Weeks10-11HWSolutions

0 i j i j e fn 1 m k t 2 f0 e

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Unformatted text preview: unctions. 4) On slide 32 of Lecture 24 (Lecture 15, Fall 2011) expressions for the tensors, ! ij and [s ] T ij , were given. Determine the corresponding expression for [! 0 ]ij . Solution: The heat current is J Qi = 1 ∑ E − Fn υ iδ f Ω k ( ) Note that E is the total energy, ! ! E = EC r + E k () () (i) (ii) (iii) The solution to the BTE is: % \$f ( ! f = " m ' # 0 * + jF j & \$E ) ECE ­656 7 Fall 2013 Mark Lundstrom 10/27/13 where the generalized force is: !! ! !1 F j = ! " j Fn ! # EC ( r ) + E k , r ! Fn ( r ) % " j T \$ &T !1 = ! " j Fn ! # E ! Fn ( r ) % " j T \$ &T () (iv) ⎛ ∂f0 ⎞ ⎧ ⎫ 1 1 ∑τ m ⎜ − ∂E ⎟ υ iυ j E − Fn ⎨−∂ j Fn − ⎡ E − Fn r ⎤ T ∂ jT ⎬ ⎣ ⎦ Ω k ⎝ ⎠ ⎩ ⎭ (v) Using (ii) and (iii), (i) becomes J Qi = ( ) () The second term in curly brackets gives the thermal conductivity. κ0 ) ( i, j υ iυ j E − Fn 1 = ∑τ m Ω k T ) 2 ⎛ ∂f0 ⎞ ⎜ − ∂E ⎟ ⎝ ⎠ 5) Solve the two equations below for the electric fields in the x ­ and y ­directions and show that the results on slide 23 of Lecture 25 (Lecture 16, Fall 2011) are correct.. J x = ! nE x " (! n µ n rH )E y Bz J y = ! nE y + (! n µ n rH )E x Bz Solution: Re ­arrange (i) and (ii) E x = ! n J x + ( µ n rH )E y Bz E y = ! n J y " ( µ n rH )E x Bz (i) (ii) (iii) (iv) (v) Insert (iv) in (iii) E x = ! n J x + ( µ n rH ) # ! n J y " ( µ n rH )E x Bz % Bz \$ & ( ) 2 E x 1 + ( µn rH ) BZ = ! n J x + ! n ( µn rH ) J y Bz ECE ­656 2 8 Fall 2013 Mark Lundstrom 10/27/13 Assume a low magnitude B ­field: (µ r B ) 2 << 1 (vi) (vii) E y = ! " n ( µ n rH Bz ) J x + " n J y...
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This document was uploaded on 01/15/2014.

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