1
$\begingroup$

I'm using magnetometers near electric fences and I'm wondering if the electric fence should be detectable and if so, how close do I need to be to the fence for there to be a "measurable" effect (assuming my device has perfect sensitivity)?

My current thinking is that given that the animal isn't touching the fence and that there is no vegetation touching the fence, there shouldn't be a magnetic field generated.

This is my reasoning: It's my understanding that a typical cattle fence is wired up to a battery so that the positive side is wired to the fence and the negative side is wired to the ground. In this set up, while there may be voltage in the wire, there will be no current until the animal touches the wire. Thus, there should be no magnetic field generated until the animal completes the circuit. However, I've often heard these devices described as emitting a certain "pulse per second". If these devices are pulsing the electricity, isn't that current and shouldn't that be generating a magnetic field?

Here's a similar device to what I'm using: I believe the device is outputting a minimum of 2000V over ~100m. premier1supplies.com/img/product/pdf/FenceOptionsHowFenceWorksRiskLiability16.pdf

Another similar question, if there was flowing current, say from the animal completing the circuit or vegetation growing and touching the wire, what is the maximum expected magnetic field that I should be able to detect?

Thanks!

$\endgroup$
2
  • $\begingroup$ I did some more research and have some better estimates. Assuming voltage=10,000V, with an output of 5J, discharged over 0.0003s with a pulse rate of 1Hz. Since the formula that (I think) I need is B = u0*I/(2pi*r) to calculate mag. field, wouldn't this be a current of Vv=Ej/Qc; Qc = 5/10,000 = 0.0005c; I = Qc/t; I = 0.0005c/0.0003s = 1.67A. With the analogy of volts = pressure and amps = current, is it reasonable to assume that the existing "volts/pressure" drains to 0 between pulses? Also, 1.67A does not seem strong enough to contain a cow. Am I missing something? $\endgroup$
    – Jade131621
    Commented May 20, 2022 at 17:55
  • $\begingroup$ Wait. That's way too high. That'd be 1670mA. Police tasers are on the order of 2-4mA and are enough to deter charging moose. Also, I'm actually working on bears so the electric fences are probably a little more souped-up than a typical cattle fence. adfg.alaska.gov/… $\endgroup$
    – Jade131621
    Commented May 20, 2022 at 18:06

2 Answers 2

2
$\begingroup$

An electric fence is powered by a supply circuit which produces a high-voltage output which is fed to the fence wire. The current accompanying the high voltage is limited by that circuit to a low enough value so a shock from the fence wire will be annoying but not lethal. To save power, the circuit turns on briefly every one or two seconds, so if you get shocked by the wire you will feel bzzzt (YOW!!!) bzzzt (YOW!!!) bzzzt (YOW!!!) bzzzt (YOW!!!).

The only time that the fence would radiate a significant magnetic field is when current is flowing through the wire i.e., when there's something getting shocked by the wire, and even then since the current is limited, the magnetic field will be very weak and have only a very short range (of order ~inches).

During the sudden turn-on and turn-off of the voltage pulses, there will be an extra bit of radiated electromagnetic interference generated, but this too will be very small effect in terms of the magnetic field produced by the wire.

Note that whenever the high voltage is on, there will also be an extra bit of current in the wire even if you are not touching it because the high voltage tends to leak off the (uninsulated) wire- but this is again a very very small effect.

To predict the size of the field you would need to know 1) the voltage on the wire and 2) the current flowing during a pulse, and apply a formula I can no longer remember.

$\endgroup$
0
$\begingroup$

Thank you Niels, your answer along with some more searching as to what the equations might be and thus the parameters I need to fill these equations allowed me to "simulate" a range of plausible values or at least, I think so.

Correct me if I'm wrong but the following might be the equations to calculate the strength of the magnetic field (B) in gauss from voltage (V), current (I), time (t), and energy output (E).

I called the company and they gave me a range of 10,000 to 12,000 V as the likely voltage for any of their electric fence chargers, since 10,000V would produce a stronger magnetic field I assumed V = 10,000V.

I was able to find a rating of 5 Joules as the maximum output for this particular device. Assuming this means that every second a pulse of approximately 5 Joules is produced then E = 5J.

I found somewhere on the internet that 3/10000 s is a potential time over which the pulse would happen thus t = 0.0003s.

I also found that a pretty sensitive magnetometer should be able to sense ~0.01 Ga.

I couldn't find any information as to what the likely current would be so I tried:

  1. to calculate it from the energetic output and the voltage and

  2. used a range of values (15 mA being the low range expected for a fence and 120 mA being the higher range expected for a fence).

Using the equations and the information I found for my device:

Vvolts = Ejoules/Qcoloumbs

10,000 = 5/Q

Q = 0.0005 coloumbs

Iamps = Qcoloumbs/tseconds

I = 0.0005/0.0003

I = 1.67 A

This seems high?? But it is for bears in a small enclosure.

Btesla = mu0IA/2pi*rmeters

mu0 = 4 * pi * 10^-7 T* m/A

I simplified the constant to:

constant = 4 * pi * 10^-7 T*m/A / 2 * pi

constant = 2 * 10^-7 T*m/A * 10,000 ga/T

constant = 0.002 ga*m/A

Bgauss = (0.002m/A * Iamps)/rmeters

B = 0.00334/r

Mag Field (ga) Distance (m)
0.3333 0.01
0.0667 0.05
0.0333 0.10
0.0167 0.20
0.0067 0.5

Thus, with a very sensitive magnetometer and the maximum likely voltage we would likely not see an effect past ~20cm.

Using a range of plausible currents:

Bgauss = (0.002m/A * Iamps)/rmeters

For I = 15 mA B = 0.002 (ga*m/A) * 0.015(A)/r B = 0.00003/r

For I = 120 mA B = 0.00024/r

Mag Field (ga) Current (mA) Distance (m)
0.003 15 0.01
0.0006 15 0.05
0.024 120 0.01
0.012 129 0.02
0.0048 120 0.05

So it looks like for a 15 mA current the magnetic field generated would be undetectable with a magnetometer of +/- 0.01ga at 1cm and the 120 mA current would be undetectable after about 2 cm from the wire.

Now, I believe the equation B = constant * I/r is the simplified equation assuming a "very long wire" and there's another equation, the Biot-Savart equation, that I don't understand. In the case of a zoo enclosure, we don't have a very long wire ~100m. As this wire is rather short, would this serve to "increase" the magnetism generated on the wire?

Also this is also assuming that all the voltage/charge on the line is drained between each pulse, is that a reasonable assumption?

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.