/*! 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 1 A single loop of wire with an ar... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

A single loop of wire with an area of \(0.0900 \mathrm{~m}^{2}\) is in a uniform magnetic field that has an initial value of \(3.80 \mathrm{~T},\) is perpendicular to the plane of the loop, and is decreasing at a constant rate of \(0.190 \mathrm{~T} / \mathrm{s}\)(a) What emf is induced in this loop? (b) If the loop has a resistance of \(0.600 \Omega,\) find the current induced in the loop.

Short Answer

Expert verified
The emf induced in the loop is 0.0171 V and the current induced in the loop is 0.0285 A. The current is in a direction to support the decreasing magnetic field.

Step by step solution

01

Identify Given Variables

First, identify given variables. The area \(A\) of the loop is \(0.0900 \mathrm{~m}^{2}\), initial magnetic field \(B_{0}\) is \(3.80 \mathrm{~T}\), rate of change of magnetic field \(\frac{dB}{dt}\) is \(0.190 \mathrm{~T/s}\), and resistance \(R\) of the loop is \(0.600 \Omega\).
02

Calculate Induced EMF

Use Faraday's law to find the induced emf: \[ǀ\(\epsilon\) ǀ = ǀ\(-N \frac{\Delta \Phi}{\Delta t}\) ǀ\]. Here, \(N\) is the number of turns in the loop, \(\Delta \Phi\) is the change in magnetic flux and \(\Delta t\) is the change in time. Since there is only one turn (single loop) and the magnetic field is decreasing perpendicularly to the loop, the change in flux can be represented as \(\Delta \Phi = A \Delta B\), where \(A\) is the area of the loop and \(\Delta B) is change in magnetic field. Substituting and rearranging, we get \[ǀ\(\epsilon\) ǀ = A ǀ \(\frac{\Delta B}{\Delta t}\) ǀ\]. Substituting with given values, we get \[ǀ\(\epsilon\) ǀ = 0.0900 *(0.190)\] m2/Ts = 0.0171 V.
03

Calculate Induced Current

Next, use Ohm's law to find the induced current: \(I=\frac{V}{R}\), where \(V\) is voltage (the induced EMF in this case) and \(R\) is resistance. Substituting with given values, we find \(I=\frac{0.0171}{0.600} = 0.0285 \mathrm{~A}\).
04

Determine Direction of Current

To find the direction of the induced current in the loop, apply Lenz's law. Since the magnetic field is decreasing, the induced current generates a field to support the decreasing field. So, the direction of the induced current is such that it creates a magnetic field in the same direction as the original field. This part is not quantitatively calculable from the given data, but is a significant detail to note.

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.

Faraday's Law of Electromagnetic Induction
Faraday's Law of Electromagnetic Induction is the scientific principle that explains how electric current can be generated in a conductor through a changing magnetic field. The law mathematically expresses the relationship between the induced electromotive force (EMF) in a coil and the rate at which the magnetic flux through that coil is changing.

To use Faraday's Law for induced EMF calculation, you need to apply the following formula: \[ |\epsilon| = |N \frac{\Delta \Phi}{\Delta t}| \] where \(N\) is the number of turns in the coil, \(\Delta \Phi\) is the change in magnetic flux, and \(\Delta t\) is the time over which this change occurs. The negative sign, often ignored in magnitude calculations, indicates the direction of the induced EMF and ultimately the direction of the current flow based on Lenz's Law, which we will discuss shortly.

Breaking this down, if the magnetic environment of a loop changes—be it from a moving magnet or a changing current inducing a magnetic field—the loop experiences an induced EMF. As the exercise demonstrates, knowing the area of the loop and the rate of change of the magnetic field allows you to calculate the EMF produced.
Magnetic Flux
Magnetic flux is a measure of the quantity of magnetism, considering the strength and extent of a magnetic field. It can be thought of as the 'amount' of magnetic field lines passing through a given area, and mathematically, it is defined as the product of the magnetic field strength, the area it penetrates, and the cosine of the angle between the magnetic field and the normal (perpendicular) to the surface: \[ \Phi = B \cdot A \cdot \cos(\theta) \]

In our case, since the magnetic field is perpendicular to the plane of the loop, \(\theta = 0\) degrees, and \(\cos(\theta) = 1\). Thus, the calculation simplifies to \(\Phi = B \cdot A\). When the flux changes, either through a change in the magnetic field strength (\(B\)) over time or via the movement of the loop relative to the magnetic field, an EMF is induced in the loop. This change in magnetic flux is at the heart of Faraday's law and is directly proportional to the induced EMF.
Ohm's Law
Ohm's Law is the relationship between voltage (V), current (I), and resistance (R) in an electrical circuit: it states that the current through a conductor between two points is directly proportional to the voltage across the two points, and inversely proportional to the resistance between them. Mathematically, this is expressed as: \[ I = \frac{V}{R} \]

Ohm's Law is fundamental to the study of electricity and is crucial for solving various problems, like finding the current resulting from an induced EMF in a circuit. With the EMF calculated using Faraday's Law, you just divide this value by the circuit's resistance to determine the current flow. In the exercise, after calculating the induced EMF, Ohm's Law is used to calculate the induced current by knowing the resistance of the wire loop.
Lenz's Law
Lenz's Law is an essential concept that complements Faraday's Law and clarifies the direction of the induced EMF and current in response to a changing magnetic flux. It states that the direction of any induced current will be such that the magnetic field it creates opposes the change that produces it. In simpler terms, an induced current will always work against the change in flux—this is nature's way of resisting changes to the status quo.

In the exercise context, since the magnetic field is decreasing, Lenz's Law tells us that the induced current will flow in a direction that attempts to keep the magnetic field strength constant. While the law does not give us a way to calculate the magnitude of the effect, it provides the necessary insight to predict the direction of current flow, which is a critical piece of understanding in analyzing electromagnetic phenomena.

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 battery has emf \(24.0 \mathrm{~V}\) and internal resistance \(3.00 \Omega .\) A resistor of resistance \(R\) is connected to the battery. What are the two values of \(R\) for which \(21.0 \mathrm{~W}\) of electrical power is consumed in the resistor?

Displacement Current in a Dielectric. Suppose that the parallel plates in Fig. 29.23 have an area of \(3.00 \mathrm{~cm}^{2}\) and are separated by a 2.50 -mm- thick sheet of dielectric that completely fills the volume between the plates. The dielectric has dielectric constant \(4.70 .\) (You can ignore fringing effects.) At a certain instant, the potential difference between the plates is \(120 \mathrm{~V}\) and the conduction current \(i_{\mathrm{C}}\) equals \(6.00 \mathrm{~mA} .\) At this instant, what are (a) the charge \(q\) on each plate; (b) the rate of change of charge on the plates; (c) the displacement current in the dielectric?

(a) Estimate the maximum volume of water the hot-water heater in your home can hold. (b) How much heat would be required to raise the temperature of that water from \(20^{\circ} \mathrm{C}\) to a standard household hot-water temperature of \(45^{\circ} \mathrm{C}\) ? (Water has a density of \(1.00 \mathrm{~kg} / \mathrm{L}\) and a heat capacity of \(4190 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K} .\) ) (c) Suppose the water should be fully heated in \(1.5 \mathrm{~h}\). To what power output does this correspond? (d) If the element has a potential difference of \(220 \mathrm{~V},\) what current is required? (e) What should be the resistance of the element?

Two cylindrical cans with insulating sides and conducting end caps are filled with water, attached to the circuitry shown in Fig. \(\mathbf{P} 25.69,\) and used to determine salinity levels. The cans are identical, with radius \(r=5.00 \mathrm{~cm}\) and length \(L=3.00 \mathrm{~cm} .\) The battery supplies a potential of \(10.0 \mathrm{~V},\) has a negligible internal resistance, and is connected in series with a resistor \(R=15.0 \Omega .\) The left cylinder is filled with pure distilled water, which has infinite resistivity. The right cylinder is filled with a saltwater solution. It is known that the resistivity of the saltwater solution is determined by the relationship \(\rho=\left(s_{0} / s\right) \Omega \cdot \mathrm{m},\) where \(s\) is the salinity in parts per thousand \((\mathrm{ppt})\) and \(s_{0}=6.30\) ppt. (a) The ammeter registers a current of \(484 \mathrm{~mA}\). What is the salinity of the saltwater solution? (b) The left cylinder acts as a capacitor. Use Eq. (24.19) for its capacitance. How much charge is present on its upper plate? Note that pure water has a dielectric constant of \(80.4 .\) (c) At what rate is energy dissipated by the saltwater? (d) For what salinity level would the \(15.0 \Omega\) resistor dissipate half the power supplied by the battery?

Light Bulbs. The power rating of a light bulb (such as a \(100 \mathrm{~W}\) bulb is the power it dissipates when connected across a \(120 \mathrm{~V}\) potential difference. What is the resistance of (a) a \(100 \mathrm{~W}\) bulb and (b) a \(60 \mathrm{~W}\) bulb? (c) How much current does each bulb draw in normal use?

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.