Question

When 120 V is applied across a 20 m long wire, the magnitude of the current density was found to be 2 * 10 ^ 8 * A / (m ^ 2) . The resistivity of this wire is

Answer 1

Answer:

$\rho=3\times 10^{-8}\ \Omega-m$

Explanation:

Given that,

Voltage, V = 120 V

The length of the wire, l = 20 m

The current density of the wire, $\dfrac{I}{A}=2\times 10^8\ A/m^2$

We need to find the resistivity of this wire. We know that,

$R=\rho \dfrac{l}{A}$

Where

$\rho$ is the resistivity of wire

Also, $R=\dfrac{V}{I}$

So,

$\dfrac{V}{I}=\rho \dfrac{l}{A}\\\\\rho=\dfrac{V}{l\dfrac{I}{A}}$

Put all the values,

$\rho=\dfrac{120}{20\times 2\times 10^8}\\\\=3\times 10^{-8}\ \Omega-m$

So, the resistivity of this wire is equal to $3\times 10^{-8}\ \Omega-m$.