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 $D$ . $\oint\overset{ \rightarrow }{B}.d\overset{ \rightarrow }{l}$ 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,
$\oint \overrightarrow{\boldsymbol{B}} \cdot \overrightarrow{d l}=\mu_0\left(\begin{array}{ll}\Sigma i & \text { inside }\end{array}\right)$
$=\mu_0\left(i_1+i_2+i_3-i_3\right)$
$=\mu_0\left(i_1+i_2\right)$