Question

The diagram show cross-section of several conductors that carry currents perpendicular to the plane of the diagram. The currents have magnitudes $I_1=3.5\text{A},$$I_2=6.5\text{A},I_3=1.5\text{A}$ and their directions are as shown. Four closed paths labelled $a$ to $d$ are shown. The line integral $\oint \overrightarrow{B}\cdot \overrightarrow{dl}$ for the paths $a,b,c$ and $d$ will be respectively (Each integral going around the path in counterclockwise direction).

• A. $0,-3.5{\mu }_o,3{\mu }_o,4.5{\mu }_o$
• B. $0,3.5{\mu }_o,-3{\mu }_o,-4.5{\mu }_o$
• C. $0,3{\mu }_o,-3.5{\mu }_o,11.5{\mu }_o$
• D. All integrals will be equal to zero.

Answer 1

The correct option is A $0,-3.5{\mu }_o,3{\mu }_o,4.5{\mu }_o$

According to Ampere's law,

$\oint \overrightarrow{B}\cdot \overrightarrow{dl}={\mu }_o⟮\sum I⟯$

From the right-hand thumb rule, when going around the path in a counter-clockwise direction, currents out of the page are positive and currents into the page are negative.

For path $a$, $\sum I=0$

For path $b$, $\sum I=-I_1=-3.5\text{A}$

For path $c$, $\sum I=-I_1+I_2=3\text{A}$

For path $d$, $\sum I=-I_1+I_2+I_3=4.5\text{A}$

$\Rightarrow {\oint }_a\overrightarrow{B}\cdot \overrightarrow{dl}=0$

$\Rightarrow {\oint }_b\overrightarrow{B}\cdot \overrightarrow{dl}=-3.5{\mu }_0$

$\Rightarrow {\oint }_c\overrightarrow{B}\cdot \overrightarrow{dl}=3{\mu }_0$

$\Rightarrow {\oint }_d\overrightarrow{B}\cdot \overrightarrow{dl}=4.5{\mu }_0$

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