Single Choice

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 $$D$$. $$\oint { \vec { B } .d\vec { l } } $$ for this path according to Ampere's law will be :

A$$\left( { I }_{ 1 }-{ I }_{ 2 }+{ I }_{ 3 } \right) { \mu }_{ 0 }$$
B$$\left( -{ I }_{ 1 }+{ I }_{ 2 } \right) { \mu }_{ 0 }$$
C$${ I }_{ 3 }{ \mu }_{ 0 }$$
D$$\left( { I }_{ 1 }+{ I }_{ 2 } \right) { \mu }_{ 0 }$$
Correct Answer

Solution

$$\mu_0(I_{enc})=\oint \bar B.\bar dl$$
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, $$\mu_0(I_1+I_3+I_2-I_3) = \oint \bar B.\bar dl = \mu_0 (I_1+I_2)$$

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