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g 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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okpalawalter8

Answer:

The value is [tex]V = 2 V[/tex]

Explanation:

From the question we are told that

The length of the wire is [tex]l = 10 \ m[/tex]

The current density is [tex]J = 4*10^6 \ A/m^2[/tex]

The conductivity is [tex]\sigma = 2*10^{7} \ S/m[/tex]

Generally conductivity is mathematically represented as

[tex]\sigma = \frac{l}{RA}[/tex]

Here R is the resistance which is mathematically represented as

[tex]R = \frac{V}{I}[/tex]

Here I is the current which is mathematically represented as

[tex]I = J * A[/tex]

So

[tex]R = \frac{V}{ J * A}[/tex]

And

[tex]\sigma = \frac{l}{\frac{V}{ J * A} * A}[/tex]

=> [tex]\sigma = \frac{l}{\frac{V}{J}}[/tex]

=> [tex]V = \frac{l * J}{\sigma }[/tex]

=> [tex]V = \frac{10 * 4*10^6}{2*10^{7} }[/tex]

=> [tex]V = 2 V[/tex]

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