What force moves electrons through a conductor that is rotating in a magnetic field [closed]

Question

Is it the magnetic force ( F= qv +B ) or the electromagnetic force (F= q(E+vxB) that acts on the electrons of a conductor that is moving in a magnetic field? Thanks.

Answer 1

Let’s focus on force. Electric current in a wire is cause by net motion of electrons in the wire. The electrons move (in net, overall) due to the fundamental force known as electromagnetism, specifically the electrostatic attraction or repulsion between charged particles. That is the basis for electric fields and electric potential; those are just ways of characterizing and analyzing this attractive and/or repulsive force between charged particles/objects. In the simple situation you describe in your question, magnetism is not part of it. (Although in your answer to your own question, which was never really a question but a point to make, there is a situation where magnetism generates the current)

Coulomb’s law:

|Force| = k_e (|q1 q2|) / r^2

is the magnitude, and the direction is along a line between them, attractive if q1 and q2 have opposite sign. The constant k_e is called Coulomb's constant and is equal to 1/4πε0 , where ε0 is the electric constant; k_e = 8.988×109 N⋅m2⋅C−2.

From wikipedia “Electromagnetism is the force that acts between electrically charged particles. This phenomenon includes the electrostatic force acting between charged particles at rest, and the combined effect of electric and magnetic forces acting between charged particles moving relative to each other.”

Current flows parallel to the electric field, so even though net charge is moving, it is the former of the above two (electrical not magnetic) that does it. You need orthogonal (perpendicular) net motion of charges for magnetism to be involved.

Answer 2

The force doing work on electrons in a conductor is always an electric force $\vec{F}=q\vec{E}$. Magnetic forces can do no work since $\vec{F}=q(\vec{v}\times\vec{B})$ is always perpendicular to the velocity $\vec{v}$.

One thing to remember is that EMF is calculated at one particular instant, and is NOT the same as the work done on a physical charge as it moves through the conductor. To calculate the EMF, you take a snapshot of the situation, compute the integral $\varepsilon = \oint{\vec{f}\cdot d\vec{r}}$, where $\vec{f}=\vec{F}/q$ is the "force per unit charge" on the electron at that instant in time. Don't think of this as a work integral, because to calculate the work, you need information about the actual motion of the electrons and the forces that act on them as they move. You don't use that information to compute the EMF. In practice, for most situations you can think of $\vec{f}$ as and electric field $\vec{E}$, which is comprised of two contributions:

(1) a conservative part $\vec{E}_c$ which satisfies $\vec{\nabla}\times \vec{E}_c=0$, or equivalently (by Stokes theorem) $\oint{\vec{E}_{c}\cdot d\vec{r}}=0$.

(2) a nonconservative part $\vec{E}_{nc}$ which has nonzero curl: $\vec{\nabla}\times \vec{E}_{nc} \ne 0$. The nonconservative electric field in a generator arises due to a time-dependent magnetic field (in the rest frame of the conducting electron.) It has $\oint{\vec{E}_{nc}\cdot d\vec{r}}\ne 0$ and leads to an EMF.