Assume $E_{n}$ is a non-conservative field due to varying magnetic flux. If this is applied over a circuit clearly the Kirchhoff's voltage law won't be obeyed (if $E_{c}$ is completely absent). What will happen to Kirchhoff's current law? Shouldn't that be obeyed so that conservation of charges still holds?
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1$\begingroup$ A non-conservative field can not create new charges. It only gets existing charges moving, so the KCL holds. $\endgroup$– FlatterMannCommented Oct 23, 2022 at 7:59
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$\begingroup$ see my answer to physics.stackexchange.com/q/733015 $\endgroup$– hyportnexCommented Oct 23, 2022 at 14:38
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1$\begingroup$ KVL applies as long as you account for all your EMFs. Indeed, I tell students that when you employ KVL you are implicitly pledging to account for EMFs. $\endgroup$– John DotyCommented Oct 24, 2022 at 12:16
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1$\begingroup$ KVL is obeyed irrespective of presence of induced electric field, because it refers to differences of electric potential, not to integrals of total electric field. For example, KVL is obeyed in AC circuits with inductors, or with transformers. However, if the induced field in circuit is due to external sources not part of the circuit (transformer), KVL, even though true, is not easily applicable and one has to to go back to its original form, Kirchhoff's second circuital law, which refers to emfs, not potential differences. $\endgroup$– Ján LalinskýCommented Oct 29, 2022 at 16:52
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1 Answer
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KCL comes from the continuity equation for electric charge.
If nodes have no capability of storing electric charges, KCL (sum of currents = 0) holds.
If a geometric volume in space has the capability of storing electric charge, the net flux of electric charge (sum of current) equals both:
- the time derivative of the electric charge contained in the volume
- the time derivative of the flux of the displacement field
Some details
Some details with differential and integral equations for a steady domain:
continuity equation of electric charge:
differential equation: $0 = \dfrac{\partial \rho}{\partial t} + \nabla \cdot \mathbf{j}$
integral equation: $0 = \displaystyle\dfrac{d}{dt} \underbrace{ \int_V \rho}_{=Q} + \underbrace{\oint_{\partial V} \mathbf{j} \cdot \mathbf{\hat{n}}}_{= \Phi_{\partial V}(\mathbf{j}) =\sum_k i_k} = \dot{Q} + \Phi_{\partial V}(\mathbf{j})$
Maxwell-Ampére equation:
differential equation: $\nabla \times \mathbf{h} = \mathbf{j} + \dfrac{\partial \mathbf{d}}{\partial t}$
integral equation: $\displaystyle \oint_{\partial S} \mathbf{h} \cdot \mathbf{\hat{t}} = \int_S \mathbf{j} \cdot \mathbf{\hat{n}} + \dfrac{d}{dt} \int_S \mathbf{d} \cdot \mathbf{\hat{n}}$
If $S$ is the closure of a volume $V$, $S = \partial V$, the closure of $S$ is an empty set, so that $|\partial S| = 0$ and the integral over $\partial S$ is identically 0,
$\displaystyle 0 = \underbrace{\int_{\partial V} \mathbf{j} \cdot \mathbf{\hat{n}}}_{=\Phi_{\partial V}(\mathbf{j})} + \dfrac{d}{dt} \underbrace{\int_{\partial V} \mathbf{d} \cdot \mathbf{\hat{n}}}_{=\Phi_{\partial V}(\mathbf{d})} = \Phi_{\partial V}(\mathbf{j}) + \dot\Phi_{\partial V}(\mathbf{d})$.
Thus, under the assumptions above (steady domain $V$), we get
$\displaystyle \sum_k i_k = \Phi_{\partial V}(\mathbf{j}) = - \dot{Q} = - \dot\Phi_{\partial V}(\mathbf{d})$
