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Ch 30: Electromagnetic Induction
Knight Calc - Physics for Scientists and Engineers 5th Edition
Knight Calc5th EditionPhysics for Scientists and EngineersISBN: 9780137344796Not the one you use?Change textbook
Chapter 30, Problem 55b

INT A 20-cm-long, zero-resistance slide wire moves outward, on zero-resistance rails, at a steady speed of 10 m/s in a 0.10 T magnetic field. (See Figure 30.26.) On the opposite side, a 1.0 Ω carbon resistor completes the circuit by connecting the two rails. The mass of the resistor is 50 mg. How much force is needed to pull the wire at this speed?

Verified step by step guidance
1
Identify the key components of the problem: The slide wire is moving in a magnetic field, which induces an electromotive force (EMF) due to Faraday's law of electromagnetic induction. This EMF drives a current through the circuit, and the magnetic field exerts a force on the current-carrying wire. The force needed to pull the wire at constant speed must counteract this magnetic force.
Calculate the induced EMF using Faraday's law of induction. The formula for EMF is: ε = B l v, where B is the magnetic field strength (0.10 T), l is the length of the wire (0.20 m), and v is the velocity of the wire (10 m/s). Substitute these values into the formula to find the EMF.
Determine the current in the circuit using Ohm's law: I = εR, where R is the resistance of the circuit (1.0 Ω). Use the EMF calculated in the previous step to find the current.
Calculate the magnetic force acting on the wire using the formula: F = I l B. Here, I is the current, l is the length of the wire, and B is the magnetic field strength. Substitute the values to find the force.
Recognize that the force needed to pull the wire at constant speed is equal in magnitude and opposite in direction to the magnetic force. Therefore, the pulling force is equal to the magnetic force calculated in the previous step.

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Faraday's Law of Electromagnetic Induction

Faraday's Law states that a change in magnetic flux through a circuit induces an electromotive force (EMF) in the circuit. In this scenario, as the slide wire moves through the magnetic field, it cuts through magnetic lines of force, generating an EMF that drives current through the resistor. The induced EMF can be calculated using the formula EMF = B * L * v, where B is the magnetic field strength, L is the length of the wire, and v is the velocity.
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Ohm's Law

Ohm's Law relates the voltage (V), current (I), and resistance (R) in an electrical circuit, expressed as V = I * R. In this problem, the induced EMF creates a current that flows through the 1.0 Ω resistor. By applying Ohm's Law, we can determine the current produced by the induced EMF, which is essential for calculating the force required to maintain the wire's motion against the magnetic force.
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Lorentz Force

The Lorentz Force describes the force experienced by a charged particle moving through a magnetic field, given by the equation F = q(v × B). In this context, the current flowing through the wire creates a magnetic field interaction that results in a force opposing the motion of the wire. This force must be countered by an external force to maintain the wire's constant speed, making it crucial for determining the required pulling force.
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Related Practice
Textbook Question

INT A 20-cm-long, zero-resistance slide wire moves outward, on zero-resistance rails, at a steady speed of 10 m/s in a 0.10 T magnetic field. (See Figure 30.26.) On the opposite side, a 1.0 Ω carbon resistor completes the circuit by connecting the two rails. The mass of the resistor is 50 mg. If the wire is pulled for 10 s, what is the temperature increase of the carbon? The specific heat of carbon is 710 J/kg K.

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Textbook Question

CALC Your camping buddy has an idea for a light to go inside your tent. He happens to have a powerful (and heavy!) horseshoe magnet that he bought at a surplus store. This magnet creates a 0.20 T field between two pole tips 10 cm apart. His idea is to build the hand-cranked generator shown in FIGURE P30.57. He thinks you can make enough current to fully light a 1.0 Ω lightbulb rated at 4.0 W. That's not super bright, but it should be plenty of light for routine activities in the tent. Find an expression for the induced current as a function of time if you turn the crank at frequency f. Assume that the semicircle is at its highest point at t = 0 s.

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Textbook Question

CALC The L-shaped conductor in FIGURE P30.54 moves at 10 m/s across and touches a stationary L-shaped conductor in a 0.10 T magnetic field. The two vertices overlap, so that the enclosed area is zero, at t = 0 s. The conductor has a resistance of 0.010 ohms per meter. a. What is the direction of the induced current?

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Textbook Question

INT A 20-cm-long, zero-resistance slide wire moves outward, on zero-resistance rails, at a steady speed of 10 m/s in a 0.10 T magnetic field. (See Figure 30.26.) On the opposite side, a 1.0 Ω carbon resistor completes the circuit by connecting the two rails. The mass of the resistor is 50 mg. What is the induced current in the circuit?

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Textbook Question

A rectangular metal loop with 0.050 Ω resistance is placed next to one wire of the RC circuit shown in FIGURE P30.53. The capacitor is charged to 20 V with the polarity shown, then the switch is closed at t = 0 s. What is the current in the loop at t = 5.0 μs? Assume that only the circuit wire next to the loop is close enough to produce a significant magnetic field.

124
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Textbook Question

CALC Your camping buddy has an idea for a light to go inside your tent. He happens to have a powerful (and heavy!) horseshoe magnet that he bought at a surplus store. This magnet creates a 0.20 T field between two pole tips 10 cm apart. His idea is to build the hand-cranked generator shown in FIGURE P30.57. He thinks you can make enough current to fully light a 1.0 Ω lightbulb rated at 4.0 W. That's not super bright, but it should be plenty of light for routine activities in the tent. With what frequency will you have to turn the crank for the maximum current to fully light the bulb? Is this feasible?

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