/*! 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 87 A square wire loop \(20 \mathrm{... [FREE SOLUTION] | 91Ó°ÊÓ

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A square wire loop \(20 \mathrm{~cm}\) on a side, with resistance \(20 \mathrm{~m} \Omega\), has its plane normal to a uniform magnetic field of magnitude \(B=2.0 \mathrm{~T}\). If you pull two opposite sides of the loop away from each other, the other two sides automatically draw toward each other, reducing the area enclosed by the loop. If the area is reduced to zero in time \(\Delta t=0.20 \mathrm{~s}\), what are (a) the average emf and (b) the average current induced in the loop during \(\Delta t\) ?

Short Answer

Expert verified
Average emf is 0.40 V, and average current is 20 A.

Step by step solution

01

Understanding the problem

We have a square wire loop with side length 20 cm and resistance 20 mΩ, placed in a magnetic field of 2.0 T. The area of the loop is reduced to zero in 0.20 s, and we need to calculate the average emf and current induced in the loop during this process.
02

Calculate the initial area of the loop

The side length of the loop is 20 cm, so the initial area of the loop is \( A = (20 \, \text{cm})^2 \). Converting to meters, we have \( A = (0.20 \, \text{m})^2 = 0.04 \, \text{m}^2 \).
03

Calculate the change in magnetic flux

The magnetic flux \( \Phi \) is given by \( \Phi = B \times A \), where \( B = 2.0 \, \text{T} \). Initially, \( \Phi_i = 2.0 \, \text{T} \times 0.04 \, \text{m}^2 = 0.08 \, \text{Wb} \). Finally, the area is zero, so \( \Phi_f = 0 \, \text{Wb} \). The change in flux is \( \Delta \Phi = \Phi_f - \Phi_i = 0 - 0.08 = -0.08 \, \text{Wb} \).
04

Calculate the average induced emf

The average emf is given by Faraday's Law of Induction: \( \varepsilon_{\text{avg}} = -\frac{\Delta \Phi}{\Delta t} \). Substituting the values, \( \varepsilon_{\text{avg}} = -\frac{-0.08 \, \text{Wb}}{0.20 \, \text{s}} = 0.40 \, \text{V} \).
05

Calculate the average current induced in the loop

Using Ohm's Law, the average current \( I_{\text{avg}} \) is given by \( I_{\text{avg}} = \frac{\varepsilon_{\text{avg}}}{R} \), where \( R = 20 \, \text{m} \Omega = 20 \times 10^{-3} \, \Omega \). Thus, \( I_{\text{avg}} = \frac{0.40 \, \text{V}}{20 \times 10^{-3} \, \Omega} = 20 \, \text{A} \).

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Key Concepts

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

Faraday's Law of Induction
When we talk about electromagnetic induction, one of the key principles is Faraday's Law of Induction. This law explains how a changing magnetic environment can produce an electric current or electromotive force (emf) within a conductor. In simpler terms, if the magnetic field around a wire loop changes, it can "induce" an electrical signal in the loop. This is precisely what was happening in the given exercise.

Faraday's Law can be mathematically expressed as: \[\varepsilon = - \frac{d\Phi}{dt} \]where:
  • \(\varepsilon\) is the induced emf
  • \(d\Phi\) is the change in magnetic flux
  • \(dt\) is the change in time
The negative sign in the equation is significant. According to Lenz's Law, it indicates that the induced emf will always work in a direction to oppose the change in magnetic flux. In the exercise, as the area of the loop was reduced, the magnetic flux decreased from its initial value of 0.08 Wb to zero, leading to an induced emf of 0.40 V. This example illustrates Faraday's Law in action.
Ohm's Law
Ohm's Law is another fundamental concept in electrical circuits. It relates voltage, current, and resistance within an electric circuit. The basic idea is that, if you know two of these quantities, you can easily find the third. Ohm's Law is expressed as:\[I = \frac{\varepsilon}{R} \]where:
  • \(I\) is the current through the conductor
  • \(\varepsilon\) is the voltage across the conductor
  • \(R\) is the resistance of the conductor
In the problem, once we found the average induced emf to be 0.40 V and knew the resistance of the loop was \(20 \, \text{m}\Omega\), using Ohm's Law was straightforward. By dividing the emf by the resistance, we computed the average induced current, which turned out to be 20 A.

This demonstrates how Ohm's Law helps bridge the gap between knowing just the electrical characteristics and determining how an electrical circuit behaves.
Magnetic Flux
Magnetic flux is a concept that quantifies the amount of magnetic field passing through an area. It's fundamental to understanding electromagnetic induction. Magnetic flux (\(\Phi\)) depends on several factors:
  • Strength of the magnetic field \(B\)
  • Area the field is passing through \(A\)
  • The angle between the field lines and the perpendicular to the surface
It is given by the formula:\[\Phi = B \times A \times \cos(\theta) \]For simplicity, when the magnetic field is perpendicular to the surface (as in our exercise, \(\theta = 0\)), \(\cos(0) = 1\) and the formula simplifies to \(\Phi = B \times A\).

In the exercise, the magnetic flux initially was \(0.08 \, \text{Wb}\), calculated using \(B = 2.0 \, \text{T}\) and \(A = 0.04 \, \text{m}^2\). Once the area reduced to zero, the magnetic flux also became zero, illustrating how crucial the size of the loop's area is in determining the flux. This change in magnetic flux over time is what generated the emf according to Faraday's Law, showing the interconnectedness of these core concepts.

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Most popular questions from this chapter

Two solenoids are part of the spark coil of an automobile. When the current in one solenoid falls from \(6.0 \mathrm{~A}\) to zero in \(2.5 \mathrm{~ms}\), an emf of \(30 \mathrm{kV}\) is induced in the other solenoid. What is the mutual inductance \(M\) of the solenoids?

Shows a closed loop of wire that consists of a pair of equal semicircles, of radius \(3.7 \mathrm{~cm}\), lying in mutually perpendicular planes. The loop was formed by folding a flat circular loop along a diameter until the two halves became perpendicular to each other. A uniform magnetic field \(\vec{B}\) of magnitude \(76 \mathrm{~m} \mathrm{~T}\) is directed perpendicular to the fold diameter and makes equal angles ( of \(45^{\circ}\) ) with the planes of the semicircles. The magnetic field is reduced to zero at a uniform rate during a time interval of \(4.5\) ms. During this interval, what are the (a) magnitude and (b) direction (clockwise or counterclockwise when viewed along the direction of \(\vec{B}\) ) of the emf induced in the loop?

Two straight conducting rails form a right angle. A conducting bar in contact with the rails starts at the vertex at time \(t=0\) and moves with a constant velocity of \(5.20 \mathrm{~m} / \mathrm{s}\) along them. A magnetic field with \(B=0.350 \mathrm{~T}\) is directed out of the page. Calculate (a) the flux through the triangle formed by the rails and bar at \(t=3.00 \mathrm{~s}\) and \((\mathrm{b})\) the emf around the triangle at that time. (c) If the emf is \(\mathscr{8}=a t^{n}\), where \(a\) and \(n\) are constants, what is the value of \(n ?\)

Coil 1 has \(L_{1}=25 \mathrm{mH}\) and \(N_{1}=100\) turns. Coil 2 has \(L_{2}=40\) \(\mathrm{mH}\) and \(N_{2}=200\) turns. The coils are fixed in place; their mutual inductance \(M\) is \(3.0 \mathrm{mH}\). A \(6.0 \mathrm{~mA}\) current in coil 1 is changing at the rate of \(4.0 \mathrm{~A} / \mathrm{s}\). (a) What magnetic flux \(\Phi_{12}\) links coil 1, and \((\mathrm{b})\) what self-induced emf appears in that coil? (c) What magnetic flux \(\Phi_{21}\) links coil 2 , and \((\mathrm{d})\) what mutually induced emf appears in that coil?

A small loop of area \(6.8 \mathrm{~mm}^{2}\) is placed inside a long solenoid that has 854 turns \(/ \mathrm{cm}\) and carries a sinusoidally varying current \(i\) of amplitude \(1.28\) A and angular frequency \(212 \mathrm{rad} / \mathrm{s}\). The central axes of the loop and solenoid coincide. What is the amplitude of the emf induced in the loop?

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