# One end of an aluminum wire of 0.125 cm in diameter is welded to a copper wire of 0.25 cm in...

## Question:

One end of an aluminum wire of 0.125 cm in diameter is welded to a copper wire of 0.25 cm in diameter. The composite wire carries a current of 8 mA. What is the current density in each wire?

## Current Density:

While an electric current passes through the wire, its current density depends on the cross sectional area, i.e. current density is the ratio of electric current to the cross sectional area of the conductor.

Given Data

• diameter of aluminium wire, {eq}d_{Al}\ = 0.125\ cm\ = 1.25\times 10^{-3}\ m{/eq}
• diameter of copper wire, {eq}d_{Cu}\ = 0.25\ cm\ = 2.5\times 10^{-3}\ m{/eq}
• Electric current, {eq}I\ = 8\ mA\ = 8\times 10^{-3}\ A{/eq}

Finding the current density in each wire

The current density in aluminium wire is calculated as:

• {eq}J_{Al}\ = \dfrac{I}{\dfrac{\pi\times d_{Al}^2}{4}} {/eq}
• {eq}J_{Al}\ = \dfrac{8\times 10^{-3}}{\dfrac{\pi\times (1.25\times 10^{-3})^2}{4}} {/eq}
• {eq}J_{Al}\ = 6.52\times 10^3\ A/m^2{/eq}

The current density in copper wire is calculated as:

• {eq}J_{Cu}\ = \dfrac{I}{\dfrac{\pi\times d_{Cu}^2}{4}} {/eq}
• {eq}J_{Cu}\ = \dfrac{8\times 10^{-3}}{\dfrac{\pi\times (2.5\times 10^{-3})^2}{4}} {/eq}
• {eq}J_{Cu}\ = 1.63\times 10^3\ A/m^2{/eq} 