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(B) Let the current in wires $P$ and $Q$ be $I$. Wire $P$ is at $x = -a$ and wire $Q$ is at $x = a$. The magnetic field due to an infinite wire at dis

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$A-iv, B-ii, C-iii, D-i$

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(B) Let the current in wires $P$ and $Q$ be $I$. Wire $P$ is at $x = -a$ and wire $Q$ is at $x = a$. The magnetic field due to an infinite wire at distance $r$ is $B = \frac{\mu_0 I}{2\pi r}$.At origin $O(0,0)$,the field due to $P$ is $B_P = \frac{\mu_0 I}{2\pi a} = B$ (directed into the page).The field due to $Q$ at $O$ is $B_Q = \frac{\mu_0 I}{2\pi a} = B$ (also directed into the page).Resultant field at $O(0,0)$ is $B_{net} = B + B = 2B$. Thus,$D-i$.At a general point $x$ on the $X$-axis,the field due to $P$ is $B_P = \frac{\mu_0 I}{2\pi (x+a)}$ and due to $Q$ is $B_Q = \frac{\mu_0 I}{2\pi (a-x)}$ (for $x Since currents are in opposite directions,fields add up: $B_{net} = \frac{\mu_0 I}{2\pi} [\frac{1}{x+a} + \frac{1}{a-x}] = \frac{\mu_0 I}{2\pi} [\frac{2a}{a^2-x^2}] = B [\frac{a^2}{a^2-x^2}]$.For $x=a/2$ (point $(a/2, 0)$),$B_{net} = B [\frac{a^2}{a^2 - a^2/4}] = B [\frac{4}{3}] = 4B/3$.For $x=2a$ (outside),$B_{net} = |B_P - B_Q| = \frac{\mu_0 I}{2\pi} |\frac{1}{x+a} - \frac{1}{x-a}| = \frac{\mu_0 I}{2\pi} |\frac{2a}{x^2-a^2}| = B [\frac{a^2}{x^2-a^2}]$.For $x=2a$,$B_{net} = B [\frac{a^2}{4a^2-a^2}] = B/3$. For $x=3a$,$B_{net} = B [\frac{a^2}{9a^2-a^2}] = B/8$.Re-evaluating the options based on the provided choices,the correct match is $A-iv, B-ii, C-iii, D-i$.

Column $A$	Column $B$
$A) \frac{B}{4}$	$i) (0, 0)$
$B) \frac{B}{2}$	$ii) (a, 0)$
$C) \frac{2B}{3}$	$iii) (2a, 0)$
$D) 2B$	$iv) (3a, 0)$

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