Answer: [tex]0.0039\ A[/tex]
Explanation:
Given
Diameter of the rod [tex]d=2.54\ cm[/tex]
length of rod is [tex]l=20\ cm[/tex]
Resistivity of silicon is [tex]\rho=6.4\times 10^2\ \Omega-m[/tex]
cross-section of the rod [tex]A[/tex]
[tex]\Rightarrow A=\dfrac{\pi d^2}{4}\\\\\Rightarrow A=\dfrac{3.142\times 2.54^2\times 10^{-4}}{4}\\\\\Rightarrow A=5.067\times 10^{-4}\ m^2[/tex]
Resistance of rod is R
[tex]\Rightarrow R=\dfrac{\rho l}{A}[/tex]
[tex]\Rightarrow R=\dfrac{640\times 0.20}{5.067\times 10^{-4}}\\\\\Rightarrow R=25.26\times 10^4\ \Omega[/tex]
Current is given by
[tex]\Rightarrow I=\dfrac{V}{R}\\\\\Rightarrow I=\dfrac{1000}{25.26\times 10^4}\\\\\Rightarrow I=0.0039\ A[/tex]
