Answer:
0.04594 cm
Explanation:
So, we need the wire melt when the max current density its [tex]350 \frac{A}{cm^2}[/tex]. Now, we got our limit current, 0.58 A.
The current density its [tex]\frac{current}{area}[/tex], so, with our data, we can obtain the cross area of the wire.
For a cylinder, the area its given by:
[tex]area =\pi r^2[/tex]
We can put all this in the equation for the max current density:
[tex]density_{max} =\frac{current_{limit}}{area}[/tex]
[tex]density_{max} =\frac{current_{limit}}{\pi r^2}[/tex]
And now, we can work it a little:
[tex]r^2 =\frac{current_{limit}}{\pi * density_{max}}[/tex]
[tex]r =\sqrt{ \frac{current_{limit}}{\pi * density_{max}}[/tex]
using our values, max current density = [tex]350 \frac{A}{cm^2}[/tex] , Â and limit current = 0.58 A,
[tex]r =\sqrt{ \frac{0.58 A}{\pi * 350 \frac{A}{cm^2} }}[/tex]
And, of course, the diameter its two times the radius:
[tex]d = 2 * r = 2 * 0.02297 cm[/tex]
[tex]d = 0.04594 cm[/tex]
