P32_037 - area enclosed by the path area of each plate = (2...

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
37. (a) At any instant the displacement current i d in the gap between the plates equals the conduction current i in the wires. Thus i d = i =2 . 0A. (b) The rate of change of the electric Feld is dE dt = 1 ε 0 A µ ε 0 d Φ E dt = i d ε 0 A = 2 . 0A (8 . 85 × 10 12 ± / m)(1 . 0m) 2 =2 . 3 × 10 11 V m · s . (c) The displacement current through the indicated path is i 0 d = i d × µ
Background image of page 1
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: area enclosed by the path area of each plate = (2 . 0 A) . 50 m 1 . 0 m 2 = 0 . 50 A . (d) The integral of the Feld around the indicated path is I ~ B d~s = i d = (1 . 26 10 6 H / m)(0 . 50 A) = 6 . 3 10 7 T m ....
View Full Document

This note was uploaded on 11/12/2011 for the course PHYS 2001 taught by Professor Sprunger during the Fall '08 term at LSU.

Ask a homework question - tutors are online