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A man was found laying near a powerline

  1. Nov 30, 2013 #1
    1. The problem statement, all variables and given/known data
    A man was found laying unconscious beneath high tension power lines conveying electricity. He was conveyed to the hospital where he was pronounced dead upon arrival. An autopsy showed that he died from cardiac arrest. The investigation finds that there was indeed an electrical fault at the time of the incident, for 1.0s, a current of 100 amperes leaked onto the ground from a vertical conducting rod whose tip was found below the ground 10m away from the man. The resistivity of the moist ground was about 100Ωm.


    1. Under any suitable approximation obtain a reasonable estimate of the potential difference between the man's feet.
    2. It is reasonable to suppose that the current went up one leg of the man, across the torso and down the other leg. The commonly accepted value for resistance of a leg is 300Ω and that of the torso is 1000Ω. From these data, estimate the current across the victim's torso.
    3.Using the expression for E(r), obtain the potential difference from the tip of the rod and a point at distance r. Take the tip of the rod to have radius a. If the integral is a problem, you can get the answer by noticing the similarity between E(r) and a well known charge configuration.
    4. Currents ranging from 0.10A and 1.0A across the torso can trigger fibrillation. Did the accident kill the man?

    2. Relevant equations
    [itex]A_(sphere)=4\pi r^2[/itex]
    [itex]R=ρl/A[/itex]
    This is one of the equations I calculated for one of the earlier questions:
    [itex]E_(r)=Iρ/(\pi r^2)[/itex]


    3. The attempt at a solution

    1. I was told by my physics teacher that current travels in a hemisphere since we are calculating it in ground, where it spreads out. I based this on that.

    Assuming the difference between the man's feet as 1m. 1 foot is 10m away, the other is 11m away from the rod.

    [itex]V=IR=Iρl/A[/itex]
    I assumed that for l, I can use the distance between the man's foot and the rod as it essentially is the distance between one end of the resistor compared to the other if it was in a wire.
    My calculations:
    [itex]v_10=100A*100Ωm*10m/(2*10m^2*\pi)=159V[/itex]
    [itex]v_11=100A*100Ωm*11m/(2*11m^2*\pi)=145V[/itex]
    [itex]ΔV=V_10-V_11=14V[/itex]

    2.
    Total Resistance: 300Ω+300Ω+1000Ω=1600Ω
    [itex]I=V/R=14V/1600Ω=0.00875A[/itex]

    3.I have no idea how to do this. Here's my attempt at the problem.
    [itex]E=ΔV/R=ΔV/(ρa/A)=VA/ρa=(V2\pi r^2)/ρa[/itex]
    [itex]E=Iρ/2\pi r^2[/itex]
    Now I solve these as equal equations.
    [itex]Iρ/2\pi a^2=V\pi a^2/ρr[/itex]
    [itex]Iρ=V\pi ^2a^2/ρr[/itex]
    [itex]Iρ^2r=V\pi^2a^2[/itex]
    [itex]V=Iρ^2r/ \pi ^2a^2[/itex]

    4. Since 0.00875A<0.10A, he didn't die from the accident.

    I'm just wondering if I did these questions right as I couldn't find anything on resistances that are outside wires, nor electric fields on current electricity.
     
  2. jcsd
  3. Nov 30, 2013 #2

    Simon Bridge

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    resistances outside wires work exactly the same as those inside wires.

    Do I read this right: you are modelling the man as three resistors in series (legs and torso) in parallel with another resistor (the ground between his feet)?

    Is 1m a reasonable approximation for someones standing foot separation?
    Stand up and compare. Try standing with your feet 1m apart. How do you feel?
    But - what is reasonable depends on what you want to achieve ... in a long answer you should try to justify the choice for this distance but not worry about it much otherwise.


    It says "using the expression for E(r)" - which one is that?
     
  4. Nov 30, 2013 #3
    1. Yes, I am modelling the man as 3 resistors, as the current would travel up one leg, through the torso and down the other leg. I haven't really considered the ground's resistivity as I was getting the potential difference between the two feet, which I used as an EMF.

    2.We are doing an approximation on the potential difference, and I picked a 1m difference because it's easy to work with.

    3. Using the expression for E(r) would be the expression I derived for the another question.
    [itex]E(r)=Iρ/(\pi r2) [/itex]
    Derivation:
    [itex]E=V/r/[/itex]
    [itex]E=IR/r[/itex]
    [itex]E=I(ρr/A)/r[/itex]
    I changed l which would be in wires into r, which is the distance between the two points in an electric field.
    [itex]E=Iρ/A[/itex]
    [itex]E=Iρ/2\pi r^2[/itex]
     
  5. Dec 1, 2013 #4

    Simon Bridge

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    How come you didn't do an integration?
    Considering the hint (a well known charge configuration) it would seem you've had an example of this sort of calculation before.
     
  6. Dec 1, 2013 #5
    I didn't do an integration because the I'm in AP Physics B, which doesn't require calculus. The teacher said that it's possible to get it without calculus, but I have no idea how to do it.
     
  7. Dec 1, 2013 #6

    haruspex

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    Some confusion here. For the first step to be right, R is a distance. For the second step to be right, R is a resistance. This has led you to some equations that are not even dimensionally correct.
    This is the key equation (and it doesn't require calculus to obtain it).
    From this, and the distance between the man's feet you can get the potential difference.
    (I agree with Simon that 1m is rather a lot, but since you are asked whether it could have been the cause of death you should take the maximum physically possible.)
    Without checking the details, I think you managed to get the right answer by not actually using your erroneous equations.
     
  8. Dec 1, 2013 #7
    Oh shoot, you are right in that I confused distance with resistance. Thanks for catching that. Redoing the derivation gets me this:
    [itex]E=\frac{ΔV}{r}[/itex] and [itex]E=\frac{Iρ}{2\pi r^2}[/itex]
    I checked over the derivation for the second one and it appeared to be a working derivation.
    [itex]\frac{ΔV}{r}=\frac{Iρ}{2\pi r^2}[/itex]
    [itex]ΔV=\frac{Iρ}{2\pi r}[/itex]

    So looking over it, I did the potential difference correct? It just seems really odd to me how low it is so I wasn't sure if I did it correct.
     
  9. Dec 1, 2013 #8

    haruspex

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    It seems reasonable. Consider him as striding a rectangular block of turf 1m long by 0.2m x 0.2m. The resistance would be 2500Ω, similar to his own. But the current through it at that distance would only be 100A 0.22/(2π102), or about 0.007A (very close to your result).
     
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