Question

According to the modified Ampere's circuital law ($i_D=$ displacement current)

• A. $\oint \overrightarrow{B}\cdot \overrightarrow{dl}={\mu }_0(i_C+{\epsilon }_0\frac{d{\varphi }_E}{dt})$
• B. $\oint \overrightarrow{B}\cdot \overrightarrow{dl}={\mu }_0{\epsilon }_0\frac{d{\varphi }_E}{dt}$
• C. $\oint \overrightarrow{B}\cdot \overrightarrow{dl}={\mu }_0{i}_C$
• D. $\oint \overrightarrow{B}\cdot \overrightarrow{dl}={\mu }_0(i_C\frac{d{\varphi }_E}{dt}+i_D)$

Answer 1

The correct option is A $\oint \overrightarrow{B}\cdot \overrightarrow{dl}={\mu }_0(i_C+{\epsilon }_0\frac{d{\varphi }_E}{dt})$

According to the modified Ampere's circuital law,

$\oint \overrightarrow{B}\cdot \overrightarrow{dl}={\mu }_0(i_C+i_D)={\mu }_0(i_C+{\epsilon }_0\frac{d{\varphi }_E}{dt})$

Where,

$\overrightarrow{B}\to$ the magnetic induction at any point on the path.

$\overrightarrow{dl}\to$ is the length element of the path,

$i_c\to$ is the total conduction current through a surface enclosed by the closed path.

$I_d={\epsilon }_0\frac{d\varphi }{dt}\to$ the fictitious displacement current arising from the changing electric flux ${\varphi }_e$.

Hence, option $\left(A\right)$ is correct.