Question

A potential difference of 1.20 V will be applied to a 33.0 m length of 18-gauge copper wire (diameter = 0.0400 in.). Calculate (a) the current, (b) the magnitude of the current density, (c) the magnitude of the electric field within the wire, and (d) the rate at which thermal energy will appear in the wire.

Answer 1

Answer:A) Current = 1.739A, B)current density, J = 2.147x10^6 A/m2

magnitude of electric field , E =  0.036 N/C

)rate of thermal energy, P  =2.086W

Explanation:

Resistance  = R =   ρL/A

But the cross-section area of the wire. is given as

Diameter / 2 = 0.04/2 =0.02in to m = 0.02 / 39.37= 0.000508

A = πr^2 = π x  0.000508^2 = 8.10 x 10^-7

since resistivity of copper,ρ= 17x10-9 ohm.m

so resistance is   R =   ρL/A

17x10-9  x 33 / 8.1x10-7

= 0.69 ohm.

A) Current =    I = Voltage /Resistance =1.20/0.69 =1.739A

B)current density, J = Current /Area

= 1.739/8.1x10-7

= 2.147x10^6 A/m2

c)magnitude of electric field , E =  Current density x resistivity =J ρ

E = 2.147 x 10^6  x 17  x 10^-9

E = 0.036 N/C

D)rate of thermal energy, P  = I² R =1.739² X 0.69

=2.086W