P29_013 - 13. (a) In Chapter 27, the electric field...

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

View Full Document Right Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: 13. (a) In Chapter 27, the electric field (called EC in this problem) which “drives” the current through the resistive material is given by Eq. 27-11, which (in magnitude) reads EC = ρJ . Combining this with Eq. 27-7, we obtain EC = ρnevd . Now, regarding the Hall effect, we use Eq. 29-10 to write E = vd B . Dividing one equation by the other, we get E/Ec = B/neρ. (b) Using the value of copper’s resistivity given in Chapter 27, we obtain E B 0.65 T = = = 2.84 × 10−3 . 28 / m3 )(1.60 × 10−19 C)(1.69 × 10−8 Ω · m) Ec neρ (8.47 × 10 ...
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