/*! 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 30 A conducting coil of 1850 turns ... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

A conducting coil of 1850 turns is connected to a galvanometer, and the total resistance of the circuit is \(45.0 \Omega\). The area of each turn is \(4.70 \times\) \(10^{-4} \mathrm{m}^{2} .\) This coil is moved from a region where the magnetic field is zero into a region where it is nonzero, the normal to the coil being kept parallel to the magnetic field. The amount of charge that is induced to flow around the circuit is measured to be \(8.87 \times 10^{-3} \mathrm{C} .\) Find the magnitude of the magnetic field.

Short Answer

Expert verified
The magnitude of the magnetic field is approximately 0.250 Tesla.

Step by step solution

01

Calculate the Total Magnetic Flux Change

First, we'll identify that we need to use Faraday's Law of Induction, which relates the change in magnetic flux to the induced EMF and thus the charge. The induced EMF \( \varepsilon \) can be expressed as \( \varepsilon = -N \frac{\Delta \Phi}{\Delta t} \) where \( N \) is the number of turns and \( \Delta \Phi \) is the change in magnetic flux. Since charge \( Q \) is related to EMF by \( Q = \frac{\varepsilon}{R} \), we substitute this to express the change in flux in terms of charge: \( \Delta \Phi = \frac{QR}{N} \). Substituting the known values, \( Q = 8.87 \times 10^{-3} \mathrm{C} \), \( R = 45.0 \Omega \), and \( N = 1850 \), we calculate: \[ \Delta \Phi = \frac{8.87 \times 10^{-3} \times 45.0}{1850} = 2.16 \times 10^{-4} \, \mathrm{Wb} \]
02

Use the Relation Between Flux and Magnetic Field

Next, we use the relationship between magnetic flux \( \Phi \) and magnetic field \( B \), which is given by \( \Phi = B \cdot A \cdot N \) where \( A \) is the area per turn. Knowing \( \Delta \Phi = B \cdot A \cdot N \) because the initial magnetic field \( B_0 = 0 \), solve for \( B \): \( B = \frac{\Delta \Phi}{A \cdot N} \). Substitute \( \Delta \Phi = 2.16 \times 10^{-4} \mathrm{Wb} \), \( A = 4.70 \times 10^{-4} \mathrm{m}^2 \), and \( N = 1850 \) into the equation to find \( B \): \[ B = \frac{2.16 \times 10^{-4}}{4.70 \times 10^{-4} \times 1850} \approx 0.250 \, \mathrm{T} \]
03

Conclude with the Final Magnetic Field

Finally, compile our findings to state the magnitude of the magnetic field developed through the process. By using the calculated change in flux and the area per turn multiplied by the number of turns, we've determined the magnetic field to be approximately 0.250 Tesla.

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.

Magnetic Flux
Magnetic flux is a fundamental concept in understanding Faraday's Law of Induction. It represents the quantity of magnetic field lines passing through a given area. You can think of it like the flow of water through a net – the greater the flow, the more lines pass through. Magnetic flux, often symbolized as \( \Phi \), is calculated using the formula:\[ \Phi = B \cdot A \cdot \cos(\theta) \]where:
  • \( B \) is the magnetic field
  • \( A \) is the area through which the field lines pass
  • \( \theta \) is the angle between the normal to the area and the magnetic field direction
In scenarios like the exercise, where the coil is parallel to the magnetic field, \( \theta \) is zero, making \( \cos(\theta) = 1 \). Hence, the magnetic flux becomes simply \( \Phi = B \cdot A \). Understanding changes in magnetic flux helps us grasp how electric currents are induced in coils and circuits. It is essential to note that any change in the number of field lines passing through a coil directly results in changes in flux.
EMF (Electromotive Force)
Electromotive force (EMF) is a concept closely associated with Faraday’s Law of Induction. It refers to the voltage generated by moving a conductor in a magnetic field or by changing the magnetic field around a conductor. According to Faraday’s Law, the induced EMF in a closed circuit is equal to the negative rate of change of magnetic flux through the circuit:\[ \varepsilon = -N \frac{\Delta \Phi}{\Delta t} \]where:
  • \( \varepsilon \) is the induced EMF
  • \( N \) is the number of turns in the coil
  • \( \Delta \Phi \) is the change in magnetic flux
  • \( \Delta t \) is the time over which the change occurs
In many practical applications, such as the one in the problem, we relate it to the current or charge using the formula \( Q = \frac{\varepsilon}{R} \), where \( R \) is resistance. This shows how the circuit's resistance impacts the movement of charge when EMF is induced. Thus, a higher resistance means a smaller current for a given EMF.
Magnetic Field
A magnetic field is the invisible force around magnets represented by field lines outside and inside a magnet. These fields can affect compounds like iron and cause electromagnetism in conductors. The strength and direction of a magnetic field are typically symbolized by \( B \) and measured in Teslas (T).The interplay between magnetic fields and conductors lies at the heart of many technologies, including electric generators and inductors. When a conductor, such as a coil, moves through a magnetic field, or when the field itself changes, an EMF is created, a phenomenon crucially explained by Faraday's Law.To comprehend the magnitude of the magnetic field from the exercise solution, we observe that when a coil moves into an area with a magnetic field changing from zero to non-zero, it helps calculate the magnetic field using:\[ B = \frac{\Delta \Phi}{A \cdot N} \]where:
  • \( \Delta \Phi \) is the change in magnetic flux
  • \( A \) is the area per loop
  • \( N \) is the number of coil turns
Thus, knowing what magnetic field represents helps us understand how inductive procedures can charge objects and circuits.

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

The magnetic flux that passes through one turn of a 12 -turn coil of wire changes to 4.0 from \(9.0 \mathrm{Wb}\) in a time of \(0.050 \mathrm{s}\). The average induced current in the coil is 230 A. What is the resistance of the wire?

The coil of a generator has a radius of \(0.14 \mathrm{m} .\) When this coil is unwound, the wire from which it is made has a length of \(5.7 \mathrm{m}\). The magnetic field of the generator is \(0.20 \mathrm{T},\) and the coil rotates at an angular speed of \(25 \mathrm{rad} / \mathrm{s} .\) What is the peak emf of this generator?

A rectangular loop of wire with sides 0.20 and \(0.35 \mathrm{m}\) lies in a plane perpendicular to a constant magnetic field (see part \(a\) of the drawing). The magnetic field has a magnitude of \(0.65 \mathrm{T}\) and is directed parallel to the normal of the loop's surface. In a time of 0.18 s, one-half of the loop is then folded back onto the other half, as indicated in part \(b\) of the drawing. Determine the magnitude of the average emf induced in the loop.

A long, current-carrying solenoid with an air core has 1750 turns per meter of length and a radius of \(0.0180 \mathrm{m} .\) A coil of 125 turns is wrapped tightly around the outside of the solenoid, so it has virtually the same radius as the solenoid. What is the mutual inductance of this system?

A \(120.0-\mathrm{V}\) motor draws a current of \(7.00 \mathrm{A}\) when running at normal speed. The resistance of the armature wire is \(0.720 \Omega .\) (a) Determine the back emf generated by the motor. (b) What is the current at the instant when the motor is just turned on and has not begun to rotate? (c) What series resistance must be added to limit the starting current to \(15.0 \mathrm{A} ?\)

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.