Question

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)

Answer 1

Answer:

The value is $V = 2 V$

Explanation:

From the question we are told that

The length of the wire is $l = 10 \ m$

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

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

Generally conductivity is mathematically represented as

$\sigma = \frac{l}{RA}$

Here R is the resistance which is mathematically represented as

$R = \frac{V}{I}$

Here I is the current which is mathematically represented as

$I = J * A$

So

$R = \frac{V}{ J * A}$

And

$\sigma = \frac{l}{\frac{V}{ J * A} * A}$

=> $\sigma = \frac{l}{\frac{V}{J}}$

=> $V = \frac{l * J}{\sigma }$

=> $V = \frac{10 * 4*10^6}{2*10^{7} }$

=> $V = 2 V$