Question

the resistivity of gold is 2.44×10−8Ω⋅m at room temperature. A gold wire that is 0.9 mm in diameter and 14 cm long carries a current of 940 mA. What is the electric field in the wire

Answer 1

0.03605 V/m is the electric field in the gold wire.

Step-by-step explanation:

Resistivity of the gold =
`\rho = 2.44* 10^(-8) \Omega.m`

Length of the gold wire = L = 14 cm = 0.14 m ( 1 cm = 0.01 m)

Diameter of the wire = d = 0.9 mm

Radius of the wire = r = 0.5 d = 0.5 × 0.9 mm = 0.45 mm =
`0.45* 0.001 m`

( 1mm = 0.001 m)

Area of the cross-section =
`A=\pi r^2=\pi r^(0.45* 0.001 m)^2`

Resistance of the wire = R

Current in the gold wire = 940 mA = 0.940 A ( 1 mA = 0.001 A)

`R=\rho* (L)/(A)`

`V(voltage)=I(current)* R(Resistance)` ( Ohm's law)

`(V)/(I)=\rho* (L)/(A)`

We know, Electric field is given by :

`E=(dV)/(dr)`

`E=(V)/(L)`

`E=(V)/(L)=\rho* (I)/(A)`

`E=2.44* 10^(-8) \Omega.m* (0.940 A)/(\pi r^(0.45* 0.001 m)^2)=0.03605 V/m`

0.03605 V/m is the electric field in the gold wire.