Question

An aluminum wire consists of the three segments shown in the following figure. The current in the top segment is 13A

Part B

For each of these three segments, find the current density J.

Part C

For each of these three segments, find the electric field E.

Part D

For each of these three segments, find the drift velocity vd.

Part E

For each of these three segments, find the electron current i.

Answer 1

a)

Current through each segment

I_top=I_mid=I_bottom=13 A

b)

radius r=d/2

Current density

J=I/A =I/pir²

J_top=13/(pi*(1*10^-3)²=4.14*10⁶ A/m²

J_mid=13/(pi*(0.5*10^-3)²=1.66*10⁷ A/m²

J_bottom=13/(pi*(1*10^-3)²=4.14*10⁶ A/m²

c)

E=J/sigma

E_top=(4.14*10⁶)/(3.5*10⁷)=0.1183 V/m

E_mid=(1.66*10⁷)/(3.5*10⁷)=0.474 V/m

E_bottom=0.1183 V/m

d)

Drift velocity

V_d=J/ne

V_top=(4.14*10⁶)/(6*10²⁸)(1.6*10^-19) =4.31*10^-4 m/s

V_mid=(1.66*10⁷)/(1.6*10^-19)(6*10²⁸)=1.73*10^-3 m/s

V_bottom=4.31*10^-4 m/s

e)

electron current

I_e=I/e =13/(1.6*10^-19) = 8.13*10¹⁹ electrons

I_top=8.13*10¹⁹ electrons

I_mid=8.13*10¹⁹ electrons

I_bottom=8.13*10¹⁹ electrons