Question

Figure shows an Amperian path $ABCDA$. Part $ABC$ is in verical plane $PSTU$ while part $CDA$ is in horizontal plane $PQRS$. Direction of circulation along the path is shown by an arrow near point $B$ and $D$.
$\oint \overrightarrow{B}.d\overrightarrow{l}$ for this path according to Ampere's 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

Answer: (D)

$(I_1+I_2){\mu }_0$

${\mu }_0\left(I_{enc}\right)=\oint \overline{B}.\overline{d}l$

Area vector of ABC is along the right direction, while ACD is along the upward direction.

Now currents passing through ABC are $I_1,I_3$ along the area directions.

Now through ACD, $I_2,I_3;I_2$ along the area vector while $I_3$ is opposite to it.

Thus, ${\mu }_0(I_1+I_3+I_2-I_3)=\oint \overline{B}.\overline{d}l={\mu }_0(I_1+I_2)$