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A very long conducting wire is bent in a semi circular shape from A to B as shown in figure. The magnetic field at point P for steady current configuration is given by

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a

$\frac{μ_{0} i}{4 R}$ pointed into the page

b

$\frac{μ_{0} i}{4 R} [1 - \frac{2}{π}]$ pointed into the page

c

$\frac{μ_{0} i}{4 R} [1 - \frac{2}{π}]$ pointed away from page

d

$\frac{μ_{0} i}{4 R}$ pointed away from the page

answer is D.

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$\begin{array}{l} {\vec{B}}_{P} = {\vec{B}}_{A} + {\vec{B}}_{A B} + {\vec{B}}_{B} \\ \Rightarrow {\vec{B}}_{P} = \frac{μ_{0} i}{4 π R} (\otimes) + \frac{μ_{0} i}{4 R} (∙) + \frac{μ_{0} i}{4 π R} (\otimes) \end{array}$
$\Rightarrow {\vec{B}}_{p} = \frac{μ_{0} i}{4 R} [1 - \frac{2}{π}]$ away from the page

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