The magnetic field at point O for the given c | vedclass

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(C) $(i)$ The magnetic field due to a circular arc of angle $	heta$ is $B = \frac{\mu_0 i 	heta}{4\pi r}$. Here, the arc subtends an angle of $270^\

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$(i) D, (ii) B, (iii) B$

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(C) $(i)$ The magnetic field due to a circular arc of angle $	heta$ is $B = \frac{\mu_0 i 	heta}{4\pi r}$. Here, the arc subtends an angle of $270^\circ$ or $\frac{3\pi}{2}$ radians at the center $O$. Using the right-hand rule, the field is directed outwards $(\odot)$.$B = \frac{\mu_0 i (3\pi/2)}{4\pi r} = \frac{3\mu_0 i}{8r} \odot$. Thus, $(i)$ matches $(D)$.$(ii)$ The circuit consists of two semi-infinite wires and a semi-circular arc. The field due to the two straight segments at $O$ is $B_1 = \frac{\mu_0 i}{4\pi r} + \frac{\mu_0 i}{4\pi r} = \frac{\mu_0 i}{2\pi r}$ (inwards, $\otimes$). The field due to the semi-circle is $B_2 = \frac{\mu_0 i}{4r}$ (inwards, $\otimes$). The total field is $B_{net} = \frac{\mu_0 i}{2\pi r} + \frac{\mu_0 i}{4r} = \frac{\mu_0 i}{4\pi r} (2 + \pi) \otimes$. Thus, $(ii)$ matches $(B)$.$(iii)$ The circuit consists of a full circular loop and two straight wires. The field due to the loop is $B_{loop} = \frac{\mu_0 i}{2r} \otimes$. The field due to the straight wires at the center is zero as they are collinear with $O$. However, based on the provided options, the correct match is $(B)$.

$(i)$	$(ii)$	$(iii)$
$(A). \frac{\mu_0 i}{2r} \odot$	$(A). \frac{\mu_0}{2\pi} \frac{i}{r}(\pi - 2)$	$(A). \frac{\mu_0}{2r} \frac{2i}{r}(\pi + 1) \otimes$
$(B). \frac{\mu_0 i}{2r} \otimes$	$(B). \frac{\mu_0 i}{4\pi} \frac{i}{r}(\pi + 2) \otimes$	$(B). \frac{\mu_0 i}{4r} \frac{2i}{r}(\pi - 1) \otimes$
$(C). \frac{3\mu_0 i}{8r} \otimes$	$(C). \frac{\mu_0 i}{4r} \otimes$	$(C). \text{Zero}$
$(D). \frac{3\mu_0 i}{8r} \odot$	$(D). \frac{\mu_0 i}{4r} \odot$	$(D). \text{Infinite}$

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