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A hollow 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?

Respuesta :

skyluke89
The current density is equal to the current intensity divided by the cross-sectional area through which the current passes:
[tex]J= \frac{I}{A} [/tex]

The inner radius of the wire is 
[tex]r_i = \frac{0.50 mm}{2}=0.25 mm [/tex]
while the outer radius is
[tex]r_o = \frac{1.80 mm}{2}=0.90 mm [/tex]
Therefore the cross-sectional area of the wire is
[tex]A= \pi (r_o^2 - r_i^2)=\pi ((0.9mm)^2-(0.25 mm)^2)=2.35 mm^2 [/tex]
[tex]= 2.35 \cdot 10^{-6} m^2[/tex]

So the current density in the wire is
[tex]J= \frac{I}{A}= \frac{15 A}{2.35 \cdot 10^{-6} m^2}=6.38 \cdot 10^6 A/m^2 [/tex]

Lanuel

The current density in the hollow copper wire is [tex]2.59[/tex] × [tex]10^{6} \;A/m^2[/tex]

Given the following data:

  • Inner diameter of wire = 0.50 mm
  • Outer diameter of wire = 1.8 mm
  • Current = 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.

[tex]Inner \; radius = \frac{Inner\;diameter}{2} \\\\Inner \; radius = \frac{0.5}{2}[/tex]

Inner radius = 0.25 mm

[tex]Outer \; radius = \frac{Outer\;diameter}{2} \\\\Outer \; radius = \frac{1.8}{2}[/tex]

Outer radius = 0.9 mm

Next, we would determine its cross-sectional area:

[tex]A = \pi (O^2 - I^2)[/tex]

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;

[tex]A = \pi (0.9^2 - 0.25^2)\\\\A = \pi (0.81 - 0.625)\\\\A = 3.142(0.185)[/tex]

A = 0.58 [tex]mm^2[/tex]

In square meters;

Area = [tex]0.58[/tex] × [tex]10^{-6} \;m^2[/tex]

Mathematically, current density is given by the formula:

[tex]J = \frac{I}{A}\\\\J = \frac{15}{0.58(10^{-6})}[/tex]

Current density, J = [tex]2.59[/tex] × [tex]10^{6} \;A/m^2[/tex]

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