The change in B divided by the time (0.45 s), multiplied by the loop area in square meters, is the average voltage in the loop during the interval. The wire resistance is 2 pi R * 3.3 10-2 ohms(?)/m.
Coumpute the average current (Vav/R) and multiply it by the average voltage, Vav, computed by the method above.
Average Power =
[(delta B)(pi R^2)/(delta t)]^2/(2 pi R)
= (1/2) pi* R^3 [(delta B)/(delta t)]^2
Check my math and my logic. I could be wrong
A piece of copper wire is formed into a single circular loop of radius 13 cm. A magnetic field is oriented parallel to the normal to the loop, and it increases from 0 to 0.70 T in a time of 0.45 s. The wire has a resistance per unit length of 3.3 10-2 /m. What is the average electrical energy dissipated in the resistance of the wire.
2 answers
A thin tube stretch across a street counts the number of pairs of wheels that pass over it. A vehicle classified as Type A with two axles registers two counts. A vehicle classified as Type B with three axles registers three counts. During a 1 hour period a traffic counter registered 35 counts. How many type A vehicles and type b vehicles passed over the traffic?
0 answers