Which one of the following options represents the magnetic field →...

Magnetic field due to a straight wire will be zero if the line of the wire passes through the point of interest. Therefore, In the above figure,

${B}_{BC} + {B}_{DE} + {B}_{FG} = 0$

Therefore,

$\overset{\to}{B}net = \overset{\to}{B}AB + \overset{\to}{B}CD + \overset{\to}{B}EF$

Magnetic field due to a finite wire is given by,

$B = \dfrac{{\mu}_{0}I}{4\pid}\left(sin\alpha + sin\beta\right)$

Using above formula,

\overset{\to}{B}AB = \dfrac{{\mu}_{0}I}{4\piL} \times \dfrac{1}{\sqrt{2}}\left(- \overset{^}{k}\right)

Magnetic field due to a circular arc is,

$B = \dfrac{{\mu}_{0}I}{2R} \times \dfrac{\theta}{2\pi}$

Therefore,

$\overset{\to}{B}CD = \dfrac{{\mu}_{0}I}{4\left(\dfrac{L}{2}\right)}$\left(- \overset{^}{k}\right) and \overset{\to}{B}EF = \dfrac{{\mu}_{0}I}{8 \times \left(\dfrac{L}{4}\right)}\left(- \overset{^}{k}\right)

Therefore, net magnetic field

\overset{\to}{B}net = \dfrac{{\mu}_{0}I}{4\pi \times L}\left(\dfrac{1}{\sqrt{2}}\right) + \dfrac{{\mu}_{0}I}{4\left(\dfrac{L}{2}\right)} + \dfrac{{\mu}_{0}I}{8 \times \left(\dfrac{L}{4}\right)}\left(- \overset{^}{k}\right)

⇒ \overset{\to}{B}net = \left(\dfrac{{\mu}_{0}I}{2L} + \dfrac{{\mu}_{0}I}{2L} + \dfrac{{\mu}_{0}I}{4\sqrt{2}\piL}\right)\left(- \overset{^}{k}\right)

⇒ \overset{\to}{B}net = \dfrac{{\mu}_{0}I}{L}\left(1 + \dfrac{1}{4\sqrt{2}\pi}\right)\left(- \overset{^}{k}\right)