Magnetic field at point O will be

\frac{μ_{0} l}{2 R} i n t e r i o r

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\frac{μ_{0} l}{2 R} e x t e r i o r

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\frac{μ_{0} l}{2 R} (1 - \frac{l}{π}) i n t e r i o r

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\frac{μ_{0} l}{2 R} (1 + \frac{l}{π}) e x t e r i o r

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Solution

The correct option is A $\frac{μ_{0} l}{2 R} i n t e r i o r$
Magnetic field by Biot Savart law,
$d B = \frac{μ_{0}}{4 π} \frac{i d l s i n θ}{r^{2}}$
Due to the straight wires, $θ = 0$ i.e. it lies on the same line as the centre.
Due to curved portion, lets consider infinitesimal length dl where dl and R are always at 90 degree.
Hence $d B = \frac{μ_{0}}{4 π} \frac{i d l}{R^{2}}$
Integrating from 0 to $π R$
$d B = \frac{μ_{0} i}{4 R}$

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