0
$\begingroup$

Griffiths's Introduction to Electrodynamics states

$$\mathcal E = \oint \mathbf f \cdot d\mathbf l$$

In which

$$\mathbf f = \mathbf f_s + \mathbf E$$

Where Griffiths describes the summation as

the source, $\mathbf f_s$, which is ordinarily confined to one portion of the loop (a battery, say), and an electrostatic force, which serves to smooth out the flow and communicate the influence of the source to distant parts of the circuit

Is $\mathbf E$ the force that prevents charge from clumping up and producing a buildup of charge in a part of the circuit (which would produce a non-steady current)?

Anyway, in magnetostatic situations, I agree with the following:

$$\oint \mathbf f \cdot d\mathbf l = \oint \mathbf f_s \cdot d\mathbf l$$

However, suppose $\frac{\partial \mathbf B}{\partial t} \ne 0$.

This would imply:

$$\mathcal E = \oint \mathbf f \cdot d\mathbf l = \oint (\mathbf f_s + \mathbf E) \cdot d\mathbf l = \oint \mathbf f_s \cdot d\mathbf l + \oint \mathbf E \cdot d\mathbf l$$

Which implies

$$\mathcal E = \oint \mathbf f_s \cdot d\mathbf l - \frac{\partial \mathbf B}{\partial t}$$

I don't know what this means though, if it's significant. What does this mean, if what I did was sensible? In addition, is my mistake that, since the sole purpose of $\mathbf E$ in the equation for $\mathbf f$ was to prevent charge clumping up, that for non-steady currents this $\mathbf E$ term cannot exist and thus in that case $\mathbf f = \mathbf f_s$?

$\endgroup$

Your Answer

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

Browse other questions tagged or ask your own question.