In the absence of a nearby metal object, the two inductances (LA and LB) in a heterodyne metal detector are the same, and the resonant frequencies of the two oscillator circuits have the same value of 580.0 kHz. When the search coil (inductor B) is brought near a buried metal object, a beat frequency of 7.0 kHz is heard. By what percentage does the buried object reduce the inductance of the search coil?
%
To calculate the percentage by which the buried object reduces the inductance of the search coil, we need to first understand the relationship between inductance and the resonant frequency in a metal detector.
In a metal detector, the resonant frequency is determined by the inductance and capacitance of the oscillator circuit. When a metal object is brought near the search coil, it affects the inductance of the coil, which in turn affects the resonant frequency.
The beat frequency is the difference between the resonant frequencies of the two oscillator circuits. In this case, the beat frequency is 7.0 kHz.
To find the percentage reduction in inductance, we can use the formula:
percentage reduction in inductance = (beat frequency / resonant frequency) * 100
Given:
Beat frequency = 7.0 kHz
Resonant frequency = 580.0 kHz
Substituting these values into the formula, we have:
percentage reduction in inductance = (7.0 kHz / 580.0 kHz) * 100 = 1.21%
Therefore, the buried object reduces the inductance of the search coil by approximately 1.21%.