Answer :
Answer:
0.0360531138247 V/m
Explanation:
[tex]\rho[/tex] = Resistivity of gold = [tex]2.44\times 10^{-8}\ \Omega .m[/tex] (General value)
I = Current = 940 mA
d = Diameter = 0.9 mm
A = Area = [tex]\dfrac{\pi}{4}d^2[/tex]
E = Electric field
Resistivity is given by
[tex]\rho=\dfrac{EA}{I}\\\Rightarrow E=\dfrac{\rho I}{A}\\\Rightarrow E=\dfrac{2.44\times 10^{-8}\times 940\times 10^{-3}}{\dfrac{\pi}{4}(0.9\times 10^{-3})^2}\\\Rightarrow E=0.0360531138247\ V/m[/tex]
The electric field in the wire is 0.0360531138247 V/m
Answer:
Explanation:
Current in the wire, i = 940 mA = 0.94 A
Length of the wire, l = 14 cm = 0.14 m
diameter of wire = 0.9 mm
radius of wire, r = 0.45 mm = 0.45 x 10^-3 m
resistivity of gold, ρ = 2.44 x 10^-8 ohm metre
Let R is the resistance of the wire.
[tex]R = \rho \times \frac{l}{A}[/tex]
[tex]R = 2.44\times 10^{-8}\times \frac{0.14}{3.14\times 0.45\times 0.45\times 10^{-6}}[/tex]
R = 5.37 x 10^-3 ohm
By the Ohm's law
V = i x R
V = 0.94 x 5.37 x 10^-3
V = 5.05 x 10^-3 V
E = V / l = (5.05 x 10^-3) / 0.14 = 0.036 V/m