Answer with Explanation:
We are given that
Inner radius,[tex]r_1=a[/tex]
Outer radius,[tex]r_2=b[/tex]
Current=I
a.If r<a
Consider a loop of radius r where r<a
Then, current enclosed by loop=[tex]I'=0[/tex]
Magnetic field,B=0
b.if a<r
Then, [tex]I'=I\frac{(r^2-a^2)}{b^2-a^2}[/tex]
[tex]B=\frac{\mu_0I'}{2\pi r}=\frac{\mu_0I(r^2-a^2)}{2\pi r(b^2-a^2)}[/tex]
c.r>b
Then, [tex]I'=I[/tex]
Magnetic field,[tex]B=\frac{\mu_0I}{2\pi r}[/tex]
