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Current circuits

  1. Mar 3, 2008 #1
    1. The problem statement, all variables and given/known data
    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?

    2. Relevant 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.

    3. 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
    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?
  2. jcsd
  3. Mar 3, 2008 #2
    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
    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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