Ahollow copper wire with an inner diameter of 0.50 mm and an outer diameter of 1.8 mm carries a current of 15 a. what is the current density in the wire?

The current density is equal to the current intensity divided by the cross-sectional area through which the current passes:


The inner radius of the wire is 

while the outer radius is

Therefore the cross-sectional area of the wire is



So the current density in the wire is

The current density in the hollow copper wire is ×

Given the following data:

Inner diameter of wire = 0.50 mmOuter diameter of wire = 1.8 mmCurrent = 15 Amps.

To find the current density in the hollow copper wire:

First of all, we would determine the radius and the cross-sectional area of the hollow copper wire.

Inner radius = 0.25 mm

Outer radius = 0.9 mm

Next, we would determine its cross-sectional area:

Where:

O is the outer radius of the wire.I is the inner radius of the wire.A is the cross-sectional area.

Substituting the given parameters into the formula, we have;

A = 0.58

In square meters;

Area = ×

Mathematically, current density is given by the formula:

Current density, J = ×

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