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A long straight wire of radius 'a' carries a steady current i. The current is uniformly distributed across its cross - section. The ratio of the magnetic fields at $\frac{a}{2}$ and 2a is,

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\frac{1}{2}

\frac{1}{4}

answer is A.

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Detailed Solution

B_{1} = \frac{μ_{0} i}{2 π r} (r \geq R)

$B_{1} = \frac{μ_{0} i}{2 π \times 2 a} = \frac{μ_{0} i}{4 π a}$

$B_{2} = \frac{μ_{0} i r}{2 π R^{2}} (r \leq R)$

$B_{2} = \frac{μ_{0} i \frac{a}{2}}{2 π . a^{2}} = \frac{μ_{0} i}{4 π a}$

$\frac{B_{1}}{B_{2}} = 1$

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