Question

A potential difference of 3.80 V will be applied to a 31.00 m length of 18-gauge silver wire (diameter = 0.0400 in). Calculate the current.

Tries 0/10

Calculate the magnitude of the current density.

Tries 0/10

Calculate the magnitude of the electric field within the wire.

Tries 0/10

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

Answer 1

here,

potential difference , V = 3.8 V

length , l = 31 m

diameter , d = 0.04 in = 0.001016 m

resistivity of silver , p =1.59 * 10^-8 ohm.m

a)

the current , I = V/ R

I = V /( p * l /(pi * (d/2)^2))

I = 3.8 /( 1.59 * 10^-8 * 31 /(pi * ( 0.001016/2)^2))

I = 6.25 A

b)

the magnitude of the current density , J = I /area

J = 6.25 /(pi *( 0.001016/2)^2)

J = 7.71 * 10^6 A/m^2

c)

the magnitude of the electric field within the wire , E = V / l

E = 3.8 /31 = 0.12 N/C

d)

the rate at which thermal energy will appear in the wire , P = V * I

P = 3.8 * 6.25 W = 23.75 W