Question 26.7: A fuse in an electric circuit is a wire that is designed to ......

The cross-sectional area of wire is given by A = πr² , where r is its radius (half its thickness). The magnitude of the current density vector is

J=i / A=i / \pi r^2,

so

r=\sqrt{\frac{i}{\pi J}}=\sqrt{\frac{0.50 \,A }{\pi\left(440 \times 10^4 \,A / m ^2\right)}}=1.9 \times 10^{-4}\, m .

The diameter of the wire is therefore d=2 r=2\left(1.9 \times 10^{-4}\, m \right)=3.8 \times 10^{-4}\, m .