Ch. 30 - Inductance, Electromagnetic Oscillations, and AC Circuits

Giancoli Douglas - Physics for Scientists and Engineers 5th edition

Giancoli Douglas5th editionPhysics for Scientists and EngineersISBN: 9780137488179Not the one you use?Change textbook

Giancoli Douglas 5th edition

Ch. 30 - Inductance, Electromagnetic Oscillations, and AC Circuits

Problem 27b

Chapter 29, Problem 27b

Two tightly wound solenoids have the same length and circular cross-sectional area. But solenoid 1 uses wire that is 1.5 times as thick as solenoid 2. What is the ratio of their inductive time constants? (Assume the only resistance in the circuits is that of the wire itself.)

Name: Physics for Scientists and Engineers
Author: Giancoli Douglas
ISBN: 9780137488179

Verified step by step guidance

Step 1: Understand the inductive time constant formula. The inductive time constant (τ) is given by τ = L/R, where L is the inductance of the solenoid and R is the resistance of the wire. We need to compare the time constants of solenoid 1 and solenoid 2.

Step 2: Analyze the inductance (L) of a solenoid. The inductance of a solenoid is given by L = μ₀ * N² * A / l, where μ₀ is the permeability of free space, N is the number of turns, A is the cross-sectional area, and l is the length of the solenoid. Since both solenoids have the same length and cross-sectional area, their inductance depends only on the number of turns (N).

Step 3: Relate the number of turns (N) to the thickness of the wire. If solenoid 1 uses wire that is 1.5 times as thick as solenoid 2, the thicker wire will result in fewer turns per unit length. Specifically, the number of turns in solenoid 1 will be 1/1.5 times the number of turns in solenoid 2. Therefore, N₁ = N₂ / 1.5.

Step 4: Analyze the resistance (R) of the wire. The resistance of the wire is given by R = ρ * l / Aₐ, where ρ is the resistivity of the material, l is the length of the wire, and Aₐ is the cross-sectional area of the wire. Since solenoid 1 uses wire that is 1.5 times as thick, its cross-sectional area will be (1.5)² = 2.25 times larger than that of solenoid 2. Thus, R₁ = R₂ / 2.25.

Step 5: Combine the relationships for L and R to find the ratio of time constants. Using τ = L/R, the ratio of time constants τ₁/τ₂ = (L₁/R₁) / (L₂/R₂). Substitute the relationships for L and R derived earlier: τ₁/τ₂ = [(N₁² / R₁) / (N₂² / R₂)] = [(N₂² / (1.5)²) / (R₂ / 2.25)] = (1 / 1.5²) * (2.25) = 1.

Key Concepts

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

Inductance

Inductance is a property of an electrical circuit that quantifies the ability of a conductor to store energy in a magnetic field when an electric current flows through it. For solenoids, inductance depends on factors such as the number of turns, the cross-sectional area, and the permeability of the core material. The formula for the inductance of a solenoid is L = (μ₀ * N² * A) / l, where N is the number of turns, A is the cross-sectional area, and l is the length.

Resistance

Resistance is a measure of the opposition to the flow of electric current in a conductor. It is influenced by the material, length, and cross-sectional area of the wire. In the context of solenoids, a thicker wire has a lower resistance due to its larger cross-sectional area, which affects the overall time constant of the circuit, defined as τ = L/R, where τ is the time constant, L is inductance, and R is resistance.

Time Constant

The time constant (τ) of an RL circuit is the time it takes for the current to reach approximately 63.2% of its final value after a voltage is applied. It is determined by the ratio of inductance (L) to resistance (R). In this problem, the time constants of the two solenoids will differ due to their respective inductances and resistances, which are influenced by the thickness of the wire used in each solenoid.

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Textbook Question

Two tightly wound solenoids have the same length and circular cross-sectional area. But solenoid 1 uses wire that is 1.5 times as thick as solenoid 2.

(a) What is the ratio of their inductances?

(b) What is the ratio of their inductive time constants? (Assume the only resistance in the circuits is that of the wire itself.)

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Textbook Question

(II) A 25-mH coil whose resistance is 0.80 Ω is connected to a capacitor C and a 420-Hz source voltage. If the current and voltage are to be in phase, what value must C have?

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Textbook Question

(II) Suppose that the U-shaped conductor and connecting rod in Fig. 29–12a are oriented vertically (but still in contact) so that the rod is falling due to the gravitational force. Find the terminal speed of the rod if it has mass m = 3.6 grams, length 𝓁 = 18 cm, and resistance R = 0.0013 Ω. It is falling in a uniform horizontal field B = 0.080 T. Neglect the resistance of the U-shaped conductor, and friction.

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Textbook Question

(II) (a) Determine the energy stored in the inductor L as a function of time for the LR circuit of Fig. 30–6a. (b) After how many time constants does the stored energy reach 99.9% of its maximum value?

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Textbook Question

(II) (a) In Fig. 30–28, assume that the switch has been in position A for sufficient time so that a steady current I₀ = V₀/R flows through the resistor R. At time t = 0, the switch is quickly switched to position B and the current decays through resistor R' (which is much greater than R) according to $I = I_{0} e^{- t / τ^{'}}$ $I = I_{0} e^{- t / τ^{'}}$ . Show that the maximum emf ε_max induced in the inductor during this time period is (R'/R)V_o. (b) If R' = 45R and V_o = 145 V, determine ε_max. [When a mechanical switch is opened, a high-resistance air gap is created, which is modeled as R' here. This Problem illustrates why high-voltage sparking can occur if a current-carrying inductor is suddenly cut off from its power source. The very high voltage can produce an electric field great enough to ionize atoms of air, which emit light when electrons recombine with the ions.]

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Textbook Question

(II) A capacitor is placed in parallel with some device, B, as in Fig. 30–18b, to filter out stray high-frequency signals, but to allow ordinary 60.0-Hz ac to pass through with little loss. Suppose that circuit B in Fig. 30–18b is a resistance R = 530 Ω connected to ground, and that C = 0.35 μF. Calculate the ratio of the capacitor’s current amplitude to the incoming current’s amplitude if the incoming current has a frequency of (a) 60.0 Hz; (b) 60.0 kHz.

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