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

List-$I$	List-$II$
$(a)$ Gauss’ law for electrostatics	$(i)$ $\oint \vec{E} \cdot d\vec{A} = \frac{Q}{\epsilon_0}$
$(b)$ Gauss’ law for magnetism	(ii) $\oint \vec{B} \cdot d\vec{l} = \mu_0 [i_c + \epsilon_0 \frac{d\phi_E}{dt}]$
$(c)$ Faraday’s law	(iii) $\oint \vec{B} \cdot d\vec{A} = 0$
$(d)$ Ampere-Maxwell’s law	(iv) $\oint \vec{E} \cdot d\vec{l} = -\frac{d\phi_B}{dt}$

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

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