A small circular loop of 5 mm radius is placed 1 m away from a 60-Hz power line. The voltage induced on this loop is measured at
0.6 µV. What is the current on the power line
1 answer:
Answer:
i = 101.4A
Explanation:
The steps to the solution can be found in the attachment below.
We have been given the frequency f = 60Hz. From this we can calculate the angular frequency of the power line.
It is assumed that sinwt = –1 in order to calculate the time varying current. Although the magnitude of the current is large, consideration should also be given to the distance between the coil and the power line. The induced emf is small considering the area of the coil which is 7.85×10-⁵m².
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Answer:
3 km/h
Explanation:
Let's call the rowing speed in still water x, in km/h.
Rowing speed in upstream is: x - 2 km/h
Rowing speed in downstream is: x + 2 km/h
It took a crew 9 h 36 min ( = 9 3/5 = 48/5) to row 8 km upstream and back again. Therefore:
8/(x - 2) + 8/(x + 2) = 48/5 (notice that: time = distance/speed)
Multiplying by x² - 2², which is equivalent to (x-2)*(x+2)
8*(x+2) + 8*(x-2) = (48/5)*(x² - 4)
Dividing by 8
(x+2) + (x-2) = (6/5)*(x² - 4)
2*x = (6/5)*x² - 24/5
0 = (6/5)*x² - 2*x - 24/5
Using quadratic formula
A negative result has no sense, therefore the rowing speed in still water was 3 km/h