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

`\rho=3* 10^(-8)\ \Omega-m`

Step-by-step explanation:

Given that,

Voltage, V = 120 V

The length of the wire, l = 20 m

The current density of the wire,
`(I)/(A)=2* 10^8\ A/m^2`

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

`R=\rho (l)/(A)`

Where

`\rho` is the resistivity of wire

Also,
`R=(V)/(I)`

So,

`(V)/(I)=\rho (l)/(A)\\\\\rho=(V)/(l(I)/(A))`

Put all the values,

`\rho=(120)/(20* 2* 10^8)\\\\=3* 10^(-8)\ \Omega-m`

So, the resistivity of this wire is equal to
`3* 10^(-8)\ \Omega-m`.