Question

An aluminum wire with a diameter of 0.105 mm has a uniform electric field of 0.325 V/m imposed along its entire length. The temperature of the wire is 35.0°C. Assume one free electron per atom.

(a) Use the information in this Table of Resistivities and Temperature Coefficients to determine the resistivity (in Ω · m) of aluminum at this temperature.

ρ = _______Ω · m

(b) What is the current density (in MA/m²) in the wire?

J = ________MA/m²

(c) What is the total current (in mA) in the wire?

I = __________ mA

(d)What is the drift speed of the conduction electrons?

v_d = _________ µm/s

(e) What potential difference must exist between the ends of a 2.30 m length of the wire to produce the stated electric field?

ΔV = __________ V

Answer 1

We are taking the room temperature to be 20 degree Celsius.

a)T1=20

for aluminum, resistivity is

p =2.82*10^-8 ohm-m

and temperature coefficient is

alpha =3.9*10^-3 /C

p=pi[1+alpha*{T2-T1)]

p=2.82*10^-8[1+3.9*10^-3(35-20)]

p= 2.98*10^-8 ohm-m

b)

J=E/p = 0.325/(2.98*10^-8)

J=10.9*10^6 A/m^2

or 10.9 MA/m^2

c)

r=d/2 =0.0525mm

A=pir^2 =pi*(0.0525*10^-3)^2 =8.66*10^-9 m^2

I=JA =10.9*10^6*8.66 *10^-9

I= 94.4 mA

d)

n=pNA/Mal =2700*6.02*10^23/(27*10^-3)

n=6.02*10^28

vd =J/ne = 10.9*10^6/(6.02*10^28*1.6*10^-19)

vd= 1.13 *10^-3 m/s

vd = 1132 um/s

e)

dV=EL =0.325*2.30 = 0.7475 V