Answer:
0.04594 cm
Explanation:
So, we need the wire melt when the max current density its
. Now, we got our limit current, 0.58 A.
The current density its
, so, with our data, we can obtain the cross area of the wire.
For a cylinder, the area its given by:
![area =\pi r^2](https://tex.z-dn.net/?f=area%20%3D%5Cpi%20r%5E2)
We can put all this in the equation for the max current density:
![density_{max} =\frac{current_{limit}}{area}](https://tex.z-dn.net/?f=density_%7Bmax%7D%20%3D%5Cfrac%7Bcurrent_%7Blimit%7D%7D%7Barea%7D)
![density_{max} =\frac{current_{limit}}{\pi r^2}](https://tex.z-dn.net/?f=density_%7Bmax%7D%20%3D%5Cfrac%7Bcurrent_%7Blimit%7D%7D%7B%5Cpi%20r%5E2%7D)
And now, we can work it a little:
![r^2 =\frac{current_{limit}}{\pi * density_{max}}](https://tex.z-dn.net/?f=r%5E2%20%3D%5Cfrac%7Bcurrent_%7Blimit%7D%7D%7B%5Cpi%20%2A%20density_%7Bmax%7D%7D)
![r =\sqrt{ \frac{current_{limit}}{\pi * density_{max}}](https://tex.z-dn.net/?f=r%20%3D%5Csqrt%7B%20%5Cfrac%7Bcurrent_%7Blimit%7D%7D%7B%5Cpi%20%2A%20density_%7Bmax%7D%7D)
using our values, max current density =
, and limit current = 0.58 A,
![r =\sqrt{ \frac{0.58 A}{\pi * 350 \frac{A}{cm^2} }}](https://tex.z-dn.net/?f=r%20%3D%5Csqrt%7B%20%5Cfrac%7B0.58%20A%7D%7B%5Cpi%20%2A%20350%20%5Cfrac%7BA%7D%7Bcm%5E2%7D%20%7D%7D)
And, of course, the diameter its two times the radius:
![d = 2 * r = 2 * 0.02297 cm](https://tex.z-dn.net/?f=d%20%3D%202%20%2A%20r%20%3D%202%20%2A%200.02297%20cm)
![d = 0.04594 cm](https://tex.z-dn.net/?f=d%20%3D%200.04594%20cm)