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 $\displaystyle\sf D .\oint \overrightarrow{B}.d\overrightarrow{L}$for this path according to Ampere’s law will be
$\bullet \: \sf| 1 | \: (i_{1} - i_{2} + i_{3}) \mu_{0}$
$\bullet \: \sf| 2 | \: ( - i_{1} + i_{2} ) \mu_{0}$
$\bullet \: \sf| 3 | \: i_{3} \mu_{0}$
$\bullet \: \sf| 4 | \: (i_{1} + i_{2} ) \mu_{0}$



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Answer 1

Consider the amperian loop ABCDA, when it is reshaped to fit the image of a circular loop. You would notice that $\sf i_1$ and $\sf i_2$ will be in the same direction and $\sf i_3$ will nullified. [Refer to the Diagram]

Using Ampere's Law,

$\sf \displaystyle \sf \oint B.dl = \mu_o I_{enclosed}$

According to the given question,

$\sf \displaystyle \sf \oint B.dl = \mu_o (i_1 + i_2 )$

Option (4) is correct.

Answer 2

$\huge\bf\fbox\red{Answer:-}$

$\bullet \: \sf| 4 | \: (i_{1} + i_{2} ) \mu_{0}$

Step-by-step explanation:

It is the correct answer.

Hope this attachment helps you.