Ampere's circuital law

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Question

A long cylindrical wire of radius ‘a’ carries a current distributed uniformly over its cross section if the magnetic fields at distances 2 cm (< a) and 8 cm (> a) have equal magnitude, then

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Solution

At ${r=2cm, B}_{1} = \frac{μ_{0}}{2} (\frac{i}{{πa}^{2}}) \times 2 unit$
and at $r = 8 cm, B_{2} = \frac{μ_{0} i}{2 π \times 8}$
Since $B_{1} = B_{2}, \frac{μ_{0} i}{2 {πa}^{2}} \times 2 = \frac{μ_{0} i}{2 π \times 8}$
$\Rightarrow a = 4 cm$

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