Question

Eight wires cut the page perpendicularly at the points shown in figure. A wire labelled with the integer $k$, $(k=1,2,....8)$ bears the current $k{i}_0$. For those with odd $k,$ the current flows up out of the page; for those with even $k,$ current flows down into the page. The value of $\oint \overrightarrow{B}\cdot d\overrightarrow{l}$ along the close path (as shown in the figure) in the direction indicated by the arrow is,
[Take out of the page as positive and into the page as negative]

• A. $10{\mu }_0{i}_0$
• B. $-10{\mu }_0{i}_0$
• C. $-4{\mu }_0{i}_0$
• D. $4{\mu }_0{i}_0$

Answer 1

The correct option is B $-10{\mu }_0{i}_0$

According to Ampere's circuital law,

$\oint \overrightarrow{B}\cdot \overrightarrow{dl}={\mu }_0(I_{net}{)}_{encl}$

$\oint \overrightarrow{B}\cdot \overrightarrow{dl}={\mu }_0\sum I$

$={\mu }_0(i_1+i_3+i_6+i_8)[\because i_2,i_4,i_5,i_7\text{lies outside the loop}]$
$={\mu }_0(1+3-6-8)i_0$

$=-10{\mu }_0{i}_0$

<!--td {border: 1px solid #ccc;}br {mso-data-placement:same-cell;}--> Hence, $\left(B\right)$ is the correct answer.