An aluminum wire with a diameter of 0.1 mm has a uniform electric field with a magnitude of 0.2 V/m imposed along its entire length.

Question

The temperature of the wire is 50 degrees (Celsius). Given that at 20 degrees, ρo = 2.82x10^-8 Ωm and α = 3.9x10^-3 /C. Determine:

a) the resistivity of the wire

b) the current density in the wire

c) the total current in the wire

d) the potential different that must exist between the ends of a 2m length of wire if the given electric field is to be produced.

Answer 1

Hi Claire!

 

This is basically a lot of equation hopping, so let's line them up! :)

 

a)  To get the resistivity ρ at 50 Celsius, given the resitstivity at 20 Celsisus, use:

 

ρ = ρ_o(1 + α(T - T_o))

 

where T_o = 20 Celsius

 

b)  Knowing the resistivity at 50 Celsius, and the (uniform) electric field E, you can determine the current density J using:

 

E = ρJ

 

(which is actually a density-averaged version of V = IR)

 

c)  Assuming the current is uniform (which is should be in a uniform electric field and constant-diameter wire), the current i can be calculated using:

 

J = i/A -->  i = JA

 

where A is the cross-sectional area of the wire (given by πr²); make sure to convert the given diameter to a radius, and the radius to base units

 

d)  Since the electric field is given in volts per meter, and you have two meters of length in the wire, you can determine directly from that how many volts difference you need at the ends of the wire to get 0.2 volts per meter.

 

0.2 =  V/d

 

with d = 2 m.  This corresponds to a uniform electric field being related to voltage by V = Ed, where d is distance along the field line.

 

I hope this helps!  If you have any questions about this, or would like to check an answer, just let me know.