/*! 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 6 A uniform magnetic field is at r... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

A uniform magnetic field is at right angles to the plane of a wire loop. If the field decreases by \(0.20 \mathrm{~T}\) in \(1.0 \times 10^{-3} \mathrm{~s}\) and the magnitude of the average emf induced in the loop is \(80 \mathrm{~V},\) (a) what is the area of the loop? (b) What would be the value of the average induced emf if the field change was the same but took twice as long to decrease? (c) What would be the value of the average induced emf if the field decrease was twice as much and it also took twice as long to change?

Short Answer

Expert verified
The area of the loop is \(0.40 \, \mathrm{m^2}\). The average emf for twice the time is \(40 \, \mathrm{V}\). For twice the field change and time, the average emf is \(80 \, \mathrm{V}\).

Step by step solution

01

Understanding Faraday's Law

According to Faraday's Law, the induced electromotive force (emf) in a loop is given by the equation \( \varepsilon = \frac{-d\Phi}{dt} \), where \( \Phi \) is the magnetic flux through the loop. If the magnetic field \( B \) is uniform and perpendicular to the loop, then \( \Phi = B \cdot A \), where \( A \) is the area of the loop.
02

Calculating the Change in Magnetic Flux

Given that the change in the magnetic field \( \Delta B = 0.20 \, \mathrm{T} \) and \( \Delta t = 1.0 \times 10^{-3} \, \mathrm{s} \), the change in magnetic flux \( \Delta \Phi \) can be expressed as \( \Delta \Phi = A \cdot \Delta B \).
03

Solving for the Area of the Loop

We know from the problem that the magnitude of the average induced emf is \( 80 \, \mathrm{V} \). Substituting the values into Faraday's Law: \( \varepsilon = \frac{A \cdot \Delta B}{\Delta t} = 80 \, \mathrm{V} \). Rearranging gives \( A = \frac{\varepsilon \cdot \Delta t}{\Delta B} = \frac{80 \, \mathrm{V} \cdot 1.0 \times 10^{-3} \, \mathrm{s}}{0.20 \, \mathrm{T}} = 0.40 \, \mathrm{m^2} \).
04

Calculating Emf for Twice the Time

If the time \( \Delta t \) for the field to decrease is doubled, \( \Delta t = 2.0 \times 10^{-3} \, \mathrm{s} \). The new induced emf is calculated as \( \varepsilon_{new} = \frac{A \cdot \Delta B}{2 \cdot \Delta t} = \frac{0.40 \, \mathrm{m^2} \cdot 0.20 \, \mathrm{T}}{2.0 \times 10^{-3} \, \mathrm{s}} = 40 \, \mathrm{V} \).
05

Calculating Emf for Double \( \Delta B \) and Time

If both the change in field \( \Delta B = 0.40 \, \mathrm{T} \) and the time \( \Delta t = 2.0 \times 10^{-3} \, \mathrm{s} \) are doubled, the new induced emf is \( \varepsilon_{new} = \frac{A \cdot 2 \cdot \Delta B}{2 \cdot \Delta t} = \frac{0.40 \, \mathrm{m^2} \cdot 0.40 \, \mathrm{T}}{2.0 \times 10^{-3} \, \mathrm{s}} = 80 \, \mathrm{V} \).

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 key concept in understanding Faraday's Law of Electromagnetic Induction. It represents the total magnetic field passing through a given area, like a loop of wire. It is calculated by multiplying the magnetic field (B) with the area (A) it passes through and the cosine of the angle (\theta) between the field and the normal to the surface: \[\Phi = B \times A \times \cos(\theta)\]When the magnetic field is perpendicular to the loop, as in our problem, \(\cos(\theta)\) becomes 1, simplifying the expression to \(\Phi = B \times A\).- The unit for magnetic flux is the Weber (Wb).- If the magnetic field changes, the magnetic flux through the loop changes, leading to the concept of induced emf.- A greater area (A) or stronger magnetic field (B) increases the magnetic flux.
Induced Electromotive Force (emf)
Induced electromotive force (emf) arises when there is a change in magnetic flux, and it is a core result described by Faraday's Law. The induced emf is the voltage generated in a loop due to the changing magnetic flux. According to Faraday's Law:\[\varepsilon = -\frac{d\Phi}{dt}\]Where:- \(\varepsilon\) is the induced emf,- \(d\Phi/dt\) is the rate of change of magnetic flux.In the context of our exercise:- The sign of \(-\) in Faraday's Law indicates the direction of the induced emf which acts in a direction to oppose the change in flux (Lenz's law).- When factors causing the change in magnetic flux (i.e., magnetic field strength or time) vary, the magnitude of the induced emf is affected.- Halving the time doubles the emf, and doubling the change in magnetic field doubles the emf.
Uniform Magnetic Field
A uniform magnetic field has the same strength and direction at all points in the space it occupies. In this problem, the magnetic field acts perpendicularly to the wire loop. This uniformity simplifies the calculations related to magnetic flux and induced emf.- In a uniform magnetic field, the magnetic flux is easily found as the product of the field and the area: \(\Phi = B \times A\).- Changing the magnitude of a uniform magnetic field affects the magnetic flux through any closed loop within it.- The uniformity eliminates the need to account for variations in field strength across different positions in the loop.- Any change in a uniform magnetic field throughout the loop area can directly induce an emf according to Faraday's Law.
Area of the Loop
The area of the loop (A) is the surface enclosed by the loop through which the magnetic flux passes. It plays a critical role in determining the amount of magnetic flux and, consequently, the induced emf. In the given exercise:- The area is part of the calculation for the magnetic flux as \(\Phi = B \times A\).- A larger loop area allows more magnetic field lines through it, increasing the flux and potential induced emf.- The area can be found by rearranging the formula for induced emf: \(A = \frac{\varepsilon \cdot \Delta t}{\Delta B}\).- In our exercise, substituting \(\varepsilon = 80 \, \mathrm{V}\), \(\Delta t = 1.0 \times 10^{-3} \, \mathrm{s}\), and \(\Delta B = 0.20 \, \mathrm{T}\), the area of the loop is calculated as \(0.40 \, \mathrm{m^2}\).Understanding the area helps determine how efficient the loop can be in generating an emf based on its exposure to a changed magnetic field.

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

Electrical power from a generator is transmitted through a power line \(175 \mathrm{~km}\) long with a resistance of \(1.2 \Omega / \mathrm{km}\). The generator's output is \(50 \mathrm{~A}\) at its operating voltage of \(440 \mathrm{~V}\). This output is increased by a single step-up for transmission at \(44 \mathrm{kV}\). (a) How much power is lost as joule heat during the transmission? (b) What must be the turn ratio of a transformer at the delivery point in order to provide an output voltage of \(220 \mathrm{~V} ?\) (Neglect the voltage drop in the line.)

The plane of a conductive loop with an area of \(0.020 \mathrm{~m}^{2}\) is perpendicular to a uniform magnetic field of \(0.30 \mathrm{~T}\). If the field drops to zero in \(0.0045 \mathrm{~s}\), what is the magnitude of the average emf induced in the loop?

A metal airplane with a wingspan of \(30 \mathrm{~m}\) flies horizontally along a north-south route in the northern hemisphere at a constant speed of \(320 \mathrm{~km} / \mathrm{h}\) in a region where the vertical component of the Earth's magnetic field is \(5.0 \times 10^{-5} \mathrm{~T}\). (a) What is the magnitude of the induced emf between the tips of its wings? (b) If the easternmost wing tip is negatively charged, is the plane flying due north or due south? Explain.

An ideal transformer steps \(8.0 \mathrm{~V}\) up to \(2000 \mathrm{~V}\), and the 4000-turn secondary coil carries 2.0 A. (a) Find the number of turns in the primary coil. (b) Find the current in the primary coil.

An ac generator supplies \(20 \mathrm{~A}\) at \(440 \mathrm{~V}\) to a \(10000-\mathrm{V}\) power line through a step-up transformer that has 150 turns in its primary coil. (a) If the transformer is \(95 \%\) efficient (see Exercise 32), how many turns are in the secondary coil? (b) What is the current in the power line?

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.