Consider a long straight wire of a circular cross-section (radius a) carrying a steady current I. The current is uniformly distributed across this cross-section. The distances from the centre of the wire's cross-section at which the magnetic field [inside the wire, outside the wire] is half of the maximum possible magnetic field, any where due to the wire, will be

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[a/4, 3a/2]

[a/2, 2a]

[a/4, 2a]

[a/2, 3a]

answer is B.

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Detailed Solution

Maximum possible magnetic field is at the surface
$\begin{array}{l} B_{max} = \frac{μ_{0} I}{2 πa} \\ \frac{B_{max}}{2} = \frac{μ_{0} I}{4 πa} \end{array}$
It can be obtained inside as well as outside the wire
For inside,
$\begin{array}{l} \frac{μ_{0} I}{4 πa} = \frac{μ_{0} Ir}{2 {πa}^{2}} \\ \Rightarrow r = \frac{a}{2} \end{array}$
For outside
$\begin{array}{l} \frac{μ_{0} I}{4 πa} = \frac{μ_{0} I}{2 πr} \\ \Rightarrow r = 2 a \end{array}$
Correct answer $[\frac{a}{2}, 2 a]$

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