Posted by
**Mary** on
.

Please help!! I have tried everything and is still coming up with and incorrect answer.

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.

For Further Reading

Physic please help! - drwls, Tuesday, October 9, 2007 at 7:49pm

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