Answer:
Current, I = 1200 A
Explanation:
It is given that,
Length of the cylinder, l = 40 m
Radius of the metal wire, r = 4 mm = 0.004 m
Voltage source, V = 16 V
Resistivity of metal, [tex]\rho=1.68\times 10^{-8}\ \Omega-m[/tex]
We know that the resistivity of any metal is given by the formula as :
[tex]R=\rho \dfrac{l}{A}[/tex]
[tex]R=1.68\times 10^{-8}\times \dfrac{40}{\pi (0.004)^2}[/tex]
R = 0.01336 ohms
Using ohm's law :
[tex]R=\dfrac{V}{I}[/tex]
[tex]I=\dfrac{V}{R}[/tex]
[tex]I=\dfrac{16}{0.01336}[/tex]
I = 1197.60 A
or
I = 1200 A
So, the current flowing in the cylindrical wire is 1200 A. Hence, this is the required solution.