1
$\begingroup$

$$ F = qE + qv \times B$$

For the Lorentz force relevant to a current carrying wire, that is caused by the motion of a wire w.r.t to an exterior magnetic field $B$, the second term($qv \times B$) on the RHS is focused, while the first term($qE$) is negated.

Only when an electric force caused by an induced $E$ from a time-varying magnetic field is the first term in the RHS is considered.

But what about the force that caused current to flow from the potential difference voltage source? The voltage source can be a battery or capacitor, is $qE$ relevant to it?

Is the electric force within the Lorentz force equation related to the power supply in anyway?

I think of feynman's statement:

The force on an electric charge depends not only on where it is, but also on how fast it is moving. Every point in space is characterized by two vector quantities which determine the force on any charge. First, there is the electric force, which gives a force component independent of the motion of the charge. We describe it by the electric field, E.

$\endgroup$
2
$\begingroup$

The Lorentz force on an electron in the wire is $$ \vec{F} = -e\left(\vec{E} + \vec{v}\times\vec{B}\right), $$ but for small velocities (compared to the speed of light), the magnetic term is negligible and thus the force is $$ \vec{F} = -e\vec{E}. $$ This is certainly valid since the average drift speed of electrons in conductors is typically quite small, due to high frequency of collisions they experience in the metal. The current density $\vec{J}$ in the wire is proportional to force on a given electron. In fact, $$\vec{J} \propto \vec{f}$$ where $\vec{f}$ is the force per unit charge --- in this case, this is simply the electric field. So $$\vec{J} \propto \vec{E}.$$ We call the proportionality constant to satisfy this equation $\sigma$, so that $\vec{J} = \sigma \vec{E}$. This is known as Ohm's law. It turns out that $\sigma$ is the conductivity.

You may have seen Ohm's law in a different form. Consider the resistivity to be $\rho = \frac{1}{\sigma}$. Also, by definition, $J = I/A$ for cross-sectional area $A$, and the electric field is given by $E = \frac{V}{L}$ for some length $L$. Plugging it all in, we get $$ J = \sigma E $$ $$ \frac{I}{A} = \left(\frac{1}{\rho}\right)\frac{V}{L}. $$ Rearranging, we find $$ V = I \left(\frac{\rho L}{A}\right) $$ $$ V = IR. $$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.