1 j y vii according to eqn

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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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This document was uploaded on 01/15/2014.

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