10 to the 27th power particles is not so large a quantity.
One mol (a standard unit equal to 6.022 x 10 to the 23rd power particles) is equivalent to 22.4 liters of, e.g., air at standard temperature and pressure. 224,000 liters of air would thus contain about 6 times 10 to the 27th power particles.
So a mere 10 to the 27th power particles per second would be equivalent to about 37,000 liters of air per second.
However, since we are talking about charged particles here, perhaps we should convert that to amperes. 1 A = 6.24 x 10^18 charged particles per second. So 10 to the 27th power charged particles per second would equal about 166 million amps.
Regards,
you are so smrt.
Then there is the duration of the event and the cross section of the Earth’s surface that is affected. In a worse case scenario it will probably suck the paint off your house and give your family a permanent orange afro.
SHOW OFF!!!!!
Roughly, how many watts of power would that be?
Thank you for making this stuff so clear
“...So a mere 10 to the 27th power particles per second would be equivalent to about 37,000 liters of air per second...”
Have you ever had that much air blowing through your breaches, smart guy? Could hurt a fellow, you know.
/s
At what voltage?
However the current is spread over a hole bigger than the earth. They don't say how much bigger, but if you use the cross section of the earth, you'd get about 1.3 x 10-6 amps per square meter, or 1.3 x 10-12 amps per square millimeter. A wire with a cross section of 1 square millimeter would about be about 28 gage, which could handle a square meter's worth of that current with no problem, in fact it could handle a square kilometer's worth of the current, or 1.3 amps, since 28 guage is rated for 1.4 amps when used for chassis wiring.
Acolytes of the Church of St. Algore ask how many windmills would that be?