Question

A potential difference of 3.60 V will be applied to a 33.00 m length of 18-gauge platinum wire (diameter = 0.0400inches). Calculate the current.

Tries 0/20

Calculate the current density.

Tries 0/20

Calculate the electric field.

Tries 0/20

Calculate the rate at which thermal energy will appear in the wire.

Answer 1

We know, resistance in wire of length 'L' and area 'A' , R =ρ * (L/A) -----1

Resistivity of platinum wire = 10.6 * 10^-8 ohm m

Length of wire ,L = 33.0 m

Area of wire ,A = d^2 / 4 = ( (0.0400 *2.54 *10^-2) ^2 ) / 4 = 8.107 * 10^-7 m^2 [ 1 inch = 2.54*10^-2 m]

now from equation 1,

R = (10.6 *10^-8) * (33.0) / (8.107 *10^-7) = 4.31 ohm

we know current,I = V/R

potential difference,V = 3.60 v [given]

Therefore, I = 3.60 /4.31 = 0.83 Ampere. [answer part 1]

Current density, J = current / cross-section area of wire

= 0.83 / (8.107*10^-7)

= 1029200.10 A/m^2. [answer part 2]

Electric field = J / σ [where σ is electrical conductivity and is given by = 1/ resistivity = 1/ρ]

=J*ρ

= 1029200.10 *  10.6 *10^-8

= 0.109 = 0.11 V/m or 0.11 N/C. [answer part 3]

Heat generated in wire is due to resistance in wire therefore rate of thermal energy is given by W = I^2 * R = 0.83^2 * 4.31

  = 2.97 J/s or 2.97 watt.[answer part 4]