Answer:
E=0.036 V/m
Explanation:
Given that
Resistivity ,ρ=2.44 x 10⁻⁸ ohms.m
d= 0.9 mm
L= 14 cm
I = 940 m A = 0.94 A
We know that electric field E
E= V/L
V= I R
R=ρL/A
So we can say that
E= ρI/A
Now by putting the values
[tex]E=\dfrac{ 2.44\times 10^{-8}\times 0.94}{\dfrac{\pi}{4}(0.9\times 10^{-3})^2}[/tex]
E=0.036 V/m
The electric field in the gold wire is 0.036V/m.
Hence, Option E) 0.036V/m is the correct answer.
Given the data in the question;
Electric field in the wire; [tex]E =\ ?[/tex]
First we find the cross-sectional area of the gold wire;
Area; [tex]A = \pi r^2 = \pi (\frac{d}{2})^2[/tex]
We substitute in our values
[tex]A = \pi (\frac{0.0009m}{2})^2 \\\\A = 6.36*10^{-7}m^2[/tex]
Next, we calculate the current density( J ), which is the ratio of Current to Area
Current density( J ) = Current / Area
[tex]J = \frac{I}{A}[/tex]
We substitute our values into the equation
[tex]J = \frac{0.94A}{6.36*10^{-7}m^2} \\\\J = 1.478 * 10^6 A/m^2[/tex]
Now, We know that, using the expression for electric field (E) inside a current conductor:
[tex]E = pJ[/tex]
Where E is electric field, [tex]p[/tex] is the resistivity of gold and [tex]J[/tex] the current density.
We substitute our values into the equation
[tex]E = (2.44 * 10^{-8} ohms.m) * ( 1.478*10^6A/m^2)\\\\E = 0.036 V/m[/tex]
The electric field in the gold wire is 0.036V/m.
Hence, Option E) 0.036V/m is the correct answer.
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