Question

A $150m$ long metal wire connects points $A$ and $B$. The electric potential at point $B$ is $50V$ less than that at point $A$. If the conductivity of the metal is $60\times {10}^{6}mho/m$ then the magnitude of the current density in the wire is equal to?

Answer 1

Step 1: Given Data

Length of the metal wire $l=150m$

The potential difference $V=50V$

The conductivity of the metal $\sigma =60\times {10}^{6}mho/m$

Step 2: Estimate the Resistance

We know that resistivity,

$\rho =\frac{1}{\sigma }=\frac{1}{60\times {10}^6}\Omega m$

We know that resistance,

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

$=\frac{1}{60\times {10}^6}\times \frac{150}{A}$

$=\frac{2.5\times {10}^{-6}}{A}\Omega$

Step 3: Calculate the Current

According to Ohm's law, the current is given as

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

$=\frac{50\times A}{2.5\times {10}^{-6}}$

$=20A\times {10}^{6}A$

Step 4: Calculate the Current Density

We know that the current density,

$J=\frac{I}{A}$

$=\frac{20A\times {10}^6}{A}$

$=20\times {10}^{6}A/m^2$

Hence, the magnitude of the current density in the wire is equal to $20\times {10}^{6}A/m^2$.