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(A) The magnetic field produced by a long straight wire at a distance $d$ is given by $B = \frac{\mu_0 I}{2 \pi d}$.For point $A$ (located at distance

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$5$

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(A) The magnetic field produced by a long straight wire at a distance $d$ is given by $B = \frac{\mu_0 I}{2 \pi d}$.For point $A$ (located at distance $r$ from wire $1$ carrying current $I$ and distance $3r$ from wire $2$ carrying current $2I$):$B_A = \frac{\mu_0 I}{2 \pi r} + \frac{\mu_0 (2I)}{2 \pi (3r)} = \frac{\mu_0 I}{2 \pi r} + \frac{\mu_0 I}{3 \pi r} = \frac{3 \mu_0 I + 2 \mu_0 I}{6 \pi r} = \frac{5 \mu_0 I}{6 \pi r}$.For point $C$ (located at distance $3r$ from wire $1$ carrying current $I$ and distance $r$ from wire $2$ carrying current $2I$):$B_C = \frac{\mu_0 I}{2 \pi (3r)} + \frac{\mu_0 (2I)}{2 \pi r} = \frac{\mu_0 I}{6 \pi r} + \frac{2 \mu_0 I}{2 \pi r} = \frac{\mu_0 I + 6 \mu_0 I}{6 \pi r} = \frac{7 \mu_0 I}{6 \pi r}$.Taking the ratio:$\frac{B_A}{B_C} = \frac{5 \mu_0 I / 6 \pi r}{7 \mu_0 I / 6 \pi r} = \frac{5}{7}$.Given that the ratio is $\frac{x}{7}$,we find $x = 5$.

Two parallel long current-carrying wires separated by a distance $2r$ are shown in the figure. The ratio of the magnetic field at $A$ to the magnetic field produced at $C$ is $\frac{x}{7}$. The value of $x$ is $\qquad$

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