Question

With reference to Ampere's circuital law for given figure. Which option is correct for $\oint \overrightarrow{B}.\overrightarrow{dl}=0$ ?

• A. $I_1+I_2-I_3=0$
• B. $I_1-I_2-I_3=0$
• C. $I_1-I_2+I_3=0$
• D. None of the above

Answer 1

The correct option is A $I_1+I_2-I_3=0$

According to Ampere's circuital law,
$\oint \overrightarrow{B}.\overrightarrow{dl}={\mu }_0{I}_{enc}$

$I_{enc}=$ net current passing through closed loop.

Given,

$\oint \overrightarrow{B}.\overrightarrow{dl}=0$

$\therefore {\mu }_0{I}_{enc}=0$

$\Rightarrow I_{enc}=0$

As the loop is traversed in the anticlockwise sense as shown in the figure, the direction of $+ve$ current is given by right-hand curl rule i.e. perpendicular to the plane and pointing outwards.

$\therefore$ current pointing outwards $=+ve$

So, $I_1=(+ve),I_2=(+ve),I_3=(-ve)$

So, $I_1+I_2-I_3=0$

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