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(B) Let the currents in both wires be $i$. The magnetic field due to a long straight wire at a distance $r$ is $B = \frac{\mu_0 i}{2 \pi r}$.Using the

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$0 : 1 : -1$

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(B) Let the currents in both wires be $i$. The magnetic field due to a long straight wire at a distance $r$ is $B = \frac{\mu_0 i}{2 \pi r}$.Using the right-hand rule,the magnetic field at point $A$ (midpoint between wires) due to wire $1$ is directed into the page ($-k$ direction) and due to wire $2$ is directed out of the page ($+k$ direction). Since the distances are equal $(R/2)$,the magnitudes are equal,so $B_A = 0$.At point $B$,both wires produce magnetic fields directed into the page. The distance from wire $1$ is $R/2$ and from wire $2$ is $3R/2$. Thus,$B_B = -\left( \frac{\mu_0 i}{2 \pi (R/2)} + \frac{\mu_0 i}{2 \pi (3R/2)} \right) = -\frac{\mu_0 i}{\pi R} (1 + 1/3) = -\frac{4 \mu_0 i}{3 \pi R}$.At point $C$,both wires produce magnetic fields directed out of the page. The distance from wire $2$ is $R/2$ and from wire $1$ is $3R/2$. Thus,$B_C = +\left( \frac{\mu_0 i}{2 \pi (R/2)} + \frac{\mu_0 i}{2 \pi (3R/2)} \right) = +\frac{4 \mu_0 i}{3 \pi R}$.The ratio $B_A : B_B : B_C = 0 : -\frac{4 \mu_0 i}{3 \pi R} : \frac{4 \mu_0 i}{3 \pi R} = 0 : -1 : 1$. Note: The options provided in the source are $0 : 1 : -1$. Depending on the sign convention for direction,$0 : 1 : -1$ is equivalent to $0 : -1 : 1$.

Two long parallel wires are at a distance $R$ apart. They carry steady equal currents in the same directions as shown in the figure. The ratio of magnetic fields at $A, B$ and $C$ respectively,is

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