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The bolt of lightning depicted in Figure $P 20.61$ passes $200 . \mathrm{m}$ from a 100 -turn coil oriented as shown. If the current in the lightning bolt falls from $6.02 \times 10^{6} \mathrm{A}$ to zero in $10.5 \mu \mathrm{s},$ what is the average voltage induced in the coil? Assume the distance to the center of the coil determines the average magnetic field at the coil's position. Treat the lightning bolt as a long, vertical wire.

1.15 \times 10^{5} V

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Okay, So to find the, uh, induced current induced the meth we have, this equation is remember, new steam, my physical to number of turns times the rate of change of magnetic flux. So this is number times the rate of change off B times a magnetic field strength times area. Mary is not changing the time, So this will be 10 times a I'm still to be over Delta two. No, to find a Delta beers. Remember that. Um, by NPR's law be is you not I over to pi times distance D So this will turn into n times, Harry over Dr T Times and you not times Delta I because it's the change in magnetic field strength. So this is the change, and I over to Pied de Li That's deep ing the distance between the lightning bolt and the center of the coils. That's 200 meters, right? Uh oh. And also, um, this A over here is, um, high times radius of the coil squared. So you multiply all of that together. Tee times to pie comes D, right. So, uh, miss, these pies can be cancelled out. So what we have is N 100 terms our squares. That's 0.8 meters squared. You not is four pi times 10 to the negative seven in us I units change and current is 6.2 times, uh, 10 to the six pimps. All right, you divide that by Delta T, which is 10.5 times 10 to the negative. Six seconds. Templar, five microseconds times two times D, which is 200 meters. This gives you do you see a mouth off? About 100 and 15 kilovolts are 1.15 times tender there, five votes.