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(A) Let the two wires be located at $x = -d$ and $x = +d$ on the $xx'$ axis. The currents are flowing out of the plane of the paper.Using the right-ha

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(A) Let the two wires be located at $x = -d$ and $x = +d$ on the $xx'$ axis. The currents are flowing out of the plane of the paper.Using the right-hand rule, the magnetic field due to a wire carrying current $i$ at a distance $r$ is $B = \frac{\mu_0 i}{2\pi r}$.For the wire at $x = -d$, the field is $B_1 = \frac{\mu_0 i}{2\pi (x+d)}$ in the $(-\hat{j})$ direction for $x > -d$.For the wire at $x = +d$, the field is $B_2 = \frac{\mu_0 i}{2\pi (d-x)}$ in the $\hat{j}$ direction for $x In the region between the wires $(-d At the midpoint $x = 0$, $B_{net} = 0$. As $x$ approaches $d$, $B_{net} 	o \infty$ in the $\hat{j}$ direction. As $x$ approaches $-d$, $B_{net} 	o -\infty$ in the $(-\hat{j})$ direction.Outside the wires, the fields add up in magnitude. The correct graphical representation showing the field passing through zero at the midpoint and having opposite signs on either side is represented by option $A$.

List-$I$	List-$II$
$a$. Fleming's left-hand rule	$e$. Direction of induced current
$b$. Fleming's right-hand rule	$f$. South pole
$c$. Clockwise current	$g$. North pole
$d$. Anticlockwise current	$h$. Direction of force

Two long straight parallel wires are a distance $2d$ apart. They carry steady equal currents flowing out of the plane of the paper. The variation of magnetic field $B$ along the line $xx'$ is given by:

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Match the following:
List-$I$ List-$II$
$a$. Fleming's left-hand rule $e$. Direction of induced current
$b$. Fleming's right-hand rule $f$. South pole
$c$. Clockwise current $g$. North pole
$d$. Anticlockwise current $h$. Direction of force

The correct answer is:

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