/*! 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 25 A conducting coil of 1850 turns ... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 22: Problem 25

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. (Such a device can be used to measure the magnetic field strength and is called a flux meter.)

Short Answer

Expert verified
The magnetic field magnitude is 0.457 T.

Step by step solution

01

Understand the Problem

We need to find the magnitude of the magnetic field when a coil with given turns, area, and resistance is moved into it, causing a charge to flow around the circuit. The key point is to use Faraday's Law of Electromagnetic Induction, which relates the change in magnetic flux to the induced EMF and charge.
02

Recall Faraday's Law of Induction

Faraday's Law of Electromagnetic Induction states that the EMF induced in a coil is equal to the rate of change of magnetic flux through the coil, mathematically given by \( \text{EMF} = -N \frac{\Delta \Phi}{\Delta t} \), where \( N \) is the number of turns and \( \Delta \Phi \) is the change in magnetic flux.
03

Relate EMF to Charge and Resistance

Ohm鈥檚 Law in electromagnetic induction states \( \text{EMF} = IR \), where \( I \) is the current. The total charge \( Q \) that flows can be related to the current as \( Q = I\Delta t \). Thus \( \text{EMF} = \frac{Q}{\Delta t} R \). Since \( \Delta t \) cancels out, we have \( \text{EMF} = \frac{Q \cdot R}{\Delta t} \).
04

Equate and Solve for Change in Magnetic Flux

Equating the two expressions for EMF: \(-N \frac{\Delta \Phi}{\Delta t} = \frac{Q \cdot R}{\Delta t} \). Cancel \(\Delta t\) and solve for \(\Delta \Phi\): \( \Delta \Phi = \frac{Q \cdot R}{N} \).
05

Relate Change in Magnetic Flux to Magnetic Field

The change in magnetic flux is given by \( \Delta \Phi = B \cdot A \), where \( B \) is the magnetic field magnitude and \( A \) is the area of one turn. Substitute \( \Delta \Phi \) from Step 4 to get \( B \cdot A = \frac{Q \cdot R}{N} \). Solve for \( B \): \( B = \frac{Q \cdot R}{N \cdot A} \).
06

Calculate the Magnetic Field Magnitude

Substitute the values: \( Q = 8.87 \times 10^{-3} \mathrm{C} \), \( R = 45.0 \Omega \), \( N = 1850 \) turns, \( A = 4.70 \times 10^{-4} \mathrm{m}^2 \) into the formula: \( B = \frac{8.87 \times 10^{-3} \times 45.0}{1850 \times 4.70 \times 10^{-4}} \). Calculate to find \( B = 0.457 \mathrm{T} \).

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 91影视!

Key Concepts

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

Electromagnetic Induction
Electromagnetic induction is a phenomenon where a changing magnetic field induces an electromotive force (EMF) in a conductor. This process is the backbone of many applications in technology, such as generators and transformers. When a conductor such as a coil is exposed to a varying magnetic field, the magnetic flux through the coil's surface changes, inducing an EMF according to Faraday鈥檚 Law of Electromagnetic Induction. This EMF can drive an electric current if the circuit is closed.

Faraday's Law is mathematically expressed by the formula:
  • \( \text{EMF} = -N \frac{\Delta \Phi}{\Delta t} \)
where \( N \) is the number of turns in the coil, and \( \Delta \Phi \) is the change in magnetic flux over time \( \Delta t \). The negative sign indicates that the induced EMF creates a current which opposes the change in magnetic flux, according to Lenz's Law.

Understanding this principle is crucial in predicting how a coil will behave when subjected to a changing magnetic environment. This action allows us to harness electromagnetic forces, making it possible to generate electricity in power plants and other devices.
Conducting Coil
A conducting coil is an essential component in electromagnetic applications. It consists of multiple loops or turns of wire, which enhance its ability to induce EMF when interacting with a magnetic field.

Several factors determine its effectiveness:
  • **Number of Turns**: More turns increase the total EMF induced, as reflected in the formula \( \text{EMF} = -N \frac{\Delta \Phi}{\Delta t} \).
  • **Area of Each Turn**: The area directly affects the magnetic flux linked with the coil. A larger area allows more magnetic field lines to pass through, enhancing the flux.
  • **Resistance**: The wire鈥檚 resistance affects how much current can flow as a consequence of the induced EMF, described by Ohm's Law \( \text{EMF} = I \times R \).
A conducting coil finds its usage in devices like galvanometers, which measure current by detecting the induced EMF. Also, coils are integral in flux meters, which can measure the strength of magnetic fields by noting the flow of charge induced in the presence of a magnetic field.
Magnetic Field Measurement
Measuring the magnetic field is a significant application of electromagnetic induction principles. Devices like the flux meter use conducting coils to determine the field's magnitude.

When a coil is moved into a magnetic field, the change in field strength alters the magnetic flux through the coil. By measuring the induced charge \( Q \) that flows in response, the magnitude of the magnetic field \( B \) can be calculated. The relationship is given by:
  • \( B = \frac{Q \cdot R}{N \cdot A} \)
Here, \( R \) is the resistance, \( N \) is the number of turns, and \( A \) is the area of the coil. Using this formula allows us to straightforwardly determine the strength of the magnetic field.

This technique of measurement is often employed because it provides a direct and efficient way of assessing magnetic fields. Whether in a laboratory or field setting, tools that use conducting coils leverage the physical principles of electromagnetic induction to provide valuable measurements of magnetic environments.

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

A magnetic field is passing through a loop of wire whose area is \(0.018 \mathrm{~m}^{2}\). The direction of the magnetic field is parallel to the normal to the loop, and the magnitude of the field is increasing at the rate of \(0.20 \mathrm{~T} / \mathrm{s}\). (a) Determine the magnitude of the emf induced in the loop. (b) Suppose the area of the loop can be enlarged or shrunk. If the magnetic field is increasing as in part (a), at what rate (in \(\mathrm{m}^{2} / \mathrm{s}\) ) should the area be changed at the instant when \(B=1.8 \mathrm{~T}\) if the induced emf is to be zero? Explain whether the area is to be enlarged or shrunk.

ssm www The drawing shows a plot of the output emf of a generator as a function of time \(t\). The coil of this device has a cross-sectional area per turn of \(0.020 \mathrm{~m}^{2}\) and contains 150 turns. Find (a) the frequency \(f\) of the generator in hertz, (b) the angular speed \(\omega\) in \(\mathrm{rad} / \mathrm{s}\), and \((\mathrm{c})\) the magnitude of the magnetic field

A vacuum cleaner is plugged into a \(120.0-\mathrm{V}\) socket and uses 3.0 A of current in normal operation when the back emf generated by the electric motor is \(72.0 \mathrm{~V}\). Find the coil resistance of the motor.

The secondary coil of a step-up transformer provides the voltage that operates an electrostatic air filter. The turns ratio of the transformer is \(50: 1\). The primary coil is plugged into a standard \(120-V\) outlet. The current in the secondary coil is \(1.7 \times 10^{-3} \mathrm{~A}\). Find the power consumed by the air filter.

The drawing shows a type of flow meter that can be used to measure the speed of blood in situations when a blood vessel is sufficiently exposed (e.g., during surgery). Blood is conductive enough that it can be treated as a moving conductor. When it flows perpendicularly with respect to a magnetic field, as in the drawing, electrodes can be used to measure the small voltage that develops across the vessel. Suppose the speed of the blood is \(0.30 \mathrm{~m} / \mathrm{s}\) and the diameter of the vessel is \(5.6 \mathrm{~mm} .\) In a 0.60 -T magnetic field what is the magnitude of the voltage that is measured with the electrodes in the drawing?
See all solutions

Recommended explanations on Physics Textbooks

Famous Physicists

Read Explanation

Electrostatics

Read Explanation

Quantum Physics

Read Explanation

Modern Physics

Read Explanation

Fluids

Read Explanation

Physics of Motion

Read Explanation
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.