Question

1. A potential difference of 1.20 V will be applied to a 27.00 m length of 18-gauge aluminum wire (diameter = 0.0400 in). Calculate the current.

2. Calculate the magnitude of the current density.

3. Calculate the magnitude of the electric field within the wire.

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

Answer 1

(1) Diameter of th wire = 0.0400 in

So, radius, r = 0.0200 in = 5.08 x 10^-4 m

so, cross-sectional area, A = pi*r^2 = 3.141*(5.08 x 10^-4)^2 = 8.106 x 10^-7 m^2

length of the wire, L = 27.0 m

So, resistance of the wire, R = k*L / A= (2.82 x 10^-8 x 27) / (8.106 x 10^-7) = 0.939 ohm

so current, I = V / R = 1.20 / 0.939 = 1.28 A

(2) Current density = I / A = 1.28 / (8.106 x 10^-7) = 1.57 x 10^6 A/m^2

(3) Magnitude of the electric field, E = V / L = 1.20 / 27 = 0.044 V/m

(4) Rate at which thermal energy will appera = I^2 x R = 1.28^2 x 0.939 = 1.538 W.