Question

Rank the magnitudes of $\oint \overrightarrow{B}.\overrightarrow{dl}$ for the closed paths $a,b,c$ and $d$ as shown in the figure. A long wire is located at the centre, perpendicular to the plane of the closed paths and carrying a current $1\text{A}$ in outward direction.

Answer 1

According to Ampere circuital law,

$\oint \overrightarrow{B}.\overrightarrow{dl}={\mu }_0\Sigma I$

Where $\Sigma I$ is net current passing through the loop.

Given, $I=1\text{A}$ is pointing outwards,

Considering the outward direction of current to be $+ve$, we traverse through all the loops in counter-clockwise sense.

For loop $a$, $\oint \overrightarrow{B}.\overrightarrow{dl}={\mu }_0\left(1\right)={\mu }_0(\because \Sigma I=1\text{A})$

For loop $b$, $\oint \overrightarrow{B}.\overrightarrow{dl}={\mu }_0\left(0\right)=0(\because \Sigma I=0)$

For loop $c$,
$\oint \overrightarrow{B}.\overrightarrow{dl}={\mu }_0\left(1\right)={\mu }_0(\because \Sigma I=1\text{A})$

For loop $d$, $\oint \overrightarrow{B}.\overrightarrow{dl}={\mu }_0\left(I\right)={\mu }_0(\because \Sigma I=1\text{A})$

Hence,the correct order of $\oint \overrightarrow{B}.\overrightarrow{dl}$ for the four loops is,
$a=d=c>b$

Hence, option $\left(b\right)$ is the correct answer.