Answer:
[tex]i = 4.34 \times 10^{-5} A[/tex]
Explanation:
As we know that current in a wire depends on its current density and given as
[tex]i = \int J.dA[/tex]
so we have
[tex]i = \int cr^2 2\pi r dr[/tex]
so we will have
[tex]i = 2\pi c \int r^3 dr[/tex]
[tex]i = \frac{\pi c}{2} (r^4)[/tex]
now we have to find current in r = 0.5 R
so we have
[tex]i = \frac{\pi c}{2}(0.5 R)^4[/tex]
now plug in all data
[tex]i = \frac{\pi(5.25 \times 10^6)}{2}(0.5 \times 3.03 \times 10^{-3})^4[/tex]
[tex]i = 4.34 \times 10^{-5} A[/tex]
