/*! 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 10 When the current in a toroidal s... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

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 \( N \):\[ N = \frac{0.0126}{0.0000741} \approx 170 \].

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

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.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

An \(L\)-\(R\)-\(C\) series circuit has \(L = 0.450\) H, \(C = 2.50 \times 10^{-5} \, \mathrm{F}\), and resistance \(R\). (a) What is the angular frequency of the circuit when \(R = 0\)? (b) What value must \(R\) have to give a 5.0\(\%\) decrease in angular frequency compared to the value calculated in part (a)?

An \(L\)-\(C\) circuit containing an 80.0-mH inductor and a 1.25-nF capacitor oscillates with a maximum current of 0.750 A. Calculate: (a) the maximum charge on the capacitor and (b) the oscillation frequency of the circuit. (c) Assuming the capacitor had its maximum charge at time \(t = 0\), calculate the energy stored in the inductor after 2.50 ms of oscillation.

A toroidal solenoid has mean radius 12.0 cm and crosssectional area 0.600 cm\(^2\). (a) How many turns does the solenoid have if its inductance is 0.100 mH? (b) What is the resistance of the solenoid if the wire from which it is wound has a resistance per unit length of 0.0760 \(\Omega\)/m?

(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?

A 7.00-\(\mu\)F capacitor is initially charged to a potential of 16.0 V. It is then connected in series with a 3.75-mH inductor. (a) What is the total energy stored in this circuit? (b) What is the maximum current in the inductor? What is the charge on the capacitor plates at the instant the current in the inductor is maximal?

See all solutions

Recommended explanations on Physics Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.