A fuse in an electric circuit is a wire that is designed to melt, and thereby open the circuit, if the current exceeds a predetermined value. Suppose that the material to be used in a fuse melts when the current density rises to $\mathbf{440}\mathbf{}\mathit{A}\mathbf{/}\mathit{c}{\mathit{m}}^{\mathbf{2}}$. What diameter of cylindrical wire should be used to make a fuse that will limit the current to 0.50 A?

1. Current density,$J=440A/c{m}^{2}or440\times {10}^{4}A/m^2$
2. The value of the current, i = 0.50 A

The current density is electric current per unit cross-section area at a given point.

Using the concept of an area, we can get the formula of the current density in terms of the diameter. Substituting the given values, we can get the required diameter of the wire.

Formulae:

The current density of the current flowing through the area,$J=\frac{i}{A}$ …(i)

The cross-section area of the wire,$A={\mathrm{\pi r}}^2$ …(ii)

where, r is its radius, i.e. half of the thickness of the wire.

Substituting the given data and the value of the area from equation (ii) in equation (i), we can get the value of the radius of the wire as follows:

$$\begin{gathered} r^2=\frac{i}{\mathrm{\pi J}} \\ r=\sqrt{\frac{i}{\mathrm{\pi J}}} \\ =\sqrt{\frac{0.50A}{\mathrm{\pi }\times 440\times {10}^4\mathrm{A}/{\mathrm{m}}^2}} \\ =1.90\times {10}^{-4}m \end{gathered}$$

So, the diameter of the wire using the radius of the wire can be given as follows:

$\begin{array}{l}d=2r\\ =2(1.9\times {10}^{-4}m)\\ =3.8\times {10}^{-4}\mathrm{m}\end{array}$

Hence, the value of the diameter of the wire is $3.8\times {10}^{-4}\mathrm{m}$.