Question

A current carrying gold wire has diameter 0.84 mm. The electric field in the wire is 0.49 V/m. What are (a) the current carried by the wire; (b) the potential difference between two points in the wire 6.4 m apart; (c) the resistance of a 6.4-m length of this wire

Answer 1

Step-by-step explanation:

Given that,

Diameter of the gold wire, d = 0.84 mm

Radius, r = 0.42 mm

Electric field in the wire, E = 0.49 V/m

(a) Electric current density is given by :

`J=(I)/(A)`

And electric field is :

`E=\rho J`

`\rho` is resistivity of Gold wire

So,

`E=(\rhi I)/(A)\\\\I=(EA)/(\rho)\\\\I=(0.49* \pi (0.42* 10^(-3))^2)/(2.44* 10^(-8))\\\\I=11.12\ A`

(b) The potential difference between two points in the wire is given by :

`V=E* l\\\\V=0.49 * 6.4\\\\V=3.136\ V`

(c) Resistance of a wire is given by :

`R=\rho (l)/(A)\\\\R=2.44* 10^(-8)* (6.4)/(\pi (0.42* 10^(-3))^2)\\\\R=0.281\ \Omega`

Hence, this is the required solution.