Calculating % Change in Inductance of Search Coil in Metal Detector

[arod2812]

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?

 

[kamerling]

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.

 

[Moetasim]

I would like to offer some clarification and guidance on how to approach this problem. First, it is important to understand that inductance is a property of a circuit or component that determines its ability to store energy in the form of a magnetic field. When a metal object is brought near the search coil in a metal detector, it can affect the inductance of the coil due to changes in the magnetic field.

In this problem, we are given the resonant frequencies of two oscillator circuits (LA and LB) in the metal detector, which are both at 430 kHz when there is no nearby metal object. When a metal object is brought near the search coil (LB), a beat frequency of 4.1 kHz is heard. This beat frequency is the difference between the resonant frequencies of the two oscillator circuits.

To calculate the percentage change in inductance, we need to first understand the relationship between inductance and frequency. As you have correctly stated, the resonant frequency of an oscillator circuit is inversely proportional to the square root of inductance. This means that as inductance increases, frequency decreases, and vice versa.

In this case, we can use the resonant frequency of the two oscillator circuits to determine the difference in inductance. Since the resonant frequency of LB is 4.1 kHz lower than that of LA, we can say that the inductance of LB is lower than that of LA. This means that the metal object has caused a decrease in the inductance of the search coil.

To calculate the percentage change, we can use the following formula:

% change = (Lb - La) / La * 100%

Where Lb is the inductance with the metal object present and La is the inductance without the metal object. Plugging in the values you have calculated, we get:

% change = (1.51E-9 - 1.37E-13) / 1.37E-13 * 100% = 1095%

This means that the inductance of the search coil has decreased by 1095% when the metal object is present. This may seem counterintuitive, but it is important to remember that inductance and frequency have an inverse relationship.

I hope this explanation helps you better understand the problem and how to approach it. As always, it is important to double-check your calculations and units to ensure accuracy