We were all taught about the electric potential $V$, which is defined up to a constant, and can be measured with a voltmeter or an oscilloscope. On the other hand, in electromagnetism are defined the scalar potential $\varphi$ and the vector potential $\mathbf A$, and there holds everywhere inside and outside the wires$$\nabla \varphi = - {\mathbf E} - {\partial {\mathbf A}\over \partial t}.$$

I always believed that the electric potential $V$ coincides with the electromagnetic scalar potential $\varphi$ inside the electric wires, and that's what is measured by a voltmeter or an oscilloscope. But thinking about, I get in trouble with that.

Question 1: The electromagnetic potential $\varphi$ is defined only up to a gauge. So, if the electric potential $V$ between two points $A$ and $B$ coincides with $\varphi(B)-\varphi(A)$, what is it? Let assume it's $V = \int_{A}^B {\mathbf E}\cdot d{\ell}$ (curvilinear integral along the wire between $A$ and $B$). Then we have $$V = \int_A^B (\nabla \varphi - {\partial {\mathbf A}\over \partial t})\cdot d \ell = \varphi(B) - \varphi(A) - \int_A^B {\partial {\mathbf A}\over \partial t}\cdot d \ell.$$ This works for steady currents, where there exists a gauge for which $A$ does not depend on $t$ (hence the integral vanishes). But for quickly varying currents like what can be measured with an oscilloscope (say 500 MHz), this fails: there is no electromagnetic potential that fulfills $\varphi(B)-\varphi(A) = \int_A^B {\mathbf E}\cdot d\ell$. So, what is the electric potential V?

question 2: I get even more in troubles when induction is involved: Consider a single loop solenoid, of surface $S$, with no resistance, subject to a uniform varying magnetic field of amplitude $B(t)$ normal to the loop. The two terminals of the solenoid are not connected, or more precisely are connected to an oscilloscope whose input impedance is very high (10 Mega ohm say). We are more or less taught that the voltage measured by the oscilloscope is $V(t) = S{\partial B\over \partial t}$. But it suffices to have a look at the demonstration of Faraday's law to see that all what is proved is that $$\int {\mathbf E}\cdot d\ell = - {\partial B\over \partial t},$$ where the integral is the curvilinear integral over a spatial closed loop. By what magics do we deduce that the potential between the two terminals of the electrical wire loop is of the aforementioned form? (I expect a relatively mathematical justification).

question 3: To bring things to an apocalyptic state, I ask about the following problem, taken from the Wikipedia article "electromotive force". As in the previous question, assume we have an electric wire loop subject to a normal uniform varying magnetic field $B(t)$. But now, we assume that one half of the loop has a resistance $R_1=100\ \Omega$, and the other half has a resistance $R_2 = 200\ \Omega$. I believe we can suppose the induced current $I(t)$ is the same all along the loop, so, at the two terminals of the resistors , we have on one hand $V = 100\ I$, and on the other hand $V = -200\ I$. The electric potential is even not well defined in this case. I'm puzzled.

We were all taught about the electric potential $V$, which is defined up to a constant, and can be measured with a voltmeter or an oscilloscope. On the other hand, in electromagnetism are defined the scalar potential $\varphi$ and the vector potential $\mathbf A$, and there holds everywhere inside and outside the wires$$\nabla \varphi = - {\mathbf E} - {\partial {\mathbf A}\over \partial t}.$$

I always believed that the electric potential $V$ coincides with the electromagnetic scalar potential $\varphi$ inside the electric wires, and that's what is measured by a voltmeter or an oscilloscope. But thinking about, I get in trouble with that.

Question 1: The electromagnetic potential $\varphi$ is defined only up to a gauge. So, if the electric potential $V$ between two points $A$ and $B$ coincides with $\varphi(B)-\varphi(A)$, what is it? Let assume it's $V = \int_{A}^B {\mathbf E}\cdot d{\ell}$ (curvilinear integral along the wire between $A$ and $B$). Then we have $$V = \int_A^B (\nabla \varphi - {\partial {\mathbf A}\over \partial t})\cdot d \ell = \varphi(B) - \varphi(A) - \int_A^B {\partial {\mathbf A}\over \partial t}\cdot d \ell.$$ This works for steady currents, where there exists a gauge for which $A$ does not depend on $t$ (hence the integral vanishes). But for quickly varying currents like what can be measured with an oscilloscope (say 500 MHz), this fails: there is no electromagnetic potential that fulfills $\varphi(B)-\varphi(A) = \int_A^B {\mathbf E}\cdot d\ell$. So, what is the electric potential V?

question 2: I get even more in troubles when induction is involved: Consider a single loop solenoid, of surface $S$, with no resistance, subject to a uniform varying magnetic field of amplitude $B(t)$ normal to the loop. The two terminals of the solenoid are not connected, or more precisely are connected to an oscilloscope whose input impedance is very high (10 Mega ohm say). We are more or less taught that the voltage measured by the oscilloscope is $V(t) = S{\partial B\over \partial t}$. But it suffices to have a look at the demonstration of Faraday's law to see that all what is proved is that $$\oint {\mathbf E}\cdot d\ell = - {\partial B\over \partial t},$$ where the integral is the curvilinear integral over a spatial closed loop. By what magics do we deduce that the potential between the two terminals of the electrical wire loop is of the aforementioned form? (I expect a relatively mathematical justification).

question 3: To bring things to an apocalyptic state, I ask about the following problem, taken from the Wikipedia article "electromotive force". As in the previous question, assume we have an electric wire loop subject to a normal uniform varying magnetic field $B(t)$. But now, we assume that one half of the loop has a resistance $R_1=100\ \Omega$, and the other half has a resistance $R_2 = 200\ \Omega$. I believe we can suppose the induced current $I(t)$ is the same all along the loop, so, at the two terminals of the resistors , we have on one hand $V = 100\ I$, and on the other hand $V = -200\ I$. The electric potential is even not well defined in this case. I'm puzzled.

We were all taught about the electric potential $V$, which is defined up to a constant, and can be measured with a voltmeter or an oscilloscope. On the other hand, in electromagnetism are defined the scalar potential $\varphi$ and the vector potential $\mathbf A$, and there holds everywhere inside and outside the wires$$\nabla \varphi = - E - {\partial {\mathbf A}\over \partial t}.$$

I always believed that the electric potential $V$ coincides with the electromagnetic scalar potential $\varphi$ inside the electric wires, and that's what is measured by a voltmeter or an oscilloscope. But thinking about, I get in trouble with that.

Question 1: The electromagnetic potential $\varphi$ is defined only up to a gauge. So, if the electric potential $V$ between two points $A$ and $B$ coincides with $\varphi(B)-\varphi(A)$, what is it? Let assume it's $V = \int_{A}^B {\mathbf E}\cdot d{\ell}$ (curvilinear integral along the wire between $A$ and $B$). Then we have $$V = \int_A^B (\nabla \varphi - {\partial {\mathbf A}\over \partial t})\cdot d \ell = \varphi(B) - \varphi(A) - \int_A^B {\partial {\mathbf A}\over \partial t}\cdot d \ell.$$ This works for steady currents, where there exists a gauge for which $A$ does not depend on $t$ (hence the integral vanishes). But for quickly varying currents like what can be measured with an oscilloscope (say 500 MHz), this fails: there is no electromagnetic potential that fulfills $\varphi(B)-\varphi(A) = \int_A^B {\mathbf E}\cdot d\ell$. So, what is the electric potential V?

question 2: I get even more in troubles when induction is involved: Consider a single loop solenoid, of surface $S$, with no resistance, subject to a uniform varying magnetic field of amplitude $B(t)$ normal to the loop. The two terminals of the solenoid are not connected, or more precisely are connected to an oscilloscope whose input impedance is very high (10 Mega ohm say). We are more or less taught that the voltage measured by the oscilloscope is $V(t) = S{\partial B\over \partial t}$. But it suffices to have a look at the demonstration of Faraday's law to see that all what is proved is that $$\int {\mathbf E}\cdot d\ell = - {\partial B\over \partial t},$$ where the integral is the curvilinear integral over a spatial closed loop. By what magics do we deduce that the potential between the two terminals of the electrical wire loop is of the aforementioned form? (I expect a relatively mathematical justification).

question 3: To bring things to an apocalyptic state, I ask about the following problem, taken from the Wikipedia article "electromotive force". As in the previous question, assume we have an electric wire loop subject to a normal uniform varying magnetic field $B(t)$. But now, we assume that one half of the loop has a resistance $R_1=100\ \Omega$, and the other half has a resistance $R_2 = 200\ \Omega$. I believe we can suppose the induced current $I(t)$ is the same all along the loop, so, at the two terminals of the resistors , we have on one hand $V = 100\ I$, and on the other hand $V = -200\ I$. The electric potential is even not well defined in this case. I'm puzzled.

We were all taught about the electric potential $V$, which is defined up to a constant, and can be measured with a voltmeter or an oscilloscope. On the other hand, in electromagnetism are defined the scalar potential $\varphi$ and the vector potential $\mathbf A$, and there holds everywhere inside and outside the wires$$\nabla \varphi = - {\mathbf E} - {\partial {\mathbf A}\over \partial t}.$$

I always believed that the electric potential $V$ coincides with the electromagnetic scalar potential $\varphi$ inside the electric wires, and that's what is measured by a voltmeter or an oscilloscope. But thinking about, I get in trouble with that.

Question 1: The electromagnetic potential $\varphi$ is defined only up to a gauge. So, if the electric potential $V$ between two points $A$ and $B$ coincides with $\varphi(B)-\varphi(A)$, what is it? Let assume it's $V = \int_{A}^B {\mathbf E}\cdot d{\ell}$ (curvilinear integral along the wire between $A$ and $B$). Then we have $$V = \int_A^B (\nabla \varphi - {\partial {\mathbf A}\over \partial t})\cdot d \ell = \varphi(B) - \varphi(A) - \int_A^B {\partial {\mathbf A}\over \partial t}\cdot d \ell.$$ This works for steady currents, where there exists a gauge for which $A$ does not depend on $t$ (hence the integral vanishes). But for quickly varying currents like what can be measured with an oscilloscope (say 500 MHz), this fails: there is no electromagnetic potential that fulfills $\varphi(B)-\varphi(A) = \int_A^B {\mathbf E}\cdot d\ell$. So, what is the electric potential V?

question 2: I get even more in troubles when induction is involved: Consider a single loop solenoid, of surface $S$, with no resistance, subject to a uniform varying magnetic field of amplitude $B(t)$ normal to the loop. The two terminals of the solenoid are not connected, or more precisely are connected to an oscilloscope whose input impedance is very high (10 Mega ohm say). We are more or less taught that the voltage measured by the oscilloscope is $V(t) = S{\partial B\over \partial t}$. But it suffices to have a look at the demonstration of Faraday's law to see that all what is proved is that $$\int {\mathbf E}\cdot d\ell = - {\partial B\over \partial t},$$ where the integral is the curvilinear integral over a spatial closed loop. By what magics do we deduce that the potential between the two terminals of the electrical wire loop is of the aforementioned form? (I expect a relatively mathematical justification).

question 3: To bring things to an apocalyptic state, I ask about the following problem, taken from the Wikipedia article "electromotive force". As in the previous question, assume we have an electric wire loop subject to a normal uniform varying magnetic field $B(t)$. But now, we assume that one half of the loop has a resistance $R_1=100\ \Omega$, and the other half has a resistance $R_2 = 200\ \Omega$. I believe we can suppose the induced current $I(t)$ is the same all along the loop, so, at the two terminals of the resistors , we have on one hand $V = 100\ I$, and on the other hand $V = -200\ I$. The electric potential is even not well defined in this case. I'm puzzled.