Three wires are carrying same constant current ‘i’ in different directions. Four loops enclosing the wires in different manners are shown. The direction of $d\overrightarrow{l}$  is shown in the fig.

 

 

	Column – I		Column – II
A)	Along closed loop – 1	P)	$\oint \overrightarrow{B}.d\overrightarrow{l}={\mu }_{0}i$
B)	Along closed loop – 2	Q)	$\oint \overrightarrow{B}.d\overrightarrow{l}=-{\mu }_{0}i$
C)	Along closed loop – 3	R)	$\oint \overrightarrow{B}.d\overrightarrow{l}=0$
D)	Along closed loop – 4	S)	Net work done by the magnetic force to move a unit charge along the loop is zero
		T)	$\oint \overrightarrow{B}.d\overrightarrow{l}={\mu }_0\left(2i\right)$

Work done by magnetic force on the charge = 0 in any part of its motion.

$\therefore$S is matching for all parts (i), (ii), (iii), (iv)

For loop – 1, $\sum I_{in}=-i+i-i=-i \therefore \oint \overrightarrow{B}.d\overrightarrow{l}={\mu }_0(-i)$

For loop – 2, $\sum I_{in}=i-i+i=i \therefore \oint \overrightarrow{B}.d\overrightarrow{l}={\mu }_0\left(i\right)$

For loop – 3, $\sum I_{in}=-i-i+i=-i \oint \overrightarrow{B}.d\overrightarrow{l}={\mu }_0(-i)$

For loop – 4, $\sum I_{in}=+i+i-i=+i \oint \overrightarrow{B}.d\overrightarrow{l}={\mu }_0\left(i\right)$

(note : That current will be taken as positive which produces lines of magnetic field in the same sense in which $d\overrightarrow{l}$ is taken)