Unformatted text preview: (iva) J y = ! nS q" y (ivb) ECE
656 and use the results of problem 6b) to find the current densities as σn
J x = − nqυ x =
2 E x − µ n BzE y 1 + ( µn Bz ) ( J y = − nqυ y = σn 1 + ( µn Bz ) 2 (E x ) + µn BzE y 10 ) , (va) (vb) Fall 2013 Mark Lundstrom 10/27/13 which can also be written as ! Jx $
'n
#
&=
# J y & 1 + ( µn Bz )2
"
% )1
+
+ µn Bz
* ( µn Bz , ! E x
.#
1
.# E y
" $
& &
% (via)
(vib) or as J i = σ ij ( Bz )E j . Note that the magnetic field affects both the diagonal and off
diagonal components of the magnetoconductivity tensor. Explain why there is no Hall factor, rH , in the result. Solution: These results follow directly from (**). Note that we are considering the motion of an average electron with a momentum relaxation time of ! m . We are neglecting the variation of scattering time with energy and considering only an average case. 2
2
Consequently, ! m = ! m and ! m = ! m and the Hall factor is one. The algebra in this approach is simpler, but we are missing the fact that there is a Hall factor. 6d) Solve eqn. (via) for the electric field and show that !E
x
#
# Ey
" $
1
1)
+
&=
& ' n + ( µ n Bz
%
* µn Bz , ! J x $
.#
& 1 .# J y &
%
" (vii) According to eqn. (vii), the longitudinal magnetoresistivity is independent of the B
field (while the longitudinal magnetoconductivity depends on B as shown in eqn. (via). Equation (vii) shows that the Hall voltage is proportional to B. Solution: We can re
write (vi) as ! Jx $ ) '
L
#
&=+
Jy & + ' T
#
"
%* (' T , ! E x
.#
' L .# E y
" $
& &
% (***) where !L = !n 1 + ( µn Bz ) 2 !T = ! n µn Bz 1 + ( µn Bz ) 2 If we invert (***), we find ECE
6...
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 Fall '14

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