PQ and RS are long parallel conductors separa | vedclass

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(B) Let the distance of point $M$ from each wire be $r$. The magnetic field due to a long straight wire at a distance $r$ is given by $B = \frac{\mu_0

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$B$

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(B) Let the distance of point $M$ from each wire be $r$. The magnetic field due to a long straight wire at a distance $r$ is given by $B = \frac{\mu_0 I}{2\pi r}$.Let $B_1$ be the magnetic field due to wire $PQ$ (carrying $2\,A$) and $B_2$ be the magnetic field due to wire $RS$ (carrying $1\,A$).Using the right-hand thumb rule,the magnetic field at $M$ due to $PQ$ is directed into the page,and due to $RS$ is directed out of the page.$B_1 = \frac{\mu_0 (2)}{2\pi r} = 2k$ (where $k = \frac{\mu_0}{2\pi r}$).$B_2 = \frac{\mu_0 (1)}{2\pi r} = k$.The net magnetic field $B = B_1 - B_2 = 2k - k = k$.When the current $2\,A$ is switched off,the field due to $PQ$ becomes zero.The new magnetic field at $M$ is $B' = B_2 = k$.Since $B = k$,the new field $B' = B$.

$PQ$ and $RS$ are long parallel conductors separated by a certain distance. $M$ is the mid-point between them. The net magnetic field at $M$ is $B$. Now,the current $2\,A$ is switched off. The field at $M$ now becomes

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