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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. 2711, which (in magnitude) reads EC = ρJ . Combining this with
Eq. 277, we obtain
EC = ρnevd .
Now, regarding the Hall eﬀect, we use Eq. 2910 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 ...
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This note was uploaded on 11/12/2011 for the course PHYS 2001 taught by Professor Sprunger during the Fall '08 term at LSU.
 Fall '08
 SPRUNGER
 Physics, Current

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