/*! 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 28 A \(0.650-\mathrm{m}\) -long met... [FREE SOLUTION] | 91Ó°ÊÓ

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A \(0.650-\mathrm{m}\) -long metal bar is pulled to the right at a steady \(5.0 \mathrm{~m} / \mathrm{s}\) perpendicular to a uniform, \(0.750 \mathrm{~T}\) magnetic field. The bar rides on parallel metal rails connected through a \(25.0 \Omega\) resistor (Fig. \(\mathbf{E} 29.28),\) so the apparatus makes a complete circuit. Ignore the resistance of the bar and the rails. (a) Calculate the magnitude of the emf induced in the circuit. (b) Find the direction of the current induced in the circuit by using (i) the magnetic force on the charges in the moving bar; (ii) Faraday's law; (iii) Lenz's law. (c) Calculate the current through the resistor.

Short Answer

Expert verified
The magnitude of the induced emf is \(2.4375 V\), the direction of the current is to the left, and the current through the resistor is \(0.0975 A\).

Step by step solution

01

Calculate induced emf

The magnitude of the emf induced in the circuit can be calculated using Faraday's law of electromagnetic induction, which states that the induced emf is equal to the rate of change of magnetic flux. In this case, since the bar is moving at a constant speed in a uniform magnetic field, the flux is changing linearly with time. We can write this as \(ε = B l v\), where \(B = 0.750 T\) is the magnetic field strength, \(l = 0.650 m\) is the length of the bar, and \(v = 5.0 m/s\) is the speed at which the bar is being pulled. Plugging these values into the equation gives \(ε = (0.750 T)(0.650 m)(5.0 m/s) = 2.4375 V\).
02

Find direction of induced current

The direction of the induced current can be found using several methods: (i) The magnetic force on the charges in the moving bar, (ii) Faraday's law, (iii) Lenz's law. Each one ultimately would tell you that the induced current direction is such that it creates a counter magnetic field to oppose the change in the original magnetic field. In this case, since the bar is being pulled to the right, the counteracting magnetic field (and thus, the current that creates it) should be in the opposite direction. Therefore, the direction of the induced current will be to the left.
03

Calculate the current through the resistor

The current \(I\) through the \(25.0 Ω\) resistor can be calculated by dividing the induced emf by the resistance. This is based on Ohm's law (\(V = I R\)), which gives \(I = ε / R\). Substituting the values gives \(I = (2.4375 V) / (25.0 Ω) = 0.0975 A\).

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Key Concepts

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

Faraday's Law
Faraday's Law is fundamental in understanding how electric currents can be generated by changing magnetic environments. It states that an electromotive force (EMF) is induced in a conducting loop when the magnetic flux through the loop changes. Mathematically, it's expressed as \( \epsilon = -\frac{d\Phi_B}{dt} \), where \( \epsilon \) is the induced EMF, and \( \Phi_B \) is the magnetic flux. The negative sign symbolizes Lenz's law, indicating the direction of the induced EMF acts to oppose the change in flux.
In practice, moving a conductor, like a metal bar, through a magnetic field or altering the field's strength can change the magnetic flux and, as such, induce an EMF. This principle underlies the operation of electric generators and transformers.
Lenz's Law
Lenz's Law, introduced by Heinrich Lenz in the 19th century, provides a rule for determining the direction of induced currents. This law complements Faraday's Law and can be summarized as: the current induced in a circuit due to a change in magnetic flux will circulate in a direction that creates a magnetic field opposing the change. It is the reason for the negative sign in Faraday's Law.
When applying Lenz's law to the example of the moving metal bar, it implies that if the bar is pulled to the right and the original magnetic field points upwards, the induced current will flow in such a way to create its magnetic field pointing downwards to counter the increase in magnetic flux.
Ohm's Law
Ohm's law is a foundational principle in electric circuits, named after German physicist Georg Simon Ohm. It relates the voltage (V), current (I), and resistance (R) in a simple and direct equation \( V = IR \). This implies that for a given resistance, the current flowing through a circuit is directly proportional to the voltage applied across it.
Using Ohm's law, we can deduce that the current through the resistor in the metal bar problem is a result of the induced EMF divided by the resistance. It demonstrates how an induced electrical potential can create a current flow in the presence of a conductive path, with the amount of current dependent on the path's resistance.
Magnetic Flux
Magnetic flux (\( \Phi_B \)) represents the quantity of magnetic field (B) passing orthogonally through a certain area (A). It can be visualized as the number of magnetic field lines going through a loop. The formula to calculate magnetic flux is \( \Phi_B = B \cdot A \cdot \cos(\theta) \), where \( \theta \) is the angle between the magnetic field and the normal to the area.
In scenarios like a bar in a magnetic field, moving the bar alters the area enclosed by the circuit, which in turn changes the flux. A difference in flux plays a crucial role in inducing an EMF and, consequently, an electrical current.
Induced EMF
The induced electromotive force (EMF) is the voltage generated in a circuit when the magnetic flux changes. According to Faraday's Law, an EMF is produced when a conductor, like our metal bar, is moved through a magnetic field, causing a change in the amount of magnetic flux through the circuit. This induced EMF is what drives the current if a closed circuit is present.
In the example problem, the induced EMF is calculated using the expression \( \epsilon = B l v \) where B is the magnetic field strength, l is the length of the moving conductor, and v is its velocity. This provides the basis for calculating how much potential is generated to push electrons around the circuit.
Induced Current
Induced current is the result of an induced EMF within a closed circuit. The magnitude and direction of this current depend on the change in magnetic flux and the characteristics of the circuit, like resistance. The EMF sets electrons in motion against the resistive forces present in the circuit, creating a flow of electric charge.
Applying this concept to our metal bar example, once we know the induced EMF and the resistance of the circuit, we can calculate the induced current's strength using Ohm's law. The direction of this current, as predicted by Lenz's law, will oppose the change in magnetic flux, ensuring energy conservation within the system.

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Most popular questions from this chapter

According to the U.S. National Electrical Code, copper wire used for interior wiring of houses, hotels, office buildings, and industrial plants is permitted to carry no more than a specified maximum amount of current. The table shows values of the maximum current \(I_{\max }\) for several common sizes of wire with varnished cambric insulation. The "wire gauge" is a standard used to describe the diameter of wires. Note that the larger the diameter of the wire, the smaller the wire gauge. $$ \begin{array}{ccc} \text { Wire gauge } & \text { Diameter }(\mathrm{cm}) & I_{\max }(\mathrm{A}) \\\ \hline 14 & 0.163 & 18 \\ 12 & 0.205 & 25 \\ 10 & 0.259 & 30 \\ 8 & 0.326 & 40 \\ 6 & 0.412 & 60 \\ 5 & 0.462 & 65 \\ 4 & 0.519 & 85 \end{array} $$ (a) What considerations determine the maximum current-carrying capacity of household wiring? (b) A total of \(4200 \mathrm{~W}\) of power is to be supplied through the wires of a house to the household electrical appliances. If the potential difference across the group of appliances is \(120 \mathrm{~V},\) determine the gauge of the thinnest permissible wire that can be used. (c) Suppose the wire used in this house is of the gauge found in part (b) and has total length \(42.0 \mathrm{~m}\). At what rate is energy dissipated in the wires? (d) The house is built in a community where the consumer cost of electrical energy is \(\$ 0.11\) per kilowatt-hour. If the house were built with wire of the next larger diameter than that found in part (b), what would be the savings in electricity costs in one year? Assume that the appliances are kept on for an average of 12 hours a day.

A closely wound rectangular coil of 80 turns has dimensions of \(25.0 \mathrm{~cm}\) by \(40.0 \mathrm{~cm} .\) The plane of the coil is rotated from a position where it makes an angle of \(37.0^{\circ}\) with a magnetic field of \(1.70 \mathrm{~T}\) to a position perpendicular to the field. The rotation takes \(0.0600 \mathrm{~s}\). What is the average emf induced in the coil?

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?

A circular loop of wire with a radius of \(12.0 \mathrm{~cm}\) and oriented in the horizontal \(x y\) -plane is located in a region of uniform magnetic field. A field of \(1.5 \mathrm{~T}\) is directed along the positive \(z\) -direction, which is upward. (a) If the loop is removed from the field region in a time interval of \(2.0 \mathrm{~ms}\), find the average emf that will be induced in the wire loop during the extraction process. (b) If the coil is viewed looking down on it from above, is the induced current in the loop clockwise or counterclockwise?

The free-electron density in a copper wire is \(8.5 \times 10^{28}\) electrons \(/ \mathrm{m}^{3} .\) The electric field in the wire is \(0.0600 \mathrm{~N} / \mathrm{C}\) and the temperature of the wire is \(20.0^{\circ} \mathrm{C}\). (a) What is the drift speed \(v_{\mathrm{d}}\) of the electrons in the wire? (b) What is the potential difference between two points in the wire that are separated by \(20.0 \mathrm{~cm} ?\)

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