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In the absence of a nearby metal object, the two inductances \(\left(L_{\mathrm{A}}\right.\) and \(\left.L_{\mathrm{B}}\right)\) in a heterodyne metal detector are the same, and the resonant frequencies of the two oscillator circuits have the same value of \(630.0 \mathrm{kHz}\) When the search coil (inductor \(\mathrm{B}\) ) is brought near a buried metal object, a beat frequency of \(7.30 \mathrm{kHz}\) is heard. By what percentage does the buried object increase the inductance of the search coil?

Short Answer

Expert verified
The buried object increases the inductance of the search coil by approximately 2.36%.

Step by step solution

01

Understand the Problem

We have two oscillators with equal initial resonant frequencies. When one is disturbed by a metal object, the resonant frequency changes, and we hear a beat frequency. We need to find how much the inductor's frequency is affected by the object.
02

Set Up the Resonant Frequency Equation

The resonant frequency \( f \) of a circuit is given by \( f = \frac{1}{2\pi\sqrt{LC}} \). Initially, for both inductors A and B, \( f_0 = 630.0 \text{kHz} \). This implies \( L_A = L_B \).
03

Calculate New Frequency from Beat Frequency

The beat frequency is the absolute difference between the two frequencies: \( f_{beat} = |f_A - f_B| \). If \( f_A = 630.0 \text{kHz} \) and \( f_{beat} = 7.30 \text{kHz} \), then the new frequency of \( f'_B = f_A \pm f_{beat} = 630.0 \text{kHz} \pm 7.30 \text{kHz} \). Only one makes sense physically: a decrease resulting in \( 622.7 \text{kHz} \).
04

Relate Frequency to Inductance

The relationship between frequency and inductance is defined as: \( f = \frac{1}{2\pi\sqrt{L'C}} \). Since \( f'_B = 622.7 \text{kHz} \), solve for \( L' \) using \( f_{B}' = \frac{1}{2\pi\sqrt{L_B'C}} \).
05

Solve for the Percentage Change in Inductance

Using the relationship \( \left(\frac{f_0}{f_B'}\right)^2 = \frac{L'_{B}}{L_B}\), calculate the percentage increase in inductance: \( \Delta L = \left(\frac{L'_{B} - L_B}{L_B}\right) \times 100 \% \). Substituting \( \left(\frac{630.0}{622.7}\right)^2 \approx \frac{L'_{B}}{L_B} \), compute \( \Delta L \).
06

Calculate Percentage

Solve \( \Delta L = (1.02356 - 1) \times 100 \% \approx 2.36 \% \). This is the percentage increase in the inductance of the search coil due to the nearby metal object.

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Key Concepts

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

Inductance
Inductance is a key property of coils within circuits and is often denoted by the symbol \( L \). It refers to the ability of the coil to store energy in a magnetic field when an electric current flows through it. Inductors are used in various electronic devices and circuits for their ability to resist changes in current.
\( L \) is measured in henries (H), which quantifies how effectively the coil can induce voltage when the current changes. High inductance means the coil stores more energy, and a change in the current will have a more significant effect on the induced voltage.
When a coil is brought near a metal object, like in a metal detector, the inductance can change. This change affects the resonant frequency of the circuit, which is defined by the relationship \( f = \frac{1}{2\pi\sqrt{LC}} \). As the exercise shows, a nearby metal increases inductance, causing a decrease in the resonant frequency.
  • Important: Inductance depends on coil properties such as number of turns, coil area, and material inside the coil.
  • Coils in close proximity to metal objects exhibit different magnetic properties, altering inductance.
Understanding how inductance works enables better manipulation of circuits to achieve desired frequencies or behaviors. It's an essential concept in designing efficient oscillators.
Oscillator Circuit
An oscillator circuit generates a repeated waveform without an external input. It's the key to producing a steady resonant frequency, which can be crucial for applications like radios, processors, or metal detectors.
Oscillator circuits usually consist of amplifying components and a frequency-determining network. The frequency of the oscillation is primarily determined by the properties of the inductance \( L \) and capacitance \( C \) with the equation \( f = \frac{1}{2\pi\sqrt{LC}} \). This formula shows how changes in inductor or capacitor values alter the frequency.
The types of oscillators include:
  • RC (Resistor-Capacitor) Oscillators: Used for low-frequency applications.
  • LC (Inductor-Capacitor) Oscillators: Used for high-frequency signals, as in radio transmissions.
  • Crystal Oscillators: Extremely accurate, used in timekeeping like in watches.
In the exercise scenario, the oscillator circuit's frequency is affected when the inductance changes due to the proximity of a metal object, demonstrating the critical interplay between circuit components.
Beat Frequency
Beat frequency occurs when two different frequencies are heard at the same time, resulting in a phenomenon known as "beats." This happens when two frequencies are very close, and the interference between their waves produces a perception of frequencies fluctuating.
In mathematical terms, the beat frequency \( f_{beat} \) is given by the absolute difference between the two frequencies \( f_1 \) and \( f_2 \): \[ f_{beat} = |f_1 - f_2| \]Beat frequency helps us measure minute deviations in frequency, which is why it is used in metal detectors to sense changes caused by nearby metal objects.
In practice:
  • A beat frequency of 7.30 kHz indicates the difference between the undisturbed oscillator and the changed oscillator frequencies.
  • Listening to beats is practically useful for tuning instruments and in radio signal processing.
This exercise demonstrates beat frequency by showing how metal influences the resonant frequency of an oscillator circuit, leading to detectable beats that tell us the metal is present.

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