A very long wire ABDMNDC is shown in figure carrying current I. AB and BC parts are straight, long and at right angle. At D wire forms a circular turn DMND of radius R. AB, BC parts are tangential to circular turn at N and D. Magnetic field at the centre of circle is :

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$\frac{μ_{0} I}{2 π R} (π + 1)$

$\frac{μ_{0} I}{2 π R} (π + \frac{1}{\sqrt{2}})$

$\frac{μ_{0} I}{2 R}$

$\frac{μ_{0} I}{2 π R} (π - \frac{1}{\sqrt{2}})$

answer is B.

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Detailed Solution

$\begin{array}{l} {\vec{B}}_{0} = {({\vec{B}}_{0})}_{1} + {({\vec{B}}_{0})}_{2} + {({\vec{B}}_{0})}_{3} + {({\vec{B}}_{0})}_{4} \\ = \frac{μ_{0} i}{4 π R} [\sin 90^{0} - \sin 45^{0}] \otimes + \frac{μ_{0} i}{2 R} ⊙ + \frac{μ_{0} i}{4 π R} (\sin 45^{0} + \sin 90^{0}) ⊙ \\ = \frac{- μ_{0} i}{4 π R} [1 - \frac{1}{\sqrt{2}}] + \frac{μ_{0} i}{2 R} + \frac{μ_{0} i}{4 π R} [\frac{1}{\sqrt{2}} + 1] ⊙ \\ = \frac{μ_{0} i}{4 π R} [- 1 + \frac{1}{\sqrt{2}} + 2 π + \frac{1}{\sqrt{2}} + 1] ⊙ \\ = \frac{μ_{0} i}{4 π R} [\sqrt{2} + 2 π] ⊙ \\ = \frac{μ_{0} i}{2 π R} [\frac{1}{\sqrt{2}} + π] ⊙ \end{array}$