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(B) Let the distance of $M$ from each wire be $r$. The magnetic field due to a long straight wire carrying current $I$ at a distance $r$ is given by $

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$B$

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(B) Let the distance of $M$ from each wire be $r$. The magnetic field due to a long straight wire carrying current $I$ at a distance $r$ is given by $B = \frac{\mu_0 I}{2 \pi r}$.In the first case,the currents are $I_1 = 2 	ext{ A}$ and $I_2 = 1 	ext{ A}$ in the same direction. Using the right-hand rule,the magnetic fields at $M$ due to these wires are in opposite directions.$B_{PQ} = \frac{\mu_0 (2)}{2 \pi r} = \frac{\mu_0}{\pi r}$$B_{RS} = \frac{\mu_0 (1)}{2 \pi r} = \frac{\mu_0}{2 \pi r}$The net magnetic field at $M$ is $B = B_{PQ} - B_{RS} = \frac{\mu_0}{\pi r} - \frac{\mu_0}{2 \pi r} = \frac{\mu_0}{2 \pi r}$.When the current $2 	ext{ A}$ is switched off,only the current $I_2 = 1 	ext{ A}$ remains.The new magnetic field at $M$ is $B' = B_{RS} = \frac{\mu_0 (1)}{2 \pi r} = \frac{\mu_0}{2 \pi r}$.Comparing the two,we get $B' = B$.

$PQ$ and $RS$ are long parallel conductors separated by a certain distance. $M$ is the midpoint between them (see the figure). The net magnetic field at $M$ is $B$. Now,the current $2 \text{ A}$ is switched off. The field at $M$ now becomes

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