Question

Figure shows an Amperian path ABCDA. Part ABC is in vertical planePSTU while part CDA is in horizontal plane PQRS. Direction of circulation along the path is shown by an arrow near point B and at $D.\int \overrightarrow{B}.\overrightarrow{dl}$ 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

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



For the given direction of circulation,
$i_3$ leaving at PSTU is positive and $i_3$ entering at PQRS is negative. $i_1$ leaving PSTU is +ve and $i_2$ leaving PQRS is positive

current is taken +ve if the direction of magnetic filed by the current along the Amperian loop is in same sense as the direction of circulation along the loop

From Ampere's law over loop ABCDA,
$\oint \overrightarrow{B}.\overrightarrow{dl}={\mu }_0{i}_{en}$

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

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

Hence, option (D) is correct.