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A long wire is bent into the hair pin-like shape shown in the figure. The magnetic induction at the point “P” is given by

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$\frac{μ_{0} i (2 + π)}{2 π r}$

$\frac{μ_{o} i (1 + π)}{4 π r}$

$\frac{μ_{0} i (2 + π)}{4 π r}$

answer is B.

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Detailed Solution

For straight parts,
$B_{1} = \frac{μ_{0} i}{2 π R}$ outwards
For semicircular,
$B_{2} = \frac{μ_{0} i}{4 R}$ outwards
$\begin{array}{l} B_{R} = B_{1} + B_{2} \\ = \frac{μ_{0} i}{4 π R} (2 + π) \end{array}$

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