Calculating % Change in Inductance of Search Coil in Metal Detector

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In a heterodyne metal detector, the inductance of the search coil decreases when a metal object is nearby, resulting in a beat frequency of 4.1 kHz. The resonant frequency of the circuits is initially 430 kHz, which corresponds to an inductance calculation of 1.37E-13 H. The new frequency with the metal object is 425.9 kHz, leading to an inductance of 1.51E-9 H. The percentage change in inductance is calculated by taking the difference between the two inductance values and dividing by the original inductance. The calculations and understanding of the relationship between frequency and inductance are crucial for solving the problem correctly.
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Homework Statement


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 430 kHz. When the search coil (inductor B) is brought near a buried metal object, a beat frequency of 4.1 kHz is heard. By what percentage does the buried object reduce the inductance of the search coil?


Homework Equations


I don't understand how the inductance REDUCES (as the question says) because the frequency decreases, so I would expect the inductance to increase. I am also confused about how to calculate percent change.

The Attempt at a Solution


f=1/(2*pi*SqRt(LC) . . . capacitance remains the same and is therefore a constant...
430000 Hz = 1/ (2*pi*SqRt(L)) --> La=1.37E-13
and
4100 Hz = 1/(2*pi*SqRt(L)) --> Lb=1.51E-9

I then took the difference between these two inductance values and divided by Lb to get the % change. I don't think I have the right answer though. IS ANY OF THIS CORRECT?
 
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arod2812 said:

Homework Statement


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 430 kHz. When the search coil (inductor B) is brought near a buried metal object, a beat frequency of 4.1 kHz is heard. By what percentage does the buried object reduce the inductance of the search coil?


Homework Equations


I don't understand how the inductance REDUCES (as the question says) because the frequency decreases, so I would expect the inductance to increase.
If a beat frequency of 4.1 KHz is heard, that means that the difference between the frequencies of the two coils is 4.1KHz. The frequency in the sense coil can be higher

The Attempt at a Solution


f=1/(2*pi*SqRt(LC) . . . capacitance remains the same and is therefore a constant...
430000 Hz = 1/ (2*pi*SqRt(L)) --> La=1.37E-13
It's clearer to keep 1/sqrt(C) in this answer or replace 2*pi/sqrt(C) by a constant. It will divide off later. The rest of your work is correct if you use the correct frequency.
 
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