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(D) The current $I$ due to a charge carrier with charge $q$,number density $n$,cross-sectional area $A$,and drift velocity $v_d$ is given by $I = n q

Vedclass · Accepted Answer

$3nev A/2$

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(D) The current $I$ due to a charge carrier with charge $q$,number density $n$,cross-sectional area $A$,and drift velocity $v_d$ is given by $I = n q A v_d$.For the positive ions $(+2e)$: The charge is $q_1 = +2e$,number density is $n$,and drift speed is $v_1 = v/4$. The current is $I_1 = n(2e)A(v/4) = nevA/2$.For the negative ions $(-e)$: The charge is $q_2 = -e$,number density is $n$,and drift speed is $v_2 = v$. Since the negative ions move in the direction opposite to the positive ions,their contribution to the conventional current is in the same direction as the positive ions. The magnitude of current is $I_2 = n|q_2|Av_2 = n(e)Av = nevA$.The total current $I_{total} = I_1 + I_2 = nevA/2 + nevA = 3nevA/2$.

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