Answer:A)0.5 and 2 mm
Explanation:
Given
radius R of wire is 1 mm
magnetic Field at edge(surface) [tex]=10^{-5} T[/tex]
Magnetic Field at a distance r' from Center is [tex]B'=5\times 10^{-6} T[/tex]
and we know
[tex]B=\frac{\mu _0I}{2\pi r}[/tex]
where [tex]I= current [/tex]
[tex]\mu _o=[/tex]Permeability of free space
[tex]r=[/tex]distance from center
For [tex]r=R[/tex]
[tex]10^{-5}=\frac{\mu _0I}{2\pi R}[/tex]---1
For [tex]r=r'[/tex]
[tex]5\times 10^{-6}=\frac{\mu _0I}{2\pi r'}[/tex]---2
Divide 1 and 2
[tex]\frac{10^{-5}}{0.5\times 10^{-5}}=\frac{r'}{R}[/tex]
[tex]r'=2 R=2 mm [/tex]
If r is inside the wire then
[tex]B=\frac{\mu _0rI}{2\pi R^2}[/tex]
for [tex]r=R[/tex]
[tex]10^{-5}=\frac{\mu _0R\cdot I}{2\pi R^2}[/tex]---3
for [tex]r=r"[/tex]
[tex]5\times 10^{-6}=\frac{\mu _0r"I}{2\pi R^2}[/tex]----4
Divide 3 and 4
[tex]2=\frac{R}{r"}[/tex]
[tex]r"=0.5 mm[/tex]
