A current i is flowing in a conductor shaped as shown in fig. The radius of the curved part is r and length
of straight portions is very large. The magnetic induction at O is

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$\frac{μ_{0} i}{4 πr} [\frac{3 π}{2} + 1]$

$\frac{μ_{0} i}{4 πr} [\frac{3 π}{2} - 1]$

$\frac{μ_{0} i}{4 πr} [\frac{π}{2} + 1]$

$\frac{μ_{0} i}{4 πr} [\frac{π}{2} - 1]$

answer is A.

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Detailed Solution

See fig.
$B_{O} = B_{EA} + B_{ABCD} + B_{DF}$
Here , $B_{EA} = 0, B_{DF} = \frac{1}{2} [\frac{μ_{0} i}{2 πr}]$
and $B_{ABCD} = \frac{μ_{0} i}{4 {πr}^{2}} \times (\frac{3}{4} \times 2 πr) = \frac{3 μ_{0} i}{8 r}$
$∴ B_{O} = \frac{μ_{0} i}{4 πr} + \frac{3 μ_{0} i}{8 r} = \frac{μ_{0} i}{4 r} [\frac{1}{π} + \frac{3}{2}] = \frac{μ_{0} i}{4 πr} [\frac{3 π}{2} + 1]$