Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Inductance and Magnetic Fields

  1. Nov 27, 2009 #1
    I am working on determining the mH needed to produce a 250 gauss (0.025 T) magnetic field by a flat coil in a 8 ohm audio circuit that is PC controlled with an NHC tone generator. I need some help to do so as the information that I am able to get online does not allow me to make the jump from classical electromagnetism equations to mH. Thank you for your help!
     
  2. jcsd
  3. Nov 27, 2009 #2
    The stored energy in an inductance L may be written
    W = ½LI2

    The stored energy in a magnetic field B, H may be written

    W = ½∫B·H dV = (1/2μ0)∫B2 dV

    using B = μ0nI for a uniform magnetic field B, where nI = amp turns per meter

    we get W = ½LI2 = (μ0/2) ∫n2I2 dV

    where ∫dV is integration over volume of magnetic field.

    Dividing by I2 and multiplying by 2 we get

    L = μ0 ∫n2 dV

    It is a little more complicated when the volume integral includes both air and permeable material with μ>1, but you get the point.

    Bob S
     
  4. Nov 27, 2009 #3
    No, I am not able to follow you. I am not a mathematician at all. I am a biologist that took Physics 40 years ago and never made it past linear algebra. I need simple math. I tried to do the simple math and found that the inductance was equal to the ohms times second. What is ohms times second? T=Wb/meter squared. Wb=V times second. L=Wb per ampere. ohms=V/A and h=v times second/ampere. Thus h=ohms times second. What is this? Do you now see my confusion?
     
  5. Nov 27, 2009 #4
    In the time domain, voltage V is

    V = L dI/dt, so

    L = V/(dI/dt) (where L is inductance) so

    inductance has units (volts/amps) x time, or ohm-seconds

    Tesla = Webers/meter squared

    Please use only the variables in this equation

    W = ½LI2 = ½∫B·H dV = (1/2μ0)∫B2 dV

    So L = 1/(μ0I2)∫B2 dV

    L = inductance (Henrys)
    I = amps
    B=Tesla
    μ0 = 4 pi x 10-7Henrys per meter
    dV = volume integral
    You can take the B2 out of the integral if it is constant throughout the integrated volume. The magnetic field is where the inductive energy is stored.

    Are you interested in a solenoid geometry without iron or ferrite?

    [added] The Biot Savart law gives exact magnetic field values everywhere. Ampere's Law is useful where magnetic fields are constant.

    Bob S
     
    Last edited: Nov 27, 2009
  6. Nov 27, 2009 #5
    Bob

    Please KISS, keep it simple stupid!

    I am not up to integration. All I want is what mH flat coil will produce a static 250 gauss magnetic field in an audio circuit of 8 ohms?
     
  7. Nov 27, 2009 #6
    It is very hard to calculate inductance without knowing anything about the coil dimensions and number of turns.
    You might look at the software at www.visimag.com
    Bob S.
     
  8. Nov 27, 2009 #7
  9. Nov 27, 2009 #8
    I guess that I am the stupid one as you do not seem to understand what I want. All I want is to know what coil to use in an audio circuit of 8 ohms to give me a static magnetic field of 250 gauss when turned on. Why is this so hard? I will use NHC tone generator software to control the frequency output to the coil from the PC. All I want is a static 250 gauss field. I care not how many turns the coil is, coils seem to come in mH. So what coil of how many mH do I buy to produce a 250 gauss field in an audio circuit of 8 ohms?

    Edward
     
  10. Nov 27, 2009 #9

    f95toli

    User Avatar
    Science Advisor
    Gold Member

    Your question is impossible to answer. The number of mH does not tell you anything about the magnetic field that is generated. The field will depend on the geometry of the coil, number of turns etc. It will obviously also depend on the current.
    Hence, you have to be more specific.
    Also, coils meant for generating magnetic fields (as oppsed to coils designed to act as inductors) do NOT "come in mH", Henry is the unit of inductance, not field. The normal way of specifiying a coil (solenoid) meant for generating a magnetic field would be to state Gauss/Ampere for specific position. One could for example have 250 Gauss/A at the centre, meaning you would get a field of 250 Gauss if you supplied a current of 1 A.








    .
     
  11. Nov 27, 2009 #10
    first determine the current in your circuit using the basic V = IR(ie, take the operating voltage of your circuit and divide it by 8)

    now keep in mind that inductance will have nothing to do with your circuit, as you want a static, ie, DC powered field(inductance only plays a significant role in AC generated fields)

    in other words, the inductance of your coil is not what determines the strength of your magnetizing field(keep in mind, this field will be measured in Ampere meters, not gauss, unless you know the magnetization value of the material you are trying to influence), but rather it will be based on the turns per unit length ratio of your coil.

    the formula relating the field which is generated by a coil to its current in general is:

    H = u(NI - LI) [1]

    where:
    N = number of coil turns / coil length(in meters)
    I = current
    L = inductance(this is zero for DC fields)
    μ = permeability of the material the coil is wrapped around(i assume in your case this will be air, so you use μ0)

    so then H = μNI [3]
    for DC fields ONLY. (keep in mind also that this is the approximate field value INSIDE of the coil.)

    Also, B = H + 4.pi.M [2]

    where M is the material's Magnetization

    in other words, if you know you want whatever the object of your experiment is to experience a field of 250 G, you first need to know the magnetization of your material(that is, the value describing how your material responds to an applied H field) so that you may determine from [2] the equivalent value of H. Once this value is determined, and using the current calculated for your circuit, its a simple process of plugging thee values into [3] and determining what value N is needed.

    To elaborate a bit further, it is essentially the thickness of the wires that make up your coil that will determine the strength of your field, the thinner the wire, the more turns that can fit per unit length, thus the higher the value of N.

    Also, i think you need to grasp the concept that a particular valued H field will result in a different B field for different materials, where this B value is dependent on the material's magnetization(think of the analogy of an electric circuit, for example a 250 V source has no defined current, the current is determined by the resistance, so while a 50 ohm and 100 ohm load may experience the same 250V, the current through them will be different)
     
    Last edited: Nov 27, 2009
  12. Nov 28, 2009 #11
    Here (I think) is the magnetic field from a spiral single-layer flat coil (without iron) with inner radius r = 0.5 cm and outer radius R = 2 cm, with N=15 turns carrying a current I=2 amps. The wire diameter is about 1 mm (18 Ga).

    For a single turn of radius r carrying a current I, the magnetic field in the center is:
    B = [0.5 u0NI][1/r] (Tesla) where u0 is the permeability of free space.

    For the coil described above,

    B = [0.5 u0NI][1/<r>]

    Where <r> = 0.0114 meters

    The result is B = 16.5 gauss.

    It looks like you will need about 15 layers to get 250 Gauss.

    The 2 amp current in this wire is about the maximum continuous recommended value for 18 Ga wire.

    The simple formula for inductance is, from
    http://en.wikipedia.org/wiki/Inductor

    L = u0 <r>2N2/(0.25<r> +0.35(Rmax – Rmin)) = 4.5 microHenrys

    Bob S
     
  13. Nov 28, 2009 #12
    OK, here is what I have been able to get as far as my PC motherboard audio output data:

    AC'97 Audio Codec A3.83
    Embedded 50mW/20ohm OP at front LINE output; 50mW/20ohm headset audio amplifier
    Power supply: digital: 3.3V; analog: 5V/3.3V; 600 Watt

    So I am guessing that I will need some sort of amplifier to boost the signal.

    Great specs provided by you, by the way. Thank you profusely.

    Edward
     
  14. Nov 28, 2009 #13
    What I am trying to do is correct the mis-alignments in a persons human body field with various sound frequencies sent into an external coil w/o the need for sound. It is the magnetic field and not the sound that the human body field reacts to in a positive manner. Of course, any factory magnetic fields due to motors and the like also put deleterious magnetic fields in to our food, processed foods, as a heretofore unknown by people result.
    Thus, geopathic stressors and electromagnetic stressors may be rectified by externally applied magnetic fields. The best magnetic field range for therapeutic use is from 0.01 to 0.04 Tesla. I picked 250 gauss (0.025 T) as midrange.
    Externally applied magnets are used in many alternative medicine applications and are known to work quite effectively. But these are simply magnets and do not have healing information encoded into them. Their effectiveness tapers off after a few days.
     
  15. Nov 29, 2009 #14
    Ok I took your project's needs into consideration and tried to obtain some values for the average susceptibility of the human body, but couldn't, however i was able to find values for human blood plasma, and as one may realise, this value changes for oxygenated and deoxygenated blood.

    assuming blood to be the major component of the body, we can use values of Xv(the magnetic susceptibility of the blood per unit volume), you can determine the required strength of the field based on an estimated volume of the body part being treated.

    If Absolutely maximum precision were desired, you could fill a gauged measuring device, fill it with water, and let them dip the body part to be tested into it so that you could determine the volume of the body part. This is only really possible for hands, feet, and joints. For other body parts i would suggest taking the length, width , and height measurements and assuming the the part to be a cuboid(better for torso, thighs, biceps). Knowing this we can apply the volume we find to the Xv value of human blood:

    Xv = -0.714 ppm
    = (-0.714 / 10^6) * Volume(in meters cubed)


    The equation relating the B field to the H field, given the Xv value is:

    B = u0(1 + Xv)H

    so H = B / u0(1+Xv)

    note X will be negative and small, so H is approximately B/u0.

    so if you want a B field within the 0.1T to 0.04 T range, lets say 0.025 T

    H = 0.025 / u0 = 19,894 Am.

    For a 50mW, 20 ohm output,

    I = 0.05 A

    Now to fit a body part into a coil, this coil is going to have a somewhat large radius, at least 7 cm to fit a hand, which means the field will be very weak. To achieve the desired field:

    H = u0NI/4pir^2
    (19894*4pi*0.07^2)/0.05u0 = N
    N = 1,234,977,883

    if you're using gauge 30 wire, thickness = 0.17 mm, therefore N = 1 / 0.00017 = 5882
    so you'd need thousands of layers of wire, which would mean huge resistances and way larger power supplies than your audio output can provide.

    Now, a more suitable power supply, say delivering 50 A

    N = 19,496,695

    since 1 layer gauge 30, N = 5882

    you would need 3315 layers of coil to produce this field
     
  16. Nov 29, 2009 #15
    Bob S.,

    If you would like to be involved in my research and possible commercialization, you are most welcome. Let me know.

    Go to this website for background into the human body field (HBF):

    http://www.journalinformationalmedicine.org/ , at the bottom of the page is an article.

    Go here for much info concerning pulsed magnetic field therapy:

    http://www.pemft.net/?gclid=CNip6IOCsJ4CFQeenAodLDu1pQ

    I am interested in building a microprocessor controlled unit that will diagnose the HBF (already being done commercially) and then treat via the coil using an audio frequency.

    Lots to do but I am getting there!

    Edward
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook