Question

In: Physics

An aluminum wire consists of the three segments shown in the following figure. The current in...

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.

Solutions

Expert Solution

Current through each segment

I_top=I_mid=I_bottom=13 A

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

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

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

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

genius_generous answered 1 year ago

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