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Magnetic field strength - unit conversions

  1. Aug 4, 2010 #1
    I have to spec a sensor to monitor the magnetic field of a DC solenoid element.

    The specification for the solenoid includes its length, DC resistance, current at the applied supply voltage, and most importantly, the expected Dipole Moment of the unit, as 42 Am^2.

    I plan to attach a solid state linear Hall Effect sensor directly to one end during test, and am trying to estimate the field strength to select a sensor.
    One family of sensors I'm looking at output a range of 2 ±2V with sensitivities of 0.5, 1.1, or 2.7 mV/Gauss, so approximate field ranges of ±5000, ±1818, or ±740 Gauss.

    My problem is that I have no idea of what field strength in Gauss to equate to 42 Am^2

    Can anyone help me solve this?
    Thanks!
    Dave
     
  2. jcsd
  3. Aug 4, 2010 #2

    diazona

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    It depends on where you're putting the sensor.

    There is a formula that gives the magnetic field strength produced by an ideal dipole:
    [tex]B(m, r, \theta) = \frac{\mu_0}{4\pi} \frac{m}{r^3} \sqrt {1+3\cos^2\theta}[/tex]
    (source: http://en.wikipedia.org/wiki/Magnetic_dipole) where m is the magnitude of the dipole moment, r is the distance between the dipole and the sensor, and θ is the angle between the dipole axis and the direction to the sensor. Unfortunately, this formula is only reasonably accurate when the distance r is larger than the size of the dipole - so in your case, larger than the solenoid. I suppose you could try using it to get a rough estimate of what range you should be looking at, but it's quite possible that the number you get may be pretty far off if you're going to be putting the sensor right up against the solenoid.
     
  4. Aug 4, 2010 #3
    Yes, sensor would be very close, say within 1% of the solenoid's length. I think that's equivalent to half the field strength at the center of the coil.

    But is there any way to convert 42Am^2 to Gauss?
     
  5. Aug 4, 2010 #4
    Re-reading diazona's reply on the big screen (instead of iPhone), I see the formula.
    This related one was (also?) in the Wiki article:
    [tex]
    B(m, r, \lambda) = \frac{\mu_0}{4\pi} \frac{m}{r^3} \sqrt {1+3\sin^2\lambda}
    [/tex]

    Since r is the distance from the center of the solenoid, this does make sense.
    sin^2(lambda) is 0 for on-axis, and m is in my Am^2 units, all I need is the value of mu0, 4Pi×10−7.

    Length of the solenoid is ~0.5 meter, so r=0.25.
    B = 1e-7/0.015625 * 42 = 0.0002688 Tesla =2.7 Gs

    Does that look correct?

    Thanks for the help!
    Dave
     
  6. Aug 4, 2010 #5

    diazona

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    Actually I would think the field strength should be nearly the same if you're at the end of the coil (in the middle of the circle) compared to the actual 3D center of the coil. An ideal solenoid creates a magnetic field that is constant throughout its interior.
    That's like asking to convert 30 miles to pounds. They're completely different physical quantities, so it can't be done.
    That's pretty much the same thing I gave you, I just changed λ to θ because it seemed to make more sense that way. In retrospect maybe I should have left it as is :wink:
    Actually I believe λ = 90° when you're on the axis, so sin2λ = 1.
    Your math looks reasonable - except that if you use λ = 90°, it'll come out to twice that (5.4 Gs), although I guess it doesn't make a difference to determine what kind of sensor you'll want to get.

    Do keep in mind, though, that that formula is only intended to apply far away from the dipole (solenoid). It might or might not work well enough to tell you what kind of sensor to get.
     
  7. Aug 4, 2010 #6

    diazona

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    Hmm... I just had a thought: A solenoid can be modeled as a stack of N current loops each carrying a current I. The magnetic field strength inside the solenoid is
    [tex]B = \mu \frac{NI}{h}[/tex]
    (for an ideal solenoid, anyway). The stack of current loops will have a total dipole moment of
    [tex]m = NI\pi r^2[/tex]
    so you could compute the magnetic field in the interior of the solenoid as
    [tex]B = \mu \frac{m}{\pi r^2h}[/tex]
    where μ is the permeability of the material in the solenoid's core (μ = μ0 if there is nothing inside the solenoid), m is the total dipole moment, r is the solenoid's radius, and h is its height (or length). The denominator πr2h is equal to the volume enclosed by the solenoid.

    I'd expect this formula to be much more applicable than the other one for your situation (unless I made a silly mistake in the math), and when I plug in some numbers here, I get a result in the thousands of Gauss, depending on the radius of your solenoid and what, if anything, is in its core. (3360 Gs for a 1 cm radius with an empty core) So you may want to get one of the larger-range sensors after all.
     
  8. Aug 4, 2010 #7
    Wow! Huge difference...

    1 cm radius is probably close, and it has a solid (possibly soft iron, possibly ferrite) core.
    2.7 Gs is pretty small, even with the most sensitive sensor of the family.
    But 3300 Gs is way up there!

    I'm thinking I will buy one on each end of the spectrum; they're cheap, if you aren't building commercial quantities...

    Thanks again!

    Dave
     
  9. Aug 5, 2010 #8

    diazona

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    Yeah, that's why I was warning about not using the first formula outside its domain of applicability :wink:

    I did a bit of hunting and found that iron has a relative permeability of about 200, which means if your solenoid has an iron core, that would increase the magnetic field by a further factor of 200, to more than half a million gauss. So if the formula B = μm/πr2h is anywhere close to being accurate, any of the three sensors you get will be overwhelmed. Still, if they're cheap, it may be worth it to just get one and try the measurement, to see if the values obtained from the formulas seem reasonable at all.
     
  10. Aug 5, 2010 #9
    Getting back to this in the morning, I must be making some horrendous math error!

    Given [tex]\mu_0 = 4 \pi 10^-17[/tex], [tex]r = 0.01[/tex], [tex]h = 0.5[/tex], [tex]m = 42[/tex]

    I get B (in Teslas) = 3.36*10^-11, or 3.36*10^-7 Gauss.

    Also, the permeability of iron looks much smaller than 200:
    http://en.wikipedia.org/wiki/Permeability_(electromagnetism)#Values_for_some_common_materials
    indicates something on the order of 100 to 4000 times µ0.
    Taking 1000x gives me something like 3.36*10^-4 Gs

    Did I mess up a major unit conversion somewhere?

    Dave
     
  11. Aug 5, 2010 #10
    I just re-read your note earlier: "iron has a relative permeability of about 200"
    I misread that as absolute permeability. 200 is a perfectly reasonable multiplier, but I'm still off somewhere, by a factor of 10^10

    Dave
     
  12. Aug 5, 2010 #11
    ... and I see where, now. Somehow, I kept reading µ0 as ~10^-17, not 10^-7.
    I'm going to blame my eyesight!

    I'll look for a considerably less sensitive device...

    Dave
     
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