Question

Figure shows an amperian path $ABCDA$ . Part $ABC$ is in vertical 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 at $D$ . $\oint \overrightarrow{B}.d\overrightarrow{l}$ for this path according to Ampere's law will be

• A. $\left({\mathrm{i}}_1-{\mathrm{i}}_2+{\mathrm{i}}_3\right){\mu }_0$
• B. $\left(-{\mathrm{i}}_1+{\mathrm{i}}_2\right){\mu }_0$
• C. ${\text{i}}_3{\mu }_0$
• D. $\left(i_1+i_2\right){\mu }_0$

Answer 1

Answer: (D)

If the loop current is coming out it should be taken as positive and if it is going in it is taken as negative.
$i_3$ is coming out and again going in so its net contribution will be zero.
According to Ampere's law,
$\oint \overrightarrow{\mathbf{B}}\cdot \overrightarrow{dl}={\mu }_0\left(\begin{array}{ll}\Sigma i& \text{ inside }\end{array}\right)$
$={\mu }_0\left(i_1+i_2+i_3-i_3\right)$
$={\mu }_0\left(i_1+i_2\right)$