Question

5. Using a surface integral calculation, find the current (in Amps) in a circular wire given a current density of)s 15(1-e-1000r)ez Amps/㎡ . That is, J has a magnitude of 15(1-e-1000r)as it is a function of distance r from the center to the outer surface. J points in the z direction. Draw the wire in a diagram with the cylindrical coordinate system and solve for I using a double integral. Hint- what is the surface area of the end of the wire where J passes through? You will find the differential surface expression as Equation A.5 on page 387 of the textbook. Also, the final answer will be in milliAmps (mA). Finally, if the resistance of the wire is 1.3 x 10 Ohms, calculate the voltage drop from one end to the other end. (20 pts.)

Answer 1

Som 군 p(3,0,2) Any point Can e J = cun unt density in wire 5)

Le. em 15(1-е 2 - JoooY along ds

ds ez So 0 Sinu en 2 So 1-e

-I οοογ -oo0 loooY .NMpe -(000γ IDoo looo

End ef wive Ro,-21) VP, Zp Resista 2.

P, (R, 0,-2) - 10002 (l000) I derends on R onl