/*! 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 16 A square loop of wire consisting... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

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 \(7.0 \times 10^{-3} \mathrm{Wb}\). What is the flux that passes through the circular loop?

Short Answer

Expert verified
The flux through the circular loop is approximately \(8.9 \times 10^{-3} \, \text{Wb}\).

Step by step solution

01

Understanding the Problem

We need to find the magnetic flux through a circular loop when we know the flux through a square loop made from the same wire in the same magnetic field.
02

Identify Key Variables

Let \( B \) be the magnetic field, \( \Phi_s = 7.0 \times 10^{-3} \, \text{Wb} \) be the flux through the square loop. The wire used is constant, so the perimeter of both the square and circle is the same.
03

Calculate Side Length of Square

Let \( s \) be the side length of the square. Since the flux \( \Phi_s = B \cdot A_s \), where \( A_s = s^2 \), we will use the perimeter condition: perimeter \( = 4s = 2\pi r \).
04

Determine Area of Square

The area of the square is \( A_s = s^2 \). Since \( B = \frac{\Phi_s}{A_s} \), we have \( Bs^2 = 7.0 \times 10^{-3} \, \text{Wb} \).
05

Relate Perimeters to Find Radius

Using the perimeter equation, we have \( 4s = 2\pi r \). Solving for \( s \), gives \( s = \frac{\pi r}{2} \), so \( s = \frac{7.0 \times 10^{-3}}{B} \frac{1}{s^2} \).
06

Calculate Area of Circle

The area of the circle is \( A_c = \pi r^2 \). From the equal perimeter, \( r = \frac{2s}{\pi} \). Thus, \( A_c = \pi \left(\frac{2s}{\pi}\right)^2 = \frac{4s^2}{\pi} \).
07

Calculate Flux in Circular Loop

Using \( \Phi_c = B \cdot A_c \), substitute \( A_c \) to find \( \Phi_c = B \cdot \frac{4s^2}{\pi} \). Substituting \( s^2 = \frac{\Phi_s}{B} \), \( \Phi_c = \frac{4 \Phi_s}{\pi} = \frac{4 \times 7.0 \times 10^{-3}}{\pi} \, \text{Wb} \).
08

Calculate Final Answer

Solve for \( \Phi_c \) to get the numeric result. The flux through the circular loop is approximately \( 8.9 \times 10^{-3} \, \text{Wb} \).

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 91Ó°ÊÓ!

Key Concepts

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

Square Loop
A square loop is a shape that has four equal sides and four right angles. Imagine a piece of wire bent to form a square. This configuration is interesting in electromagnetic problems because it often makes calculations easier. Consider that each side of the square loop has the same length, which we'll call \(s\). This means the total length of wire, or the perimeter of the square, is described by the equation
  • \( 4s \)
The perimeter is essential for comparing the square loop with other shapes, like a circular loop, that might have the same wire length.
The **area** of the square, which affects the magnetic flux, is calculated as:
  • \( A_s = s^2 \)
This formula shows how the magnetic field interacts with the square loop. The larger the area, the more magnetic flux (\( \Phi_s \)) can pass through.
Circular Loop
A circular loop is formed by bending the same piece of wire used for the square into a circular shape. Unlike a square, a circle's perimeter relates to a radius \(r\) instead of a side length. The perimeter (or circumference) of the circle is given by:
  • \( 2\pi r \)
Since the wire length is unchanged, its perimeter remains constant when reshaped from a square to a circle.
To determine the circle’s area \( A_c \), which is crucial for calculating magnetic flux, you use:
  • \( A_c = \pi r^2 \)
The circular loop allows a different interaction with the magnetic field due to its shape, altering the flux \( \Phi_c \) despite the unchanged wire length.
Perimeter Equation
The perimeter equation is central to converting shapes while keeping the wire length constant. For a square loop, the perimeter equation is:
  • \( 4s \)
For a circular loop, the equation is:
  • \( 2\pi r \)
Both shapes have crucial relationships through these equations. They tell us that when the square loop is perfectly transformed into a circle, the perimeter stays the same. Thus, setting:
  • \( 4s = 2\pi r \)
This balance allows you to solve for unknowns like the side length \(s\) or radius \(r\), which in turn helps in determining how the wire's shape affects the magnetic flux through each form.
Magnetic Field
The magnetic field, represented as \(B\), influences how much magnetic flux passes through a loop, whether square or circular. This field is uniform, meaning its strength and direction are consistent throughout the area we are examining. The magnetic flux (\( \Phi \)) through a loop is determined by the equation:
  • \( \Phi = B \cdot A \)
Here, \(A\) represents the area of the loop. For the square loop:
  • \( \Phi_s = B \cdot s^2 \)
And for the circular loop:
  • \( \Phi_c = B \cdot \frac{4s^2}{\pi} \)
Understanding \(B\)'s role highlights how the structure's area directly affects the magnetic flux experienced by different loop configurations, even when the length of the wire remains unchanged.

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

In each of two coils the rate of change of the magnetic flux in a single loop is the same. The emf induced in coil \(1,\) which has 184 loops, is \(2.82 \mathrm{V}\). The emf induced in coil 2 is 4.23 V. How many loops does coil 2 have?

A standard door into a house rotates about a vertical axis through one side, as defined by the door's hinges. A uniform magnetic field is parallel to the ground and perpendicular to this axis. Through what angle must the door rotate so that the magnetic flux that passes through it decreases from its maximum value to one-third of its maximum value?

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.

Two coils of wire are placed close together. Initially, a current of 2.5 A exists in one of the coils, but there is no current in the other. The current is then switched off in a time of \(3.7 \times 10^{-2}\) s. During this time, the average emf induced in the other coil is \(1.7 \mathrm{V}\). What is the mutual inductance of the two-coil system?

A constant magnetic field passes through a single rectangular loop whose dimensions are \(0.35 \mathrm{m} \times 0.55 \mathrm{m} .\) The magnetic field has a magnitude of \(2.1 \mathrm{T}\) and is inclined at an angle of \(65^{\circ}\) with respect to the normal to the plane of the loop. (a) If the magnetic field decreases to zero in a time of \(0.45 \mathrm{s},\) what is the magnitude of the average emf induced in the loop? (b) If the magnetic field remains constant at its initial value of \(2.1 \mathrm{T},\) what is the magnitude of the rate \(\Delta A / \Delta t\) at which the area should change so that the average emf has the same magnitude as in part (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.