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Now, let us compute the power transmitted by the electromagnetic field to $D$. Notice that only the component of the electromagnetic force parallel to the curvilinear speed $\mathbf v$ of the charges in the device provide work to the device, hence $$P = \int_D d\mathbf F\cdot \mathbf v = \int_D (\rho\, d\ell )\,[{\mathbf E + (\mathbf v+\mathbf V)\times \mathbf B}]\cdot \mathbf v.$$ Since $\mathbf v$ is orthogonal to $\mathbf v\times \mathbf B$, this expression simplifies to $$P = \int_D \rho\, (\mathbf E + \mathbf V\times \mathbf B)\cdot \mathbf v \,d\ell = \int_D(\mathbf E + \mathbf V \times \mathbf B)\cdot \mathbf j \, d\ell,$$ where we have denoted by $\mathbf j$ the current density with respect to the conductor (the curvilinear current density component of the total current density). If we assume that the current is uniform inside $D$, which is a valid assumption as far as the wave length of the electric wave is large with respect to the dimension of the device, then $||\mathbf j|| = I$, $\mathbf j = I{d\vec \ell \over d\ell}$ and $$P = I\int_D (\mathbf E + \mathbf V \times \mathbf B)\cdot d\vec\ell = I V_T.$$ This is valid for every filiform two terminal devices.

Now, let us compute the electrical power transmitted by the electromagnetic field to $D$. Notice that only the component of the electromagnetic force parallel to the curvilinear speed $\mathbf v$ of the charges in the device provides electrical work to the device, hence $$P = \int_D d\mathbf F\cdot \mathbf v = \int_D (\rho\, d\ell )\,[{\mathbf E + (\mathbf v+\mathbf V)\times \mathbf B}]\cdot \mathbf v.$$ Since $\mathbf v$ is orthogonal to $\mathbf v\times \mathbf B$, this expression simplifies to $$P = \int_D \rho\, (\mathbf E + \mathbf V\times \mathbf B)\cdot \mathbf v \,d\ell = \int_D(\mathbf E + \mathbf V \times \mathbf B)\cdot \mathbf j \, d\ell,$$ where we have denoted by $\mathbf j$ the current density with respect to the conductor (the curvilinear current density component of the total current density). If we assume that the current is uniform inside $D$, which is a valid assumption as far as the wave length of the electric wave is large with respect to the dimension of the device, then $||\mathbf j|| = I$, $\mathbf j = I{d\vec \ell \over d\ell}$ and $$P = I\int_D (\mathbf E + \mathbf V \times \mathbf B)\cdot d\vec\ell = I V_T.$$ This is valid for every filiform two terminal devices.

Unfortunately, there is some of cheating here, because the "conservative part of the E-field" is not uniquely defined by Helmholtz decomposition which needs not be unique, even if it is specified as some "ground" point. Indeed, two possible functions $V_1$ and $V_2$ which agree at some point are possible: it suffices to choose an harmonic function $\Lambda$ equal to $0$ at the specified point, and to set $V_2 = V_1+\Lambda$. Then it is easily seen that if $\mathbf E = \nabla V_1 + \nabla\times \mathbf F_1$, then $\mathbf E = \nabla V_2 + \nabla \times (\mathbf F_1- \mathbf F_2)$, where $\nabla \times \mathbf F_2 = \nabla \Lambda$ (which is possible since $\nabla^2\Lambda = 0$ by definition).

question 2 and 3: Assume we have a filiform two terminal device $D$ of some shape. We also assume here, for the sake of simplicity, that the device does not move. Let $\rho(z, t)$ be the moving charge linear density at some point $z$ of $D$ and $\mathbf v$ its speed. Denote by $V_{emf}$ the integral of $d\vec\ell \cdot {\mathbf E + \mathbf v\times \mathbf B}$ along $D$, from the extremity where the current exits to the extremity where the current enters into the wire.

Pay attention that $V_{emf}$ is not defined as a potential between two points, but as an electromotive force along a (filiform) device. Since $\mathbf v\times \mathbf B$ is orthogonal to $\mathbf v$, and hence to $d\vec\ell$, We have $$V_{emf} = \int_D E \cdot d\vec\ell .$$ Furthermore, the power transmitted by the electromagnetic field to $D$ is $$P = \int_D d\mathbf F\cdot \mathbf v = \int_D (\rho\, d\ell )\,({\mathbf E + \mathbf v\times \mathbf B})\cdot \mathbf v = \int_D \rho\, \mathbf E\cdot \mathbf v \,d\ell = \int_D\mathbf E \cdot \mathbf j d\ell.$$ If we assume that the current is uniform inside $D$ which is a valid assumption as far as the wave length of the electric wave is large with respect to the dimension of the device, then $||\mathbf j|| = I$, $\mathbf j = I{d\vec \ell \over d\ell}$ and $$P = I\int_D \mathbf E\cdot d\vec\ell = I V_{emf}.$$ This is valid for every filiform two terminal devices.

On the other hand, one of the most basic law of electrical engineering is that the power transmitted to a two terminal device is $P = VI$, where $V$ is the electric potential between the two terminals. We see that this is valid only whenever $V = V_{emf}$, that is, with Maxwell equation $E = -\nabla \varphi - \partial_t \mathbf A$, whenever $\partial_t\mathbf A = 0$ (under the Coulomb gauge). Since $\mathbf B = \nabla \times \mathbf A$, this implies $$\partial_t \mathbf B = 0.$$ The meaning of this equation is :

Conversely, if this condition holds, then $\nabla \times \partial_t \mathbf A = 0$, hence $\partial_t A$ is irrotational: $\partial_t\mathbf A = \nabla \mathbf F$. Coulomb gauge $\nabla\cdot \mathbf A = 0$ implies $$\nabla^2 \mathbf F = 0.$$
Observe that $\mathbf E$ vanishes at infinity (since the electric wires are contained in a bounded domain). From the definition of $\varphi$ and Maxwell equation $$\mathbf E = -\nabla \varphi - \partial_t \mathbf A,$$ it follows that $\partial_t \mathbf A = \nabla \mathbf F$ vanishes at infinity. Hence $\mathbf F$ is constant (uniqueness of the solution of Laplace equations with Neuman boundary conditions). Hence $\partial_t \mathbf A = 0$. We see that the above condition is also sufficient.

In general, laws of the form $V = f(I)$ for two terminal devices have to be replaced by $$V_{emf} = f(I),$$ and one has to be careful that $V_{emf}$ is not a potential between two points, that would be independant of the path joining those points, but an "electromotrive force" along a given path, which depends also on the shape of the path. If the path is oriented, then $V_{emf}$ is defined as above if the orientation is opposite to the current flow, and is defined as its opposite otherwise. With these conventions, everything holds true by replacing everywhere $V$ by $V_{emf}$, except for the path dependence. Nevertheless, at those zones of the circuit where there is no magnetic induction, the electromotive force $V_{emf}$ along the path joining two points $A$ and $B$ is equal to $V(B) - V(A)$, as explained above. So everything works, if one is careful to isolate those zones where there is some magnetic induction and to give them a special treatment.

This works in synergy with induction loops (inductor etc.), where all what is specified is the electromotive force as a function of the flux $\Phi$: $$V_{emf} = \partial_t \Phi.$$

So, the apparent paradox in question 3 can be solved as follows: $$V_{emf} = \partial_t \Phi = V_{emf}({\rm path_1}, R_1) + V_{emf}({\rm path_2}, R_2) = R_1 I + R_2 I = (R_1 + R_2)I.$$ Hence $I = V_{emf}/(R_1+R_2)$.

At the terminals of the loop, and for any inductor in general, the magnetic induction is weak and we have $$V_{emf}(\mathrm{along\ any\ path\ joining\ the\ terminals}) = \Delta V$$ (assuming the other paths joining the terminals lie in induction free zones).

Unfortunately, there is some cheating here, because the "conservative part of the E-field" is not uniquely defined by Helmholtz decomposition which needs not be unique, even if it is specified as some "ground" point. Indeed, two possible functions $V_1$ and $V_2$ which agree at some point are possible: it suffices to choose an harmonic function $\Lambda$ equal to $0$ at the specified point, and to set $V_2 = V_1+\Lambda$. Then it is easily seen that if $\mathbf E = \nabla V_1 + \nabla\times \mathbf F_1$, then $\mathbf E = \nabla V_2 + \nabla \times (\mathbf F_1- \mathbf F_2)$, where $\nabla \times \mathbf F_2 = \nabla \Lambda$ (which is possible since $\nabla^2\Lambda = 0$ by definition).

question 2 and 3: Assume we have a filiform two terminal device $D$ of some shape. The device may be moving or deforming, so, we shall denote by $\mathbf V(z)$ the speed of the device at (curvilinear) coordinate $z$. Let $\rho(z, t)$ be the moving charge linear density at some point $z$ of $D$, and $\mathbf v$ its curvilinear speed. So, the total speed of the moving charges inside the conductor is $\mathbf v + \mathbf V$. We define the quantity $V_T$ by $$V_T = \int_D (\mathbf E + \mathbf V\times \mathbf B) \cdot d\vec\ell.$$ Pay attention that $V_T$ should not be seen as a potential between two points, but as an electromotive propensity along a (filiform) device.

Now, let us compute the power transmitted by the electromagnetic field to $D$. Notice that only the component of the electromagnetic force parallel to the curvilinear speed $\mathbf v$ of the charges in the device provide work to the device, hence $$P = \int_D d\mathbf F\cdot \mathbf v = \int_D (\rho\, d\ell )\,[{\mathbf E + (\mathbf v+\mathbf V)\times \mathbf B}]\cdot \mathbf v.$$ Since $\mathbf v$ is orthogonal to $\mathbf v\times \mathbf B$, this expression simplifies to $$P = \int_D \rho\, (\mathbf E + \mathbf V\times \mathbf B)\cdot \mathbf v \,d\ell = \int_D(\mathbf E + \mathbf V \times \mathbf B)\cdot \mathbf j \, d\ell,$$ where we have denoted by $\mathbf j$ the current density with respect to the conductor (the curvilinear current density component of the total current density). If we assume that the current is uniform inside $D$, which is a valid assumption as far as the wave length of the electric wave is large with respect to the dimension of the device, then $||\mathbf j|| = I$, $\mathbf j = I{d\vec \ell \over d\ell}$ and $$P = I\int_D (\mathbf E + \mathbf V \times \mathbf B)\cdot d\vec\ell = I V_T.$$ This is valid for every filiform two terminal devices.

On the other hand, one of the most basic law of electrical engineering is that the power transmitted to a two terminal device is $P = VI$, where $V$ is the electric potential between the two terminals. We see that this is valid only whenever $V = V_T$, that is, with Maxwell equation $E = -\nabla \varphi - \partial_t \mathbf A$, whenever $\partial_t\mathbf A = 0$ (under the Coulomb gauge). Since $\mathbf B = \nabla \times \mathbf A$, this implies $$\partial_t \mathbf B = 0.$$ The meaning of this equation is :

Conversely, if this condition holds, then $\nabla \times \partial_t \mathbf A = 0$, hence $\partial_t A$ is irrotational: $\partial_t\mathbf A = \nabla \mathbf F$. Coulomb gauge $\nabla\cdot \mathbf A = 0$ implies $$\nabla^2 \mathbf F = 0.$$
Observe that $\mathbf E$ vanishes at infinity (since the electric wires are contained in a bounded domain). From the definition of $\varphi$ and Maxwell equation $$\mathbf E = -\nabla \varphi - \partial_t \mathbf A,$$ it follows that $\partial_t \mathbf A = \nabla \mathbf F$ vanishes at infinity. Hence $\mathbf F$ is constant (uniqueness of the solution of Laplace equations with Neuman boundary conditions). Hence $\partial_t \mathbf A = 0$. We see that the above condition is also sufficient. In general, laws of the form $V = f(I)$ for two terminal devices have to be replaced by $$V_T = f(I),$$ and one has to be careful that $V_T$ is not a potential between two points, that would be independant of the path joining those points, but an "electromotrive force" along a given path, which depends also on the shape of the path. If the path is oriented, then $V_T$ is defined as above if the orientation is opposite to the current flow, and is defined as its opposite otherwise. With these conventions, everything holds true by replacing everywhere $V$ by $V_T$, except for the path dependence. Nevertheless, at those zones of the circuit where there is no magnetic induction, the electromotive force $V_T$ along the path joining two points $A$ and $B$ is equal to $V(B) - V(A)$, as explained above. So everything works, if one is careful to isolate those zones where there is some magnetic induction and to give them a special treatment.

This works in synergy with induction loops (inductor etc.), where all what is specified is the electromotive force as a function of the flux $\Phi$: $$V_T = \partial_t \Phi.$$

So, the apparent paradox in question 3 can be solved as follows: $$V_T = \partial_t \Phi = V_T({\rm path_1}, R_1) + V_T({\rm path_2}, R_2) = R_1 I + R_2 I = (R_1 + R_2)I.$$ Hence $I = V_T/(R_1+R_2)$.

At the terminals of the loop, and for any inductor in general, the magnetic induction is weak and we have $$V_T(\mathrm{along\ any\ path\ joining\ the\ terminals}) = \Delta V$$ (assuming the other paths joining the terminals lie in induction free zones).

question 2 and 3: Assume we have a filiform two terminal device $D$ of some shape. Let $\rho(z, t)$ be the moving charge linear density at some point $z$ of $D$ and $\mathbf v$ its speed. Denote by $V_{emf}$ the integral of $d\vec\ell \cdot {\mathbf E + \mathbf v\times \mathbf B}$ along $D$, from the extremity where the current exits to the extremity where the current enters into the wire.

Pay attention that $V_{emf}$ is not defined as a potential between two points, but as an electromotive force along a (filiform) device. The power transmitted by the electromagnetic field to $D$ is $$P = \int_D d\mathbf F\cdot \mathbf v = \int_D (\rho\, d\ell )\,({\mathbf E + \mathbf v\times \mathbf B})\cdot \mathbf v = \int_D(\mathbf E + \mathbf v\times \mathbf B)\cdot \mathbf j d\ell.$$ If we assume that the current is uniform inside $D$ which is a valid assumption as far as the wave length of the electric wave is large with respect to the dimension of the device, then $||\mathbf j|| = I$, $\mathbf j = I{d\vec \ell \over d\ell}$ and $$P = I\int_D \mathbf (\mathbf E+\mathbf v \times \mathbf B)\cdot d\vec\ell = I V_{emf}.$$ This is valid for every filiform two terminal devices.

question 2 and 3: Assume we have a filiform two terminal device $D$ of some shape. We also assume here, for the sake of simplicity, that the device does not move. Let $\rho(z, t)$ be the moving charge linear density at some point $z$ of $D$ and $\mathbf v$ its speed. Denote by $V_{emf}$ the integral of $d\vec\ell \cdot {\mathbf E + \mathbf v\times \mathbf B}$ along $D$, from the extremity where the current exits to the extremity where the current enters into the wire.

Pay attention that $V_{emf}$ is not defined as a potential between two points, but as an electromotive force along a (filiform) device. Since $\mathbf v\times \mathbf B$ is orthogonal to $\mathbf v$, and hence to $d\vec\ell$, We have $$V_{emf} = \int_D E \cdot d\vec\ell .$$ Furthermore, the power transmitted by the electromagnetic field to $D$ is $$P = \int_D d\mathbf F\cdot \mathbf v = \int_D (\rho\, d\ell )\,({\mathbf E + \mathbf v\times \mathbf B})\cdot \mathbf v = \int_D \rho\, \mathbf E\cdot \mathbf v \,d\ell = \int_D\mathbf E \cdot \mathbf j d\ell.$$ If we assume that the current is uniform inside $D$ which is a valid assumption as far as the wave length of the electric wave is large with respect to the dimension of the device, then $||\mathbf j|| = I$, $\mathbf j = I{d\vec \ell \over d\ell}$ and $$P = I\int_D \mathbf E\cdot d\vec\ell = I V_{emf}.$$ This is valid for every filiform two terminal devices.