, and average magnetic flux, resulting in about 170 turns.

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Most popular questions from this chapter

An inductor with an inductance of 2.50 H and a resistance of 8.00 \(\Omega\) is connected to the terminals of a battery with an emf of 6.00 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 A; (c) the current 0.250 s after the circuit is closed; (d) the final steady-state current.

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 km in radius. The material in a sunspot has a density of about \(3 \times 10^{-4} \, \mathrm{kg/m}^3\). Assume \(\mu\) for the sunspot material is \(\mu_0\). If 100\(\%\) of the magnetic-field 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 ejected? Compare to the sun's escape speed, which is about \(6 \times 10^5 \, \mathrm{m/s}\). (\(Hint:\) Calculate the kinetic energy the magnetic field could supply to 1 m\(^3\) of sunspot material.)

A solenoidal coil with 25 turns of wire is wound tightly around another coil with 300 turns (see Example 30.1). The inner solenoid is 25.0 cm long and has a diameter of 2.00 cm. At a certain time, the current in the inner solenoid is 0.120 A and is increasing at a rate of \(1.75 \times 10^3\) A/s. For this time, calculate: (a) the average magnetic flux through each turn of the inner solenoid; (b) the mutual inductance of the two solenoids; (c) the emf induced in the outer solenoid by the changing current in the inner solenoid.

A small solid conductor with radius \(a\) is supported by insulating, nonmagnetic disks on the axis of a thin-walled tube with inner radius \(b\). The inner and outer conductors carry equal currents \(i\) in opposite directions. (a) Use Ampere's law to find the magnetic field at any point in the volume between the conductors. (b) Write the expression for the flux \(d\Phi_B\) through a narrow strip of length \(l\) parallel to the axis, of width \(dr\), at a distance \(r\) from the axis of the cable and lying in a plane containing the axis. (c) Integrate your expression from part (b) over the volume between the two conductors to find the total flux produced by a current \(i\) in the central conductor. (d) Show that the inductance of a length \(l\) of the cable is $$L = l \frac{\mu_0}{2\pi} \mathrm{ln}(\frac{b}{a})$$ (e) Use Eq. (30.9) to calculate the energy stored in the magnetic field for a length l of the cable.

At the instant when the current in an inductor is increasing at a rate of 0.0640 A/s, the magnitude of the self-induced emf is 0.0160 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 A?

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