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(C) The magnetic field at the centre $O$ is the vector sum of the magnetic field due to the straight wire and the circular loop.$1$. Magnetic field du

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$\frac{\mu_0 I}{2 \pi R}(\pi+1)$

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(C) The magnetic field at the centre $O$ is the vector sum of the magnetic field due to the straight wire and the circular loop.$1$. Magnetic field due to the long straight wire at distance $R$ from the centre $O$:$B_1 = \frac{\mu_0 I}{2 \pi R}$ (directed outwards, perpendicular to the plane).$2$. Magnetic field due to the circular loop of radius $R$ at its centre $O$:$B_2 = \frac{\mu_0 I}{2 R}$ (directed outwards, perpendicular to the plane).Since both fields are in the same direction (outwards), the resultant magnetic field $B$ is:$B = B_1 + B_2$$B = \frac{\mu_0 I}{2 \pi R} + \frac{\mu_0 I}{2 R}$$B = \frac{\mu_0 I}{2 \pi R} (1 + \pi)$Thus, the magnitude of the magnetic induction is $\frac{\mu_0 I}{2 \pi R}(\pi+1)$.

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

An infinitely long straight conductor is bent into the shape as shown below. It carries a current of $I$ ampere and the radius of the circular loop is $R$ metre. Then, the magnitude of magnetic induction at the centre of the circular loop is

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