Question

II You need to design a $1.0\mathrm{A}$fuse that "blows" if the current exceeds $1.0\mathrm{A}$. The fuse material in your stockroom melts at a current density of $500\mathrm{A}/{\mathrm{cm}}^2$. What diameter wire of this material will do the job?

Short answer

The diameter wire of the material is find by the given current $1.0\mathrm{A}$and density $500\mathrm{A}/{\mathrm{cm}}^2$by using current density formula $J=\frac{I}{A}$and get the diameter of the wire as$0.5mm$.

Step-by-step solution

Given information

When the fuse "blows" if the current exceeds $1.0\mathrm{A}$the fuse, material in the stockroom melts at a current density$500\mathrm{A}/{\mathrm{cm}}^2$.

Explanation

The wire will blow if the current reaches a density of $500\mathrm{A}/{\mathrm{m}}^2$.

The current density is role="math" localid="1648620008873" $J=\frac{I}{A}=\frac{4I}{\pi D^2}$
To find the diameter for a provided current density and current by,
$D=\sqrt{\frac{4I}{\pi J}}$
In numerical chance,
role="math" localid="1648620366993" $D=\sqrt{\frac{4\times 1A}{\pi \times 500\mathrm{A}/{\mathrm{cm}}^2}}$

role="math" localid="1648620375079" $D=\sqrt{\frac{4\times 1A}{22/7\times 500\mathrm{A}/{\mathrm{cm}}^2}}$

$D=\sqrt{0.0025}$

role="math" localid="1648619723131" $D=0.05cm$

The diameter wire of the material is$0.5mm$.