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If we go with conventional current sign, would it be right to say DC current is always as per the conventional current and AC current switches between conventional and actual electron flow? Not …

At a minimum it is potentially confusing to talk about a fast current. From the perspective of Maxwell’s equations there is no difference between a given current density and a current density produced …

From a microscopic perspective $J=\rho v$ holds, but from a macroscopic perspective it can fail. Consider two microscopic charge distributions, one with charge density and no velocity and the other …

in reality current is flow of electrons which flow from low to high This is not correct. Current is defined so that it points in the opposite direction of the electron flow. So current in a resistor …

Current density is a vector, $\vec J$, and it must be a vector based on where it shows up in Maxwell’s equations. Current is $I=\int_A \vec J \cdot d\vec A$, where $A$ is an area and $d\vec A$ is a di …

Remember that current density (or more precisely probability current density) is a vector field. It has a magnitude and direction at each point. Two atoms with the same magnitude current density at a …

In my bank account, withdrawals that are equal and opposite to deposits cancel out. Nevertheless, my account balance is not a vector. Mere cancellation of equal and opposite quantities does not imply …

You cannot analyze an ideal wire connecting the terminals of an ideal battery, it is an irresistible force meeting an immovable object. In the case of an ideal wire connected to a real battery, the in …

The magnetic field is determined by the conventional current, regardless of the sign of the charge carriers. It is important to understand that the conventional current is the current. In metals the c …

But the charges which are moving in the circuit are the electron moving from the negative terminal to the positive terminal This is not generally true. In a typical circuit you may have negative ele …

Normally, after a good and complete answer like the one by @Ben Crowell I would not bother to post my own answer. However, since @Md Ashraful Islam seems skeptical, I thought that it would be worthwhi …

shouldn’t the work done per unit charge, i.e, voltage, increase? The work done by the E field is $W=\int \vec F \cdot d \vec s$. And since $\vec E=\vec F/q$ we have $W/q=\int \vec E \cdot d\vec s$ b …

It is actually easier to start with current density, particularly if you want a definition based on charge density. Current density, $\vec J = \rho \vec v$ is the charge density $\rho$ times the velo …

The relationship is actually between charge density $\rho$, velocity $\vec v$, and current density $\vec j$. The relationship is exceptionally simple: $\vec j=\rho \vec v$ The charge density is the ch …

Is the book mentioning about varying cross section in the same wire, if yes then can you please explain how does current density increase in a smaller diameter wire? Yes. To see this, let’s consider …