Question

For the given circuit, at steady state, the Ampere Maxwell Law can be modified to -

• A. $\oint \overrightarrow{B}.\overrightarrow{dl}={\mu }_0{i}_c$
• B. $\oint \overrightarrow{B}.\overrightarrow{dl}={\mu }_0{i}_d$
• C. $\oint \overrightarrow{B}.\overrightarrow{dl}=0$
• D. $\oint \overrightarrow{B}.\overrightarrow{dl}={\mu }_0(i_c+i_d)$

Answer 1

The correct option is C $\oint \overrightarrow{B}.\overrightarrow{dl}=0$


We know that Ampere Maxwell Law is given by,

$\oint \overrightarrow{B}.\overrightarrow{dl}={\mu }_0(i_c+i_d)={\mu }_0(i_c+{ϵ}_0\frac{d{\varphi }_E}{dt})$

For the given circuit, at steady state, capacitor act as an open switch, therefore conduction current is zero.

Also, the electric field between the plates of the capacitor is constant at steady state, which means electric flux is constant. Therefore, displacement current is also zero.

So, the Ampere Maxwell Law will be $\oint \overrightarrow{B}.\overrightarrow{dl}=0$.

Hence, option $\left(C\right)$ is the correct answer.