Answer:
I = 4.75 A
Explanation:
To find the current in the wire you use the following relation:
[tex]J=\frac{E}{\rho}[/tex] (1)
E: electric field E(t)=0.0004t2−0.0001t+0.0004
ρ: resistivity of the material = 2.75×10−8 ohm-meters
J: current density
The current density is also given by:
[tex]J=\frac{I}{A}[/tex] (2)
I: current
A: cross area of the wire = π(d/2)^2
d: diameter of the wire = 0.205 cm = 0.00205 m
You replace the equation (2) into the equation (1), and you solve for the current I:
[tex]\frac{I}{A}=\frac{E(t)}{\rho}\\\\I(t)=\frac{AE(t)}{\rho}[/tex]
Next, you replace for all variables:
[tex]I(t)=\frac{\pi (d/2)^2E(t)}{\rho}\\\\I(t)=\frac{\pi(0.00205m/2)^2(0.0004t^2-0.0001t+0.0004)}{2.75*10^{-8}\Omega.m}\\\\I(t)=4.75A[/tex]
hence, the current in the wire is 4.75A
