A current I flows through the loop as shown i | vedclass

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(D) The loop consists of two circular arcs and two straight segments. The straight segments pass through the centre $O$,so the magnetic field due to t

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$\frac{5{\mu _0}I}{16R} \odot $

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(D) The loop consists of two circular arcs and two straight segments. The straight segments pass through the centre $O$,so the magnetic field due to them at $O$ is zero.The magnetic field at the centre due to a circular arc of radius $r$ and angle $	heta$ is $B = \frac{\mu_0 I 	heta}{4 \pi r}$.For the inner arc of radius $R$ and angle $\frac{3\pi}{2}$ (counter-clockwise),the field is $B_1 = \frac{\mu_0 I (3\pi/2)}{4 \pi R} = \frac{3\mu_0 I}{8R}$ (outwards,$\odot$).For the outer arc of radius $2R$ and angle $\frac{\pi}{2}$ (clockwise),the field is $B_2 = \frac{\mu_0 I (\pi/2)}{4 \pi (2R)} = \frac{\mu_0 I}{16R}$ (inwards,$\otimes$).The net magnetic field is $B_{net} = B_1 - B_2 = \frac{3\mu_0 I}{8R} - \frac{\mu_0 I}{16R} = \frac{6\mu_0 I - \mu_0 I}{16R} = \frac{5\mu_0 I}{16R}$ (outwards,$\odot$).

$A$ current $I$ flows through the loop as shown in the figure. The magnetic field at the centre $O$ is:

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