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It's a pity to see so many misinformed, high-ranking answers on this question - so I decided to finally open an account and contribute, after years of lurking :)

One way too see power transmission is current going through the wire - modeled as kinetic energy of particles (electrons) within. However, especially in AC installations, if one models electromagnetic energy (through Maxwell's equations), one sees the power carried in the space between and around the conductors.

So there is EM danger for anything even close to the lines. Its level depends, for a given bird-line system, on the overall power going through - tension and intensity!

This quantitative answer I found on https://van.physics.illinois.edu/qa/listing.php?id=1341 applies:

Q: Why don't birds feel a static shock? I understand that birds don't provide a path to ground, so they won't carry a steady-state current. But surely when they first land there is a current to charge up the bird's capacitance? I've read that static shocks are painful at around 10 kV. These power lines carry hundreds of kV, so wouldn't the static shock from a power line be very painful? Thanks, Ted

• Ted (age 26) Stanford, CA, USA

A: Yes, It is not strictly true that there will not be any current at all. There are currents, but they are really small, and this not limited to landing only. Perhaps most negligible of all, the humid air is not a perfect insulator, so there will be losses from the body of the bird. But as you also point out, a bird can be considered a (roughly spherical) capacitor with second shell infinitely far away and at 0 potential. Therefore the bird will be charged and discharged at f=60Hz (50 Hz in Europe), because the power lines carry AC, not DC.

Lets make a rough calculation considering the bird as a sphere with 20cm diameter, the capacitance C should then be ~10pF. The rms current is then 2πfVrmsC f. Say there is 100kV on the wires, these parameters give about 400 µA for the rms current. For comparison, for a human being AC currents of around 10 mA start to become dangerous.(https://en.wikipedia.org/wiki/Electric_shock) For a bird, somewhat lower currents presumably can be dangerous. It sounds like even for the high voltage line, however, the purely capacitative current isn't quite a problem.

Tunc + Mike W.

It's a pity to see so many misinformed, high-ranking answers on this question - so I decided to finally open an account and contribute, after years of lurking :)

One way to see power transmission is current going through the wire - modeled as kinetic energy of particles (electrons) within. However, especially in AC installations, if one models electromagnetic energy (through Maxwell's equations), one sees the power carried in the space between and around the conductors.

So there is EM danger for anything even close to the lines. Its level depends, for a given bird-line system, on the overall power going through - tension and intensity!

This quantitative answer I found on https://van.physics.illinois.edu/qa/listing.php?id=1341 applies:

Q: Why don't birds feel a static shock? I understand that birds don't provide a path to ground, so they won't carry a steady-state current. But surely when they first land there is a current to charge up the bird's capacitance? I've read that static shocks are painful at around 10 kV. These power lines carry hundreds of kV, so wouldn't the static shock from a power line be very painful? Thanks, Ted

• Ted (age 26) Stanford, CA, USA

A: Yes, It is not strictly true that there will not be any current at all. There are currents, but they are really small, and this not limited to landing only. Perhaps most negligible of all, the humid air is not a perfect insulator, so there will be losses from the body of the bird. But as you also point out, a bird can be considered a (roughly spherical) capacitor with second shell infinitely far away and at 0 potential. Therefore the bird will be charged and discharged at f=60Hz (50 Hz in Europe), because the power lines carry AC, not DC.

Lets make a rough calculation considering the bird as a sphere with 20cm diameter, the capacitance C should then be ~10pF. The rms current is then 2πfVrmsC f. Say there is 100kV on the wires, these parameters give about 400 µA for the rms current. For comparison, for a human being AC currents of around 10 mA start to become dangerous.(https://en.wikipedia.org/wiki/Electric_shock) For a bird, somewhat lower currents presumably can be dangerous. It sounds like even for the high voltage line, however, the purely capacitative current isn't quite a problem.

Tunc + Mike W.

It's a pity to see so many misinformed, high-ranking answers on this question - so I decided to finally open an account and contribute, after years of lurking :)

One way too see power transmission is current going through the wire - modeled as kinetic energy of particles (electrons) within. However, especially in AC installations, if one models electromagnetic energy (through Maxwell's equations), one sees the power carried in the space between and around the conductors.

So there is danger for anything near-by. Its level depends on the overall power going through - tension and intensity!

This quantitative answer I found on https://van.physics.illinois.edu/qa/listing.php?id=1341 applies:

Q: Why don't birds feel a static shock? I understand that birds don't provide a path to ground, so they won't carry a steady-state current. But surely when they first land there is a current to charge up the bird's capacitance? I've read that static shocks are painful at around 10 kV. These power lines carry hundreds of kV, so wouldn't the static shock from a power line be very painful? Thanks, Ted

• Ted (age 26) Stanford, CA, USA

A: Yes, It is not strictly true that there will not be any current at all. There are currents, but they are really small, and this not limited to landing only. Perhaps most negligible of all, the humid air is not a perfect insulator, so there will be losses from the body of the bird. But as you also point out, a bird can be considered a (roughly spherical) capacitor with second shell infinitely far away and at 0 potential. Therefore the bird will be charged and discharged at f=60Hz (50 Hz in Europe), because the power lines carry AC, not DC.

Lets make a rough calculation considering the bird as a sphere with 20cm diameter, the capacitance C should then be ~10pF. The rms current is then 2πfVrmsC f. Say there is 100kV on the wires, these parameters give about 400 µA for the rms current. For comparison, for a human being AC currents of around 10 mA start to become dangerous.(https://en.wikipedia.org/wiki/Electric_shock) For a bird, somewhat lower currents presumably can be dangerous. It sounds like even for the high voltage line, however, the purely capacitative current isn't quite a problem.

Tunc + Mike W.

It's a pity to see so many misinformed, high-ranking answers on this question - so I decided to finally open an account and contribute, after years of lurking :)

One way too see power transmission is current going through the wire - modeled as kinetic energy of particles (electrons) within. However, especially in AC installations, if one models electromagnetic energy (through Maxwell's equations), one sees the power carried in the space between and around the conductors.

So there is EM danger for anything even close to the lines. Its level depends, for a given bird-line system, on the overall power going through - tension and intensity!

This quantitative answer I found on https://van.physics.illinois.edu/qa/listing.php?id=1341 applies:

Q: Why don't birds feel a static shock? I understand that birds don't provide a path to ground, so they won't carry a steady-state current. But surely when they first land there is a current to charge up the bird's capacitance? I've read that static shocks are painful at around 10 kV. These power lines carry hundreds of kV, so wouldn't the static shock from a power line be very painful? Thanks, Ted

• Ted (age 26) Stanford, CA, USA

A: Yes, It is not strictly true that there will not be any current at all. There are currents, but they are really small, and this not limited to landing only. Perhaps most negligible of all, the humid air is not a perfect insulator, so there will be losses from the body of the bird. But as you also point out, a bird can be considered a (roughly spherical) capacitor with second shell infinitely far away and at 0 potential. Therefore the bird will be charged and discharged at f=60Hz (50 Hz in Europe), because the power lines carry AC, not DC.

Lets make a rough calculation considering the bird as a sphere with 20cm diameter, the capacitance C should then be ~10pF. The rms current is then 2πfVrmsC f. Say there is 100kV on the wires, these parameters give about 400 µA for the rms current. For comparison, for a human being AC currents of around 10 mA start to become dangerous.(https://en.wikipedia.org/wiki/Electric_shock) For a bird, somewhat lower currents presumably can be dangerous. It sounds like even for the high voltage line, however, the purely capacitative current isn't quite a problem.

Tunc + Mike W.

It's a pity to see so many misinformed, high-ranking answers on this question - so I decided to finally open an account and contribute, after years of lurking :)

One way too see power transmission is current going through the wire - mostly kinetic energy of particles (electrons) within. However, if one models electromagnetic energy (especially in AC installations), one sees the power carried in the space between and around the conductors.

So there is danger for anything near-by. Its level depends on the overall power going through - tension and intensity!

This quantitative answer I found on https://van.physics.illinois.edu/qa/listing.php?id=1341 applies:

Q: Why don't birds feel a static shock? I understand that birds don't provide a path to ground, so they won't carry a steady-state current. But surely when they first land there is a current to charge up the bird's capacitance? I've read that static shocks are painful at around 10 kV. These power lines carry hundreds of kV, so wouldn't the static shock from a power line be very painful? Thanks, Ted

• Ted (age 26) Stanford, CA, USA

A: Yes, It is not strictly true that there will not be any current at all. There are currents, but they are really small, and this not limited to landing only. Perhaps most negligible of all, the humid air is not a perfect insulator, so there will be losses from the body of the bird. But as you also point out, a bird can be considered a (roughly spherical) capacitor with second shell infinitely far away and at 0 potential. Therefore the bird will be charged and discharged at f=60Hz (50 Hz in Europe), because the power lines carry AC, not DC.

Lets make a rough calculation considering the bird as a sphere with 20cm diameter, the capacitance C should then be ~10pF. The rms current is then 2πfVrmsC f. Say there is 100kV on the wires, these parameters give about 400 µA for the rms current. For comparison, for a human being AC currents of around 10 mA start to become dangerous.(https://en.wikipedia.org/wiki/Electric_shock) For a bird, somewhat lower currents presumably can be dangerous. It sounds like even for the high voltage line, however, the purely capacitative current isn't quite a problem.

Tunc + Mike W.

It's a pity to see so many misinformed, high-ranking answers on this question - so I decided to finally open an account and contribute, after years of lurking :)

One way too see power transmission is current going through the wire - modeled as kinetic energy of particles (electrons) within. However, especially in AC installations, if one models electromagnetic energy (through Maxwell's equations), one sees the power carried in the space between and around the conductors.

So there is danger for anything near-by. Its level depends on the overall power going through - tension and intensity!

This quantitative answer I found on https://van.physics.illinois.edu/qa/listing.php?id=1341 applies:

Q: Why don't birds feel a static shock? I understand that birds don't provide a path to ground, so they won't carry a steady-state current. But surely when they first land there is a current to charge up the bird's capacitance? I've read that static shocks are painful at around 10 kV. These power lines carry hundreds of kV, so wouldn't the static shock from a power line be very painful? Thanks, Ted

• Ted (age 26) Stanford, CA, USA

A: Yes, It is not strictly true that there will not be any current at all. There are currents, but they are really small, and this not limited to landing only. Perhaps most negligible of all, the humid air is not a perfect insulator, so there will be losses from the body of the bird. But as you also point out, a bird can be considered a (roughly spherical) capacitor with second shell infinitely far away and at 0 potential. Therefore the bird will be charged and discharged at f=60Hz (50 Hz in Europe), because the power lines carry AC, not DC.

Lets make a rough calculation considering the bird as a sphere with 20cm diameter, the capacitance C should then be ~10pF. The rms current is then 2πfVrmsC f. Say there is 100kV on the wires, these parameters give about 400 µA for the rms current. For comparison, for a human being AC currents of around 10 mA start to become dangerous.(https://en.wikipedia.org/wiki/Electric_shock) For a bird, somewhat lower currents presumably can be dangerous. It sounds like even for the high voltage line, however, the purely capacitative current isn't quite a problem.

Tunc + Mike W.