Question

Figure shows an ampereian path $\text{ABCDA}$. Part $\text{ABC}$ lies in vertical plane $\text{PSTU}$ while part $\text{CDA}$ lies in horizontal plane $\text{PQRS}$. $\oint \overrightarrow{B}.\overrightarrow{dl}$ for this path according to the Ampere's circuital law will be:

• A. $(i_1-i_2+i_3){\mu }_0$
• B. $(-i_1+i_2){\mu }_0$
• C. $i_3{\mu }_0$
• D. $(i_1+i_2){\mu }_0$

Answer 1

The correct option is D $(i_1+i_2){\mu }_0$

Let us assume the entire loop $\text{ABCDA}$ as a combination of two separate loops $\text{ABCA}$ and $\text{ACDA}$ as shown in the figure.

Ampere's circuital law states that,

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

For the loop section $\text{ABCA}$, using Right-hand thumb rule, the direction of current $i_1$ and $i_3$ is positive.

So, For the loop $\text{ABCA}$,

$\oint \overrightarrow{B}.\overrightarrow{dl}={\mu }_0(i_1+i_3)$

For the loop $\text{ACDA}$, using Right-hand thumb rule, the direction of current $i_2$ is positive and $i_3$ is negative.

So, For loop $\text{ACDA}$,

$\oint \overrightarrow{B}.\overrightarrow{dl}={\mu }_0(i_2-i_3)$

For the complete loop $\text{ABCDA}$,

$\oint \overrightarrow{B}.\overrightarrow{dl}={\mu }_0(i_1+i_3)+{\mu }_0(i_2-i_3)$

$\oint \overrightarrow{B}.\overrightarrow{dl}={\mu }_0(i_1+i_2)$

Why this question ? Note : The current element which does not passes through the cross-sectional area of the loop will not affect $I_{\text{enc}}$ by the loop.