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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
All textbooksKnight Calc 5th EditionCh 30: Electromagnetic InductionProblem 53b
Chapter 30, Problem 53b

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.

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Step 1: Analyze the RC circuit. The capacitor is initially charged to 20 V, and when the switch is closed at t = 0 s, the capacitor begins to discharge through the resistor. The current in the circuit can be modeled using the equation for an RC circuit: \( I(t) = \frac{V_0}{R} e^{-t/(RC)} \), where \( V_0 \) is the initial voltage, \( R \) is the resistance, \( C \) is the capacitance, and \( t \) is the time.
Step 2: Calculate the time constant \( \tau \) of the RC circuit. The time constant is given by \( \tau = RC \). Substitute \( R = 2.0 \, \Omega \) and \( C = 5.0 \mu F \) into the formula to find \( \tau \).
Step 3: Determine the current in the RC circuit at \( t = 5.0 \mu s \). Use the formula \( I(t) = \frac{V_0}{R} e^{-t/(RC)} \) and substitute \( V_0 = 20 \, V \), \( R = 2.0 \, \Omega \), \( C = 5.0 \mu F \), and \( t = 5.0 \mu s \). This gives the current in the wire next to the loop.
Step 4: Calculate the magnetic field produced by the current in the wire. The magnetic field at a distance \( r \) from a long straight wire carrying current \( I \) is given by \( B = \frac{\mu_0 I}{2 \pi r} \), where \( \mu_0 \) is the permeability of free space (\( \mu_0 = 4 \pi \times 10^{-7} \, T \cdot m/A \)). Use \( r = 0.5 \, cm \) (distance from the wire to the loop).
Step 5: Determine the induced current in the rectangular loop. The changing magnetic field induces an electromotive force (EMF) in the loop according to Faraday's law: \( \text{EMF} = -\frac{d\Phi_B}{dt} \), where \( \Phi_B \) is the magnetic flux. The induced current is then \( I_{loop} = \frac{\text{EMF}}{R_{loop}} \), where \( R_{loop} = 0.050 \, \Omega \). Calculate the flux change rate and use it to find the induced current.

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Key Concepts

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

RC Circuit

An RC circuit consists of a resistor (R) and a capacitor (C) connected in series or parallel. When the switch is closed, the capacitor begins to charge or discharge, affecting the current flow. The time constant, τ = RC, determines how quickly the circuit responds to changes, influencing the current at any given time.
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Induced Current and Magnetic Fields

When a current flows through a wire, it generates a magnetic field around it. If a conductive loop is placed near this wire, the changing magnetic field can induce a current in the loop according to Faraday's Law of Electromagnetic Induction. The magnitude of the induced current depends on the rate of change of the magnetic field and the resistance of the loop.
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Ohm's Law

Ohm's Law states that the current (I) through a conductor between two points is directly proportional to the voltage (V) across the two points and inversely proportional to the resistance (R) of the conductor. It is expressed as I = V/R. This principle is essential for calculating the current in the loop once the induced voltage is determined.
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Related Practice
Textbook Question

CALC An electric generator has an 18-cm-diameter, 120-turn coil that rotates at 60 Hz in a uniform magnetic field that is perpendicular to the rotation axis. What magnetic field strength is needed to generate a peak voltage of 170 V?

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Textbook Question

A small, 2.0-mm-diameter circular loop with R = 0.020 Ω is at the center of a large 100-mm-diameter circular loop. Both loops lie in the same plane. The current in the outer loop changes from +1.0 A to −1.0 A in 0.10 s. What is the induced current in the inner loop?

Textbook Question

FIGURE P30.48 shows two 20-turn coils tightly wrapped on the same 2.0-cm-diameter cylinder with 1.0-mm-diameter wire. The current through coil 1 is shown in the graph. Determine the current in coil 2 at (a) t = 0.05 s and (b) t = 0.25 s. A positive current is into the figure at the top of a loop. Assume that the magnetic field of coil 1 passes entirely through coil 2.

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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. How much force is needed to pull the wire at this speed?

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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