Answer:
[tex]B = \mu_0(\frac{1}{3} \frac{J_0}{r_0} r^2)[/tex]
Explanation:
As the current density is given as
[tex]J = \frac{J_0}{r_0}r[/tex]
now we have current inside wire given as
[tex]i = \int J(2\pi r)dr[/tex]
[tex]i = \int \frac{J_0}{r_0} r(2\pi r)dr[/tex]
[tex]i = 2\pi \frac{J_0}{r_0} \int r^2 dr[/tex]
[tex]i = \frac{2}{3} \pi \frac{J_0}{r_0} r^3[/tex]
Now by Ampere's law we will have
[tex]\int B. dl = \mu_0 i[/tex]
[tex]B. (2\pi r) = \mu_0(\frac{2}{3} \pi \frac{J_0}{r_0} r^3)[/tex]
[tex]B = \mu_0(\frac{1}{3} \frac{J_0}{r_0} r^2)[/tex]
