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The current through a 10 m long wire has a current density of 4 cross times 10 to the power of 6 space open parentheses bevelled A over m squared close parentheses. The wire conductivity is 2 cross times 10 to the power of 7 space open parentheses bevelled S over m close parentheses. Find the voltage drop across the wire. (Answer with the numeric value, don't write the unit V)

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onyebuchinnaji

Answer:

The voltage drop across the wire is 2 V

Explanation:

Given;

length of wire, L = 10 m

current density, I/A, μ = 4 x 10⁶ (A/m²)

wire conductivity, σ = 2 x 10⁷ (S/m)

The resistivity of wire is given by;

[tex]\rho = \frac{RA}{L} \\\\But \ R = V/I\\\\\rho = \frac{VA}{IL}[/tex]

Conductivity, σ = ¹/ρ

[tex]\sigma = \frac{IL}{VA}\\ \\V = \frac{IL}{ A \sigma}\\\\V = (\frac{I}{A})\frac{L}{\sigma}\\ \\V = (\mu)\frac{L}{\sigma}\\\\V = (4*10^{6} )*\frac{10}{2*10^{7} } \\\\V = 2 \ V[/tex]

Therefore, the voltage drop across the wire is 2 V

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