Question

Find the value of $\oint \overrightarrow{B}\cdot \overrightarrow{dl}$ for the loop $L_2$ as shown in figure. The direction of magnetic field is represented as loops and direction of current element is given in the figure.

• A. ${\mu }_0(I_1-I_2)$
• B. Zero
• C. ${\mu }_0(I_1+I_3+I_4)$
• D. ${\mu }_0(I_1-I_2+I_4)$

Answer 1

The correct option is D ${\mu }_0(I_1-I_2+I_4)$

Assuming plane of loops is lying on the plane of paper.

For loop $L_2$, the direction of magnetic field line is in anticlockwise which suggests perpendicularly outward direction of current as $+ve$.

Applying Ampere's circuital law for loop $L_2$

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

$\oint \overrightarrow{B}.\overrightarrow{dl}={\mu }_0(I_4+I_1-I_2)$

Here, current $I_2$ is considered $-ve$ because it will give clockwise sense of magnetic field lines for the loop.

$\therefore \oint \overrightarrow{B}.\overrightarrow{dl}={\mu }_0(I_4+I_1-I_2)$

Hence, option $\left(d\right)$ is correct.

Why this question ? Tip: In questions involving current element and closed loop, by applying right-hand thumb rule we can figure out +ve reference for current direction as per considered closed loop.