# Current Density Calculation in Copper Wire

## What is the current density in the hollow copper wire?

The current density in the hollow copper wire is approximately 2.09 × 10⁶ A/m².

## Answer:

The current density in the hollow copper wire is approximately 2.09 × 10⁶ A/m².

Calculating the current density in a hollow copper wire involves determining the cross-sectional area and using the formula J = I/Area, where J is the current density, I is the current, and Area is the cross-sectional area of the wire. Let's break down the steps to calculate the current density:

## Step 1: Determine the Cross-Sectional Area

Given the inner diameter of 1.2 mm and outer diameter of 2.5 mm, we first need to calculate the cross-sectional area of the wire.

## Step 2: Cross-Sectional Area Calculation

1. Convert diameters to radii:

- Inner radius (r1) = 0.6 mm

- Outer radius (r2) = 1.25 mm

2. Convert radii to meters:

- r1 = 0.0006 m

- r2 = 0.00125 m

3. Calculate the cross-sectional area:

- Area = π(r2² - r1²)

- Area = π((0.00125)² - (0.0006)²) = 3.82116 × 10⁻⁶ m²

## Step 3: Calculate Current Density

Now, we can find the current density (J) using the formula J = I/Area, where I is the current (8.0 A) and Area is the cross-sectional area calculated in the previous step.

J = 8.0 A / 3.82116 × 10⁻⁶ m² ≈ 2.09 × 10⁶ A/m²

Therefore, the current density in the hollow copper wire is approximately 2.09 × 10⁶ A/m².