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Magnetic field in an AC Solenoid Coil

  1. Nov 21, 2008 #1

    I have a question about AC magnetic field, wondering if you can help me understand this.

    For a DC solenoid coil, We can say Magnetic field H=N*i/L (simplest form)
    For an AC input, H=N*isin(wt)/L ; H(rms) = N*i/L (i is RMS value)

    As long as the current i-rms is kept constant, the solenoid must produce the same magnetic
    field at any frequency. But,in my measurements, even if I keep the current constant, the field
    is still dropping as the frequency is increased (from 100 Hz to 10kHz).
    I wonder why the field decreases..
    I understand that apparent power increases because of the inductance, eddy current etc..
    Does the field really reduce with frequency?

    I wonder if there is any different formula to estimate the field at high frequency.


  2. jcsd
  3. Nov 21, 2008 #2


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    I am guessing the coil +series resistance in the circuit is simply low-pass filtering your signal.
    Try measuring the voltage across the coil and see if it changes as you increase the frequency (use a good multimeter, many hand-held multimeters won't work well at 10 kHz)
    What is R and L in you circuit?
  4. Nov 21, 2008 #3
    Thank you f95toli, My multimeter is HP make 34401A.. its decent peice.
    R of the coil is 1 ohm and L is around 0.25 milli henry.

    The measurements at 100 Hz are at good agreement with estimated but
    those at 10 kHz vary by 50%.
    thanks for your input

  5. Nov 21, 2008 #4
    For DC input, I derived the equation to tell the field along the axis of the coil.
    How can I modify that for an AC?
    Lets say, the coil is of length 2L..-L to +L.
    Does H at +L or -L same for both AC(rms) and DC of same current?
    If different, By what factor is AC field lesser?
  6. Nov 21, 2008 #5


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    Those values would give you a -3dB rolloff of something like 1 kHz if ithe coil was driven by a voltage source
    Are you using a "real" current source to drive the coil? If so, what is its output impedance?
    Also, remember that coils are capacitive as well; at high frequencies the capacitance can actually dominate and will effectively "short" the coil.

    If you I were you I would draw a simple circuit model which includes the impedances in the circuit (R, L and C)+the output impedance of the source.
    Driving a real coil at 10 kHz is not easy.

    btw, a 34401A can measure 10 kHz without problems.
  7. Nov 21, 2008 #6
    the way i am driving is .. connect command signal to car audio amplifier, and connect the coil to the coil.. V and current are measured using 2 34401As.. and magnetic field is measured by bell gaussmeter..which can go upto 12 kHz with with error within 6%.
    ..does capacitance between the coil layers and wires really matter? at 10kHz?
    See.. there can be power loss, am not worried about that... but why is field reducing?
  8. Jan 5, 2009 #7
    hi chitrageetam,
    you said you are working on an AC Solenoid.. of course the field and the current depend upon the frequency..
    the current that flows through the solenoid coils depends upon the voltage from the mains and the impedance of the coil..
    as you may re-call the impedance depends upon the frequency.. Z=R+2*pi*f*L;
    L is the inductance and f is the frequency.. so as you increase the frequency the impedance increases and so V/Z and thus the current flowing through the coils will decrease and this will bring down your field...
  9. Jan 6, 2009 #8


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    I assume from your posts that you are keeping the current the same as you increase frequency. Have you checked whether your measurement of the magnetic field is good at the higher frequencies?
  10. Jun 10, 2009 #9
    with ac magnetic field in a solenoid is= n*i sin(wt)-n*i sin(wt)*(R^2 -r^2)*w^2/(4*c^2) ,(with first approximation),in fact the magnetic field in side the solenoid is not uniform ,it has some r-variation.
  11. Jun 10, 2009 #10
    What you need to do is place a small resistor (say R= 1 ohm) in series with the solenoid, and monitor the voltage V across the resistor R to get current = V/R. If you always (per clem in post 8) adjust your current to be the same, you should get better results. Also be sure there is no metal or magnetic material nearby.
    (Added note] The fundamental relation between the current (or current density) and the magnetic field is Maxwell's equation: curl H = J, or curl B = u u0 J, so there is a direct frequency-independent relationship. For a given AC input current, the magnetic field strength should be independent of frequency. The frequency could fall off at higher frequencies due to eddy currents in material (ferrites, laminations, etc.) near or in the solenoid.
    Last edited: Jun 10, 2009
  12. Jun 11, 2009 #11
    Maxwell eqation is curl B=J+[tex]\partial[/tex]E/[tex]\partial[/tex]t (ignoring some factors)
    and also curl E=-[tex]\partial[/tex]B/[tex]\partial[/tex]t
    as J is time dependent, therefore magnetic field should depends on frequence.
  13. Jun 11, 2009 #12
    At these frequencies, and without any ferrites or displacement currents (from capacitance), the magnetic intensity H and field B everywhere should be directly proportional to J (i.e., curl H = J), the current density in the wire. The voltage on the solenoid will be largely in quadrature with the current, but not completely, due to coil resistance. There is a small effect, that will occur above 100 kHz; eddy currents in the wire will force the currents toward the surface of the wire (skin depth). This will increase the effective wire resistance. If the applied current is kept constant, this will be a very small effect.
    Last edited: Jun 11, 2009
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