Question

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

• A. $\oint \vec{B}\cdot \overrightarrow{dl}={\mu }_0\left(i_C+{\epsilon }_0\frac{d{\varphi }_E}{dt}\right)$
• B. $\oint \vec{B}\cdot \overrightarrow{dl}={\mu }_0{\epsilon }_0\frac{d{\varphi }_E}{dt}$
• C. $\oint \vec{B}\cdot \overrightarrow{dl}={\mu }_{0}i$
• D. $\oint \vec{B}\cdot \overrightarrow{d/}={\mu }_0\left(i_C\frac{d{\varphi }_E}{dt}+i_D\right)$

Answer 1

Answer: (A)

According to modified Ampere's circuital law.
$\oint \vec{B}\cdot \overrightarrow{dl}={\mu }_0\left(i_C+{\epsilon }_0\frac{d{\varphi }_E}{dt}\right)$