The magnetic field at O due to current in the infinite wire forming a loop as shown in figure is

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$\frac{μ_{0}}{4 π} \frac{I}{d} (\cos ϕ_{1} + \sin ϕ_{2})$

$\frac{μ_{0}}{4 π} \frac{I}{d} (\tan ϕ_{1} + \tan ϕ_{2})$

$\frac{μ_{0}}{4 π} \frac{I}{d} (\sin ϕ_{1} + \sin ϕ_{2})$

$\frac{μ_{0} I}{4 πd} (\cos ϕ_{1} + \cos ϕ_{2})$

answer is A.

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Detailed Solution

$\begin{array}{l} B = \frac{μ_{0} I}{4 πd} [\sin θ_{1} + \sin θ_{2}] \\ But θ_{1} = 90^{0} - φ_{1} a n d θ_{2} = 90^{0} - φ_{2} \\ \sin θ_{1} = \sin (90^{0} - ϕ_{1}) = \cos ϕ_{1} \\ Similarly \sin θ_{2} = \cos ϕ_{2} . Magnetic field on the axis of a current carrying conductor is zero \\ B = \frac{μ_{0} I}{4 πd} (\cos ϕ_{1} + \cos ϕ_{2}) \end{array}$