Elements in Column –I contains infinitely long straight conductor(s) carrying current I normally outwards from the paper and the paths shown with arrows. Column – II contains the value of $\int \vec{B} . \vec{dl}$ along the shown paths. Match the two columns.
COLUMN – I COLUMN – II
A)
P) Zero
B)
Q) $μ_{0} i$
C)
R) $\frac{μ_{0} i}{2}$
D)
S) $\frac{μ_{0} i}{4}$

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$A \to p, q; B \to s; C \to r; D \to q, s$

$A \to r; B \to p; C \to r, s; D \to p, q$

$A \to r; B \to p; C \to s; D \to q$

$A \to s; B \to r; C \to q; D \to p$

answer is D.

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Detailed Solution

A) Here 3 more similar conductors are needed to enclosed the current carrying conductor.
$∴$ Total no. of similar conductors = 4.
Similar to Gauss law in electrostatics total magnetic flux through the closed surface.
$\int \vec{B} . \vec{d l} = μ_{0} i$
Flux through the given curved surface is $= \frac{1}{4} (μ_{0} i)$
Question Image

B) Here one more conductor is needed to enclose the current carrying conductor.
$∴$ Total no. of similar conductors = 2
Magnetic flux through the closed $\int \vec{B} . \vec{d l} = μ_{0} i$
Flux through the given surface $= \frac{1}{2} μ_{0} i$
Question Image
C) Given surface is a closed surface from $\int \vec{B} . \vec{d l} = μ_{0} i$
D) Due to current in left conductor of the given path $P Q, \int \vec{B} . \vec{d l}$ will be positive

As angle between $\vec{B}, \vec{d l}$ is less than $90^{0}$ at any point along the path.
In the same manner $\int \vec{B} . \vec{d l}$ along that path due to the current in the right side of the path P Q is – ve but same amount as previous.
Hence net $\int \vec{B} . \vec{d l}$ will be zero.