/*! 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

BIO Treatment of Heart Failure. A heart defibrillator is used to enable the heart to start beating if it has stopped. This is done by passing a large current of 12 A through the body at \(25 \mathrm{~V}\) for a very short time, usually about \(3.0 \mathrm{~ms}\). (a) What power does the defibrillator deliver to the body, and (b) how much energy is transferred?

An incandescent light bulb uses a coiled filament of tungsten that is \(580 \mathrm{~mm}\) long with a diameter of \(46.0 \mu \mathrm{m} .\) At \(20.0^{\circ} \mathrm{C}\) tungsten has a resistivity of \(5.25 \times 10^{-8} \Omega \cdot \mathrm{m} .\) Its temperature coefficient of resistivity is \(0.0045\left(\mathrm{C}^{\circ}\right)^{-1},\) and this remains accurate even at high temperatures. The temperature of the filament increases linearly with current, from \(20^{\circ} \mathrm{C}\) when no current flows to \(2520^{\circ} \mathrm{C}\) at 1.00 A of current. (a) What is the resistance of the light bulb at \(20^{\circ} \mathrm{C} ?\) (b) What is the current through the light bulb when the potential difference across its terminals is \(120 \mathrm{~V} ?\) (Hint: First determine the temperature as a function of the current; then use this to determine the resistance as a function of the current. Substitute this result into the equation \(V=I R\) and solve for the current \(I .\) ) (c) What is the resistance when the potential is \(120 \mathrm{~V} ?\) (d) How much energy does the light bulb dissipate in 1 min when \(120 \mathrm{~V}\) is supplied across its terminals? (e) How much energy does the light bulb dissipate in 1 min when half that voltage is supplied?

A small, closely wound coil has \(N\) turns, area \(A\), and resistance \(R\). The coil is initially in a uniform magnetic field that has magnitude \(B\) and a direction perpendicular to the plane of the loop. The coil is then rapidly pulled out of the field so that the flux through the coil is reduced to zero in time \(\Delta t\). (a) What are the magnitude of the average \(\operatorname{emf} \mathcal{E}_{\text {av }}\) and average current \(I_{\mathrm{av}}\) induced in the coil? (b) The total charge \(Q\) that flows through the coil is given by \(Q=I_{\mathrm{av}} \Delta t .\) Derive an expression for \(Q\) in terms of \(N, A, B,\) and \(R .\) Note that \(Q\) does not depend on \(\Delta t .\) (c) What is \(Q\) if \(N=150\) turns, \(A=4.50 \mathrm{~cm}^{2}, R=30.0 \Omega,\) and \(B=0.200 \mathrm{~T} ?\)

A material with resistivity \(\rho\) is formed into a cylinder of length \(L\) and outer radius \(r_{\text {outer }}\). A cylindrical core with radius \(r_{\text {inner }}\) is removed from the axis of this cylinder and filled with a conducting material, which is attached to a wire. The outer surface of the cylinder is coated with a conducting material and attached to another wire. (a) If the second wire has potential \(V\) greater than the first wire, in what direction does the local electric field point inside of the cylinder? (b) The magnitude of this electric field is \(c / r,\) where \(c\) is a constant and \(r\) is the distance from the axis of the cylinder. Use the relationship \(V=\int \overrightarrow{\boldsymbol{E}} \cdot d \overrightarrow{\boldsymbol{l}}\) to determine the constant \(c .(\mathrm{c})\) What is the resistance of this device? (d) A \(1.00-\mathrm{cm}\) -long hollow cylindrical resistor has an inner radius of \(1.50 \mathrm{~mm}\) and an outer radius of \(3.00 \mathrm{~mm} .\) The material is a blend of powdered carbon and ceramic whose resistivity \(\rho\) may be altered by changing the amount of carbon. If this device should have a resistance of \(6.80 \mathrm{k} \Omega,\) what value of \(\rho\) should be selected?

Compact Fluorescent Bulbs. Compact fluorescent bulbs are much more efficient at producing light than are ordinary incandescent bulbs. They initially cost much more, but they last far longer and use much less electricity. According to one study of these bulbs, a compact bulb that produces as much light as a \(100 \mathrm{~W}\) incandescent bulb uses only \(23 \mathrm{~W}\) of power. The compact bulb lasts 10,000 hours, on the average, and costs \(\$ 11.00,\) whereas the incandescent bulb costs only \(\$ 0.75\), but lasts just 750 hours. The study assumed that electricity costs \(\$ 0.080\) per kilowatt-hour and that the bulbs are on for \(4.0 \mathrm{~h}\) per day. (a) What is the total cost (including the price of the bulbs) to run each bulb for 3.0 years? (b) How much do you save over 3.0 years if you use a compact fluorescent bulb instead of an incandescent bulb? (c) What is the resistance of a "100 W" fluorescent bulb? (Remember, it actually uses only \(23 \mathrm{~W}\) of power and operates across \(120 \mathrm{~V} .\) )

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