/*! 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 14 A loop of wire has the shape sho... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 22: Problem 14

A loop of wire has the shape shown in the drawing. The top part of the wire is bent into a semicircle of radius \(r=0.20 \mathrm{m}\) . The normal to the plane of the loop is parallel to a constant magnetic field \(\left(\phi=0^{\circ}\right)\) of magnitude 0.75 \(\mathrm{T}\) . What is the change \(\Delta \Phi\) in the magnetic flux that passes through the loop when, starting with the position shown in the drawing, the semicircle is rotated through half a revolution?

Short Answer

Expert verified
The change in magnetic flux is -0.0942 Wb.

Step by step solution

01

Understanding Magnetic Flux

Magnetic flux is defined as the product of the magnetic field, the area it penetrates, and the cosine of the angle between the field and the normal to the area. The formula is \( \Phi = B \cdot A \cdot \cos(\theta) \), where \(B\) is the magnetic field, \(A\) is the area, and \(\theta\) is the angle between the field and the normal to the loop.
02

Initial Position Calculation

Initially, the normal to the plane of the semicircle (\phi=0^{\circ}) is parallel to the magnetic field, so \( \theta = 0^{\circ} \). The magnetic flux is \( \Phi_1 = B \cdot A \cdot \cos(0) = B \cdot A \).
03

Area of the Semicircle

The area of the semicircle is half the area of a full circle. The formula for the area of a full circle is \( \pi r^2 \), so the area of the semicircle is \( A = \frac{1}{2} \pi r^2 \). Substitute \( r = 0.20 \text{ m} \) to get \( A = \frac{1}{2} \pi (0.20)^2 \approx 0.0628 \text{ m}^2 \).
04

Initial Magnetic Flux

Substitute \( B = 0.75 \text{ T} \) and \( A \approx 0.0628 \text{ m}^2 \) to find \( \Phi_1 = 0.75 \cdot 0.0628 = 0.0471 \text{ Wb} \).
05

Final Position Calculation

After a half revolution (180掳 or \( \theta = 180^{\circ} \)), the normal is opposite to the magnetic field, thus \( \cos(180^{\circ}) = -1 \). The magnetic flux in this position is \( \Phi_2 = B \cdot A \cdot \cos(180) = -B \cdot A \).
06

Final Magnetic Flux

Using the same area and magnetic field, \( \Phi_2 = -0.75 \cdot 0.0628 = -0.0471 \text{ Wb} \).
07

Change in Magnetic Flux

The change in magnetic flux, \( \Delta \Phi \), is \( \Phi_2 - \Phi_1 = -0.0471 \text{ Wb} - 0.0471 \text{ Wb} = -0.0942 \text{ Wb} \).

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.

Semicircle
A semicircle is a shape formed by dividing a circle into two equal halves. In this exercise, the loop of wire forms a semicircle, which plays a crucial role in calculating the magnetic flux.
The radius of the semicircle is given as 0.20 meters. Understanding the properties of a circle helps in determining the semicircle's parameters, including its area.
With the semicircle, only half of the circle's area contributes to the flux calculation through the loop. This understanding of the semicircle's geometric properties simplifies the subsequent steps in determining the magnetic flux changes.
Magnetic Field
The magnetic field is represented by the symbol \( B \) and has a magnitude given as 0.75 Tesla (T) in this exercise. A magnetic field exerts a force on charged particles in its vicinity and is a fundamental concept in electromagnetism.
In this problem, the magnetic field is constant, meaning that its strength and direction do not change over time. It runs parallel to the normal of the loop in its starting position. This information indicates that the influence on the magnetic flux depends solely on the loop's orientation change, not on the magnetic field's alteration.
Understanding these properties helps elucidate why the magnetic field remains a constant factor in the flux calculation formula.
Area Calculation
Calculating the area is an essential step in determining the magnetic flux passing through the loop. Since the loop is a semicircle, understanding how to find its area is vital.
The area of a full circle is calculated using the formula \( \pi r^2 \), with \( r \) being the radius. For a semicircle, this area is halved, leading to the formula \( A = \frac{1}{2} \pi r^2 \).
Substituting the given radius of 0.20 meters into this formula results in an area of approximately 0.0628 square meters. This numerical value is crucial, as it feeds directly into the calculation of the magnetic flux through the loop. Understanding the area calculation ensures precise results within the exercise.
Flux Change
Flux change is a key aspect of this exercise, defined as the difference between the initial and final magnetic flux through a loop. The magnetic flux, \( \Phi \), is dependent on the magnetic field \( B \), area \( A \), and the cosine of the angle \( \theta \) between the field and the loop's normal.
Initially, the angle \( \theta \) is 0 degrees, meaning the normal is aligned with the magnetic field. This results in a positive cosine value (1), leading to an initial flux \( \Phi_1 \) of 0.0471 Weber (Wb).
When the semicircle rotates through half a revolution (180 degrees), the angle becomes \( \theta = 180^{\circ} \), and the flux \( \Phi_2 \) becomes negative, due to the cosine value of -1.
The change in magnetic flux, \( \Delta \Phi \), is thus \( \Phi_2 - \Phi_1 = -0.0942 \) Weber. Understanding flux change involves recognizing how the loop's orientation affects magnetic interactions.

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 resistances of the primary and secondary coils of a transformer are 56 and \(14 \Omega,\) respectively. Both coils are made from length of the same copper wire. The circular turns of each coil have the same diameter. Find the turns ratio \(N_{s} / N_{p} .\)

A copper rod is sliding on two conducting rails that make an angle of 19 with respect to each other, as in the drawing. The rod is moving to the right with a constant speed of 0.60 \(\mathrm{m} / \mathrm{s} .\) A \(0.38-\mathrm{T}\) uniform magnetic field is perpendicular to the plane of the paper. Determine the magnitude of the average emf induced in the triangle \(A B C\) during the 6.0 -s period after the rod has passed point \(A\) .

A square loop of wire consisting of a single turn is perpendicular to a uniform magnetic field. The square loop is then re-formed into a circular loop, which also consists of a single turn and is also perpendicular to the same magnetic field. The magnetic flux that passes- through the square loop is \(7.0 \times 10^{-3}\) Wb. What is the flux that passes through the circular loop?

A generator uses a coil that has 100 turns and a \(0.50-\) T magnetic field. The frequency of this generator is 60.0 \(\mathrm{Hz}\) , and its emf has an rms value of 120 \(\mathrm{V}\) . Assuming that each turn of the coil is a square (an approximation), determine the length of the wire from which the coil is made.

Suppose there are two transformers between your house and the high-voltage transmission line that distributes the power. In addition, assume that your house is the only one using electric power. At a substation the primary coil of a step-down transformer (turms ratio \(=1 : 29\) ) receives the voltage from the high-voltage transmission line. Because of your usage, a current of 48 mA exists in the primary coil of this transformer. The secondary coil is connected to the primary of another step-down transformer (turns ratio \(=1 : 32 )\) somewhere near your house, perhaps up on a telephone pole. The secondary coil of this transformer delivers a \(240-\mathrm{V}\) emf to your house. How much power is your house using? Remember that the current and voltage given in this problem are rms values.
See all solutions

Recommended explanations on Physics Textbooks

Oscillations

Read Explanation

Astrophysics

Read Explanation

Particle Model of Matter

Read Explanation

Circular Motion and Gravitation

Read Explanation

Fields in Physics

Read Explanation

Atoms and Radioactivity

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.