Phys 181b Chapter 32 Solutions

Phys 181b Chapter 32 Solutions - 1. (a) The flux through...

Info iconThis preview shows pages 1–4. Sign up to view the full content.

View Full Document Right Arrow Icon
1. (a) The flux through the top is +(0.30 T) π r 2 where r = 0.020 m. The flux through the bottom is +0.70 mWb as given in the problem statement. Since the net flux must be zero then the flux through the sides must be negative and exactly cancel the total of the previously mentioned fluxes. Thus (in magnitude) the flux though the sides is 1.1 mWb. (b) The fact that it is negative means it is inward.
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
16. (a) From Eq. 32-10, () ( ) 54 4 00 0 0 2 12 2 2 4 2 8 4.0 10 6.0 10 6.0 10 V m s C 8.85 10 4.0 10 m 6.0 10 V m s Nm 2.1 10 A. E d d dE d iA A t A dt dt dt εε ε −− Φ ªº == × × = ×⋅ ¬¼ §· =− × × × ¨¸ ©¹ × Thus, the magnitude of the displacement current is 8 | | 2.1 10 A. d i (b) The negative sign in d i implies that the direction is downward. (c) If one draws a counterclockwise circular loop s around the plates, then according to Eq. 32-18 s d Bd s i z ⋅= < G G µ 0 0, which means that G G B ds ⋅< 0 . Thus G B must be clockwise.
Background image of page 2
18. (a) Since i = i d (Eq. 32-15) then the portion of displacement current enclosed is ii R i d R , .
Background image of page 3

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Image of page 4
This is the end of the preview. Sign up to access the rest of the document.

This note was uploaded on 07/18/2008 for the course PHYS 181 taught by Professor Stephenirons during the Spring '08 term at Yale.

Page1 / 5

Phys 181b Chapter 32 Solutions - 1. (a) The flux through...

This preview shows document pages 1 - 4. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online