The magnetic field at center ‘O’ of the arc in figure is

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$\frac{μ_{o} I}{4 π \times r} [\sqrt{2} + π]$

$\frac{μ_{o} I}{2 π r} [\frac{π}{4} + (\sqrt{2} - 1)]$

$\frac{μ_{0}}{4 π} \times \frac{I}{r} [\sqrt{2} - π]$

$\frac{μ_{0}}{4 π} \times \frac{I}{r} [\sqrt{2} + \frac{π}{4}]$

answer is B.

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Detailed Solution

Magnetic field due to straight wires and quarter circle is given by

$B_{c i r c u l a r} = \frac{1}{4} \frac{μ_{0} I}{2 r} = \frac{μ_{0} I}{8 r} B_{s t r a i g h t} = 2 \frac{μ_{0} I}{4 π (r \cos 45^{0})} [\sin 90^{0} - \sin 45^{0}] = \frac{\sqrt{2} μ_{0} I}{2 π r} (1 - \frac{1}{\sqrt{2}}) H e n c e, B_{n e t} = \frac{μ_{0} I}{8 r} + \frac{\sqrt{2} μ_{0} I}{2 π r} (1 - \frac{1}{\sqrt{2}}) = \frac{μ_{o} I}{2 π r} [\frac{π}{4} + (\sqrt{2} - 1)]$