Which is the missing term in Ampere's circuit | vedclass

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(A) Ampere's circuital law states that the line integral of the magnetic field $\vec{B}$ around any closed path is equal to $\mu_0$ times the total cu

Vedclass · Accepted Answer

$\mu_0 \epsilon_0 \frac{d\Phi_E}{dt}$

Vedclass · Answer

(A) Ampere's circuital law states that the line integral of the magnetic field $\vec{B}$ around any closed path is equal to $\mu_0$ times the total current $I$ passing through the surface enclosed by the path.Maxwell noticed that this law was inconsistent for time-varying electric fields.He introduced the concept of displacement current $I_d = \epsilon_0 \frac{d\Phi_E}{dt}$ to maintain the continuity of current.The modified Ampere's circuital law is given by $\oint \vec{B} \cdot d\vec{l} = \mu_0 (I_c + I_d) = \mu_0 I_c + \mu_0 \epsilon_0 \frac{d\Phi_E}{dt}$.Therefore, the missing term that Maxwell added to the original law is $\mu_0 \epsilon_0 \frac{d\Phi_E}{dt}$.

$A$. $\oint E \cdot dA$	$(i)$ $0$
$B$. $\oint B \cdot dA$	$(ii)$ $-\frac{d\phi_B}{dt}$
$C$. $\oint E \cdot dl$	$(iii)$ $\frac{Q}{\varepsilon_0}$
$D$. $\oint B \cdot dl$	$(iv)$ $\mu_0(i_c + i_d)$

Which is the missing term in Ampere's circuital law?

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Match the following List-$I$ with List-$II$.
$A$. $\oint E \cdot dA$ $(i)$ $0$
$B$. $\oint B \cdot dA$ $(ii)$ $-\frac{d\phi_B}{dt}$
$C$. $\oint E \cdot dl$ $(iii)$ $\frac{Q}{\varepsilon_0}$
$D$. $\oint B \cdot dl$ $(iv)$ $\mu_0(i_c + i_d)$

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