# Bird on an electric wire

1. Feb 10, 2008

### drpizza

I'd rather be very picky about this. What's the best way to explain (to bright students) why there's no current in a bird when it sits on a high tension wire? i.e. when a student points out that it creates a parallel circuit with resistance in the wire in the ballpark of 5 times 10^-7 ohms, and resistance in the bird of, say, 10,000 ohms.

Or, is there a current in the bird, (nano-amperes?) which is absolutely harmless?

2. Feb 10, 2008

### cepheid

Staff Emeritus
Both of the bird's feet (the terminals of the resistor) are in contact with portions of the wire at the same voltage. There is no potential difference across the bird, hence no current.

3. Feb 10, 2008

### mgb_phys

Actually the OP is right there is a very small current in the bird.
Unless the power line is a superconductor there is a tiny potential difference between the bird's feet 1cm apart and there is a large but finite resistance of the bird - so a very small current does flow through the bird.

(At normal frequencies, neglecting skin effects and transmission line modes)

4. Feb 10, 2008

### drpizza

But, would a "breakdown voltage" be necessary before the bird started conducting? This is the point at which my own understanding breaks down.

5. Feb 10, 2008

### mgb_phys

No the breakdown voltage really only applies to gases that need to be ionized to conduct effectively.

Some guestimates - worst case sweaty skin has a hand-hand resistance of around 15K.
A heavy duty power line has a resistance of around 0.01R /mile
so even with 1000A flowing through the line there is only a potential difference of 60uV between the birds feet and a current of a few nA flowing through it.
Probably much less since birds feet are a lot less conductive than skin and there are AC skin effects even at 60Hz.

6. Feb 10, 2008

### drpizza

Okay, thanks. It's nice to have some confirmation.

7. Feb 11, 2008

### Lojzek

Wouldn't a bird standing on one wire die because of it's electric capacity, if the (alternating) voltage was high enough?

(if we apply a voltage U0*sin(f*t) to a capacitor with capacity C, then there is a current
I=U0*f*cos(f*t)*C needed to charge the capacitor)

Last edited: Feb 12, 2008
8. Feb 11, 2008

### mgb_phys

Yes but the point is that the voltage is very very low.
Since the transmission line is a rather good conductor there is very little potential difference between two points a few cm apart.

9. Feb 11, 2008

### Lojzek

I'm not talking about the current caused by the small voltage difference over length of the wire, but the (alternating) current caused by the large voltage difference over time. This current must be present even if the bird is touching the wire only in one point.

Last edited: Feb 11, 2008
10. Feb 11, 2008

### mgb_phys

A capacitor only charges up if there is a potential differences.
The bird as at the same potential as the wire and so there is no field

11. Feb 11, 2008

### rbj

let's say it was a bat, instead of a bird. and the bat was hanging upside-down on the 765 kV line.

i s'pose you could consider the the top of the bat's head (which is the closest part of the bat to ground) sorta one of the plates of a capacitor, and the spot on the ground below the bat as the other plate. there is a time-variant potential difference between the bat's head and the ground, but with the height of these lines being 30 or 40 meters (which is the spacing between plates), i wouldn't expect there to be much capacitance.

let's guess:

$$C = \frac{\epsilon A}{d} = \frac{10^{-11}(10^{-2})^2}{30} \approx 3 \times 10^{-17} \mathrm{F}$$

$$|I| = 2 \pi f C |V| = 2 \pi 60 (3 \times 10^{-17} ) \times | 765 \times 10^3| \approx 10^{-8} \mathrm{A}$$

i dunno, might be measurable. might be noticable for a bat or bird. myabe not.

you might get about the same AC current when you stand under one of them things. for sure, the electrons in your body are sloshing around a bit 120 times each second. they don't seem to mind, though.

Last edited: Feb 11, 2008
12. Feb 11, 2008

### mgb_phys

But the bat is conducting so the top of it's head is at almost the same potential as the wire - the field 'per unit bat' is much less than the field in space below the wire.

13. Feb 12, 2008

### Lojzek

I have made similar calculation as rbj, but with different capacity estimate:

the aproximation of a plate capacitor is not good here, because the distance between the two "plates" (bird and ground) is much larger than the size of the bird. Field lines near the charged bird are not parallel, but are more similar to the field of a charged sphere.

A sphere with radious R has a capacity 4*Pi*epsilon0*R (when charged with a voltage relative to a large object infinitely far away).

If R=0.1 m then C=1.1*10^-11 As/V

With this capacity and U0=765 kV, f=50 Hz I got the result:

I0=2.7 mA

14. Feb 12, 2008

### rbj

i agree. with DC i would expect identical potential. i don't even know what the potential difference would be, but am just trying to guess at what charge will flow in and out because it's AC and there is some capacitance, even though i am groping to guess at how much.

i think a closer model is a finite plate in parallel to an infinite plate. but i'm too lazy to integrate so i think it might be in the same ballpark as an hemisphere about the size of the bat's head and another hemisphere, which is perhaps half of the capacitance of concentric spheres of radii: a < b.

$$C = \frac{2 \pi \epsilon_0}{1/a \ - \ 1/b }$$

wow, once b gets big enough, it doesn't really matter what it is, so it's a function only of the size of the bird or bat. actually that might make sense, for the field produced by an infinite plane of charge is constant no matter how far away you are from it (that infinite plane of charge still looks like an infinite plane of charge, even if you're a lightyear away from it). so now it's about 101 x 10-11 x 10-2 or about a pico farad instead of atto farads. maybe a little more juice sloshing in and outa the bird.

damn Europeans. if God had intended you to use 50 Hz for your power, she'd have given you five digits on your hand.