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(A) The magnetic field $B$ at a distance $x$ from a wire carrying current $I$ is given by $B = \frac{\mu_0 I}{2 \pi x}$.Case $1$: Currents in the same

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$3$

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(A) The magnetic field $B$ at a distance $x$ from a wire carrying current $I$ is given by $B = \frac{\mu_0 I}{2 \pi x}$.Case $1$: Currents in the same direction.The magnetic fields due to the two wires at distance $X_1$ and $(X_0 - X_1)$ are in opposite directions. The net magnetic field $B_1$ is:$B_1 = \left| \frac{\mu_0 I}{2 \pi X_1} - \frac{\mu_0 I}{2 \pi (X_0 - X_1)} \right| = \frac{\mu_0 I}{2 \pi} \left| \frac{X_0 - 2X_1}{X_1(X_0 - X_1)} \right|$.Given $\frac{X_0}{X_1} = 3$, so $X_0 = 3X_1$. Substituting this:$B_1 = \frac{\mu_0 I}{2 \pi} \left| \frac{3X_1 - 2X_1}{X_1(3X_1 - X_1)} \right| = \frac{\mu_0 I}{2 \pi} \left( \frac{X_1}{2X_1^2} \right) = \frac{\mu_0 I}{4 \pi X_1}$.Case $2$: Currents in opposite directions.The magnetic fields due to the two wires are in the same direction. The net magnetic field $B_2$ is:$B_2 = \frac{\mu_0 I}{2 \pi X_1} + \frac{\mu_0 I}{2 \pi (X_0 - X_1)} = \frac{\mu_0 I}{2 \pi} \left( \frac{X_0 - X_1 + X_1}{X_1(X_0 - X_1)} \right) = \frac{\mu_0 I}{2 \pi} \left( \frac{X_0}{X_1(X_0 - X_1)} \right)$.Substituting $X_0 = 3X_1$:$B_2 = \frac{\mu_0 I}{2 \pi} \left( \frac{3X_1}{X_1(2X_1)} \right) = \frac{3 \mu_0 I}{4 \pi X_1}$.The radius of curvature is $R = \frac{mu}{qB}$, so $R \propto \frac{1}{B}$.Therefore, $\frac{R_1}{R_2} = \frac{B_2}{B_1} = \frac{3 \mu_0 I / 4 \pi X_1}{\mu_0 I / 4 \pi X_1} = 3$.

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

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