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(B) $(i)$ The magnetic field due to straight segments is zero at $O$. The field due to the semi-circular arc is $B = \frac{\mu_0 i}{4r}$ directed into

Vedclass · Accepted Answer

$(i) C, (ii) B, (iii) A$

Vedclass · Answer

(B) $(i)$ The magnetic field due to straight segments is zero at $O$. The field due to the semi-circular arc is $B = \frac{\mu_0 i}{4r}$ directed into the page $(\otimes)$. Thus,$(i)$ matches $(C)$.$(ii)$ The magnetic field due to the two semi-circular arcs at $O$ are $B_1 = \frac{\mu_0 i}{4r_1}$ $(\otimes)$ and $B_2 = \frac{\mu_0 i}{4r_2}$ $(\otimes)$. The total field is $B_{net} = \frac{\mu_0 i}{4}(\frac{1}{r_1} + \frac{1}{r_2})$ $(\otimes)$. Thus,$(ii)$ matches $(B)$.$(iii)$ The magnetic field due to the two semi-circular arcs at $O$ are $B_1 = \frac{\mu_0 i}{4r_1}$ $(\otimes)$ and $B_2 = \frac{\mu_0 i}{4r_2}$ $(\odot)$. Since $r_1  B_2$. The net field is $B_{net} = \frac{\mu_0 i}{4}(\frac{1}{r_1} - \frac{1}{r_2})$ $(\otimes)$. Thus,$(iii)$ matches $(A)$.

$(i)$	$(ii)$	$(iii)$
$(A). \frac{\mu_0 i}{r} \otimes$	$(A). \frac{\mu_0 i}{4}(\frac{1}{r_1} - \frac{1}{r_2}) \otimes$	$(A). \frac{\mu_0 i}{4}(\frac{1}{r_1} - \frac{1}{r_2}) \otimes$
$(B). \frac{\mu_0 i}{2r} \odot$	$(B). \frac{\mu_0 i}{4}(\frac{1}{r_1} + \frac{1}{r_2}) \otimes$	$(B). \frac{\mu_0 i}{4}(\frac{1}{r_1} + \frac{1}{r_2}) \otimes$
$(C). \frac{\mu_0 i}{4r} \otimes$	$(C). \frac{\mu_0 i}{4}(\frac{1}{r_1} - \frac{1}{r_2}) \odot$	$(C). \frac{\mu_0 i}{4}(\frac{1}{r_1} - \frac{1}{r_2}) \odot$
$(D). \frac{\mu_0 i}{4r} \odot$	$(D). 0$	$(D). 0$

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

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