Answer:
0.0389 cm
Explanation:
The current density in a conductive wire is given by
[tex]J=\frac{I}{A}[/tex]
where
I is the current
A is the cross-sectional area of the wire
In this problem, we know that:
- The fuse melts when the current density reaches a value of
[tex]J=520 A/cm^2[/tex]
- The maximum limit of the current in the wire must be
I = 0.62 A
Therefore, we can find the cross-sectional area that the wire should have:
[tex]A=\frac{I}{J}=\frac{0.62}{520}=1.19\cdot 10^{-3} cm^2[/tex]
We know that the cross-sectional area can be written as
[tex]A=\pi \frac{d^2}{4}[/tex]
where d is the diameter of the wire.
Re-arranging the equation, we find the diameter of the wire:
[tex]d=\sqrt{\frac{4A}{\pi}}=\sqrt{\frac{4(1.19\cdot 10^{-3})}{\pi}}=0.0389 cm[/tex]
