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(D) The Ampere-Maxwell law is a modification of Ampere's circuital law to include the displacement current,which accounts for time-varying electric fi

Vedclass · Accepted Answer

$\oint \vec{B} \cdot d\vec{l} = \mu_0 \epsilon_0 \frac{d\phi_E}{dt} + \mu_0 i_{	ext{enclosed}}$

Vedclass · Answer

(D) The Ampere-Maxwell law is a modification of Ampere's circuital law to include the displacement current,which accounts for time-varying electric fields.The generalized Ampere-Maxwell law is given by:$\oint \vec{B} \cdot d\vec{l} = \mu_0 i_c + \mu_0 \epsilon_0 \frac{d\phi_E}{dt}$Where:$1. \oint \vec{B} \cdot d\vec{l}$ is the line integral of the magnetic field around a closed loop.$2. i_c$ is the conduction current.$3. \epsilon_0 \frac{d\phi_E}{dt}$ is the displacement current $(i_d)$.Thus,the expression is $\oint \vec{B} \cdot d\vec{l} = \mu_0 i_{	ext{enclosed}} + \mu_0 \epsilon_0 \frac{d\phi_E}{dt}$.

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

Choose the $CORRECT$ expression for the Ampere-Maxwell law.

Similar Questions

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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