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A toroidal solenoid has 2900 closely wound turns, cross-sectional area 0.450 cm\(^2\), mean radius 9.00 cm, and resistance \(R\) = 2.80 \(\Omega\). Ignore the variation of the magnetic field across the cross section of the solenoid. What is the amplitude of the current in the solenoid if it is connected to an ac source that has voltage amplitude 24.0 V and frequency 495 Hz?

Short Answer

Expert verified
The amplitude of the current is calculated by finding the impedance and using Ohm's Law for AC: \(I = \frac{24.0}{Z}\).

Step by step solution

01

Convert Cross-Sectional Area to Meters Squared

The cross-sectional area is given in cm², and for solving this problem, we need it in m². Therefore, we need to convert 0.450 cm² to m². We do this by using the conversion factor: 1 cm² = 0.0001 m². \[ 0.450 \text{ cm}^2 = 0.450 \times 0.0001 \text{ m}^2 = 4.50 \times 10^{-5} \text{ m}^2 \]
02

Calculate the Inductance (L) of the Solenoid

The inductance of a toroidal solenoid can be calculated using the formula: \[ L = \mu_0 \cdot \frac{N^2 \cdot A}{2\pi \cdot r} \] where \(\mu_0 = 4\pi \times 10^{-7} \text{ T m/A} \), \(N = 2900\) turns, \(A = 4.50 \times 10^{-5} \text{ m}^2 \), and \(r = 9.00 \text{ cm} = 0.09 \text{ m}\). Substituting these values in, we get: \[ L = 4\pi \times 10^{-7} \cdot \frac{2900^2 \cdot 4.50 \times 10^{-5}}{2\pi \cdot 0.09} \]
03

Evaluate the Inductive Reactance (X_L)

With the inductance known, we calculate the inductive reactance \(X_L\) using the formula: \[ X_L = 2\pi f L \] where \(f = 495\text{ Hz}\). We substitute \(L\) from Step 2 and the frequency to find the value of \(X_L\).
04

Calculate the Impedance (Z) of the Circuit

Using the resistance \(R = 2.80 \Omega\) from the problem and the inductive reactance \(X_L\) calculated in Step 3, compute the total impedance using: \[ Z = \sqrt{R^2 + X_L^2} \]
05

Determine the Amplitude of the Current

The amplitude of the current \(I\) can be determined using Ohm's Law for an AC circuit: \[ I = \frac{V}{Z} \] where \(V\) is the given voltage amplitude, 24.0 V.

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Key Concepts

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

Inductance Calculation
Inductance is a critical concept when working with circuits that include coils such as solenoids. In a toroidal solenoid, which is essentially a coil shaped like a donut, the inductance can be calculated using a specific formula. This gives us a measure of how effectively the solenoid can generate a magnetic field with a certain current. The formula for the inductance, denoted as \( L \), of a toroidal solenoid is: \[ L = \mu_0 \cdot \frac{N^2 \cdot A}{2\pi \cdot r} \]Here:
  • \( \mu_0 \) is the permeability of free space, which is a constant \( 4\pi \times 10^{-7} \text{ T m/A} \).
  • \( N \) is the total number of turns in the solenoid.
  • \( A \) is the cross-sectional area in square meters.
  • \( r \) is the mean radius of the solenoid in meters.
When calculating inductance, ensure all units are in the International System of Units (SI). This guarantees accuracy in your calculations. For instance, converting the area from cm² to m² is essential in this calculation. Inductance quantifies how a solenoid opposes changes in current, owing to its electromagnetic properties.
AC Circuit Analysis
AC circuit analysis involves understanding how alternating current behaves in different components of an electrical circuit. It's different from DC circuits, where the current flows only in one direction. In AC circuits, the current reverses direction periodically. This behavior is described by the frequency of the AC supply. When analyzing AC circuits that incorporate inductive components, like our toroidal solenoid, it is important to consider how inductance introduces a phase difference between the voltage and current. It suggests that the voltage leads the current by a phase angle, which is important for calculating other parameters like inductive reactance and impedance. In this exercise, the AC source voltage amplitude is given as 24.0 V with a frequency of 495 Hz. This forms a critical aspect of problem-solving in AC circuit contexts, as both parameters directly influence how energy is transferred and used within the circuit components.
Inductive Reactance
Inductive reactance is a measure of how much a solenoid or inductor resists the flow of alternating current. It is an essential variable when calculating the impedance of circuits with inductors. The reactance, denoted by \( X_L \), depicts the dependency on the frequency of the AC current and the inductance of the solenoid.The formula for inductive reactance is given by:\[ X_L = 2\pi f L \]where:
  • \( f \) is the frequency of the AC source in hertz (Hz).
  • \( L \) is the inductance in henrys (H).
A higher frequency or greater inductance leads to a larger inductive reactance, indicating a higher opposition to the flow of AC current. This reactance contributes to the circuit's overall impedance, affecting the current flow within the system. Understanding this concept is fundamental, specifically when designing or analyzing AC circuits.
Impedance Calculation
Impedance is akin to resistance but specifically applies to AC circuits. It's a combination of the circuit's resistance and its reactance, which includes both inductive and capacitive reactance. For a circuit with resistance and inductive reactance, the impedance, represented as \( Z \), can be calculated using the formula:\[ Z = \sqrt{R^2 + X_L^2} \]where:
  • \( R \) is the resistance in ohms (Ω).
  • \( X_L \) is the inductive reactance calculated previously.
The impedance provides a complete picture of how much the circuit opposes the flow of AC current. This value influences the current magnitude according to Ohm's Law, stated in the context of AC circuits as \( I = \frac{V}{Z} \). Understanding and calculating impedance is crucial for correctly predicting the behavior of AC circuits, which is necessary for tasks such as designing circuits and troubleshooting electrical systems.

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Most popular questions from this chapter

A series circuit has an impedance of 60.0 \(\Omega\) and a power factor of 0.720 at 50.0 Hz. The source voltage lags the current. (a) What circuit element, an inductor or a capacitor, should be placed in series with the circuit to raise its power factor? (b) What size element will raise the power factor to unity?

(a) What is the reactance of a 3.00-H inductor at a frequency of 80.0 Hz? (b) What is the inductance of an inductor whose reactance is 120\(\Omega\) at 80.0 Hz? (c) What is the reactance of a 4.00-\(\mu\)F capacitor at a frequency of 80.0 Hz? (d) What is the capacitance of a capacitor whose reactance is 120 \(\Omega\) at 80.0 Hz?

A transformer connected to a 120-V (rms) ac line is to supply 12.0 V (rms) to a portable electronic device. The load resistance in the secondary is 5.00 \(\Omega\). (a) What should the ratio of primary to secondary turns of the transformer be? (b) What rms current must the secondary supply? (c) What average power is delivered to the load? (d) What resistance connected directly across the 120-V line would draw the same power as the transformer? Show that this is equal to 5.00 \(\Omega\) times the square of the ratio of primary to secondary turns.

Consider an \(L-R-C\) series circuit with a 1.80-H inductor, a 0.900-\(\mu\)F capacitor, and a 300-\(\Omega\) resistor. The source has terminal rms voltage V\(_{rms}\) = 60.0 V and variable angular frequency \(\omega\). (a) What is the resonance angular frequency \(\omega_0\) of the circuit? (b) What is the rms current through the circuit at resonance, I\(_{rms}\)-0? (c) For what two values of angular frequency, \(\omega\)1 and \(\omega\)2, is the rms current half the resonance value? (d) The quantity \(\omega\)1 - \(omega\)2 defines the resonance \(width\). Calculate I\(_{rms}\)-0 and the resonance width for R = 300 \(\Omega\), 30.0 \(\Omega\), and 3.00 \(\Omega\). Describe how your results compare to the discussion in Section 31.5. the

An \(L-R-C\) series circuit is constructed using a 175-\(\Omega\) resistor, a 12.5-\(\mu\)F capacitor, and an 8.00-mH inductor, all connected across an ac source having a variable frequency and a voltage amplitude of 25.0 V. (a) At what angular frequency will the impedance be smallest, and what is the impedance at this frequency? (b) At the angular frequency in part (a), what is the maximum current through the inductor? (c) At the angular frequency in part (a), find the potential difference across the ac source, the resistor, the capacitor, and the inductor at the instant that the current is equal to one-half its greatest positive value. (d) In part (c), how are the potential differences across the resistor, inductor, and capacitor related to the potential difference across the ac source?

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