Notes on E&M

Definitions

$\vec E = \frac{\vec F}{q}$
$\phi_E = \int \vec E \cdot d\vec A$
$\phi_B = \int \vec B \cdot d\vec l$
$\varepsilon = \oint_{L} \frac{\vec F_{on \, charge \, q}}{q} \cdot d\vec l$
$\Delta V = -\int^B_A \vec E \cdot d\vec l$    if $\oint_L\vec E \cdot d\vec l= 0$      $\vec E$ created by charges

Maxwell’s Equations

Gauss’s Law

$\oint \vec E \cdot d \vec A = \frac{q_{enclosed}}{\varepsilon_0}$     → charges make $\vec E$, which is path-independent
$\oint \vec B \cdot d \vec A = 0$    → there is no magnetic monopole

Ampere’s Law

$\oint \vec B \cdot d \vec l = \mu_0I + \mu_0 \varepsilon_0 \frac{d\phi_E}{dt}$    → current and changing electric fulx make $\vec B$

Faraday’s Law

$\varepsilon=-\frac{d\phi_B}{dt}$    → changing magnetic flux makes emf

EMF

$\varepsilon = \oint_{Loop} \frac{\vec F_{on \, charge \, q}}{q} \cdot d\vec l$
If $\vec F_{on \, charge \, q}$ comes from:

1. $\vec B$, then
   it is a motional emf

   e.g.
   A rectangular conductor with length L is moving in a uniform magnetic field, then charges on the conductor would separate by magnetic for and creates a electric field inside the conductor. The charges would keep separating until $|F_E|=|F_B|$
   $F_E=F_B$
   $qE=qvB$
   $\therefore E=vB$
   So, by $\Delta V = -\int^B_A \vec E \cdot d\vec l$,
   Get $\Delta V =vBL$

2. $\vec E$, then
   it is a tranformer emf
   In which $\oint_L\vec E \cdot d\vec l \neq 0$, so by the Faraday’s Law,
   $\oint_L\vec E \cdot d\vec l = -\frac{d}{dt}(\int\vec B \cdot d\vec A)$ changing $\vec B$ makes $\vec E$ (i.e. changing $\vec B$ induces I).
   $\vec E$ here depends on path

Yet it seems that these two types of emf are different phenomena, both of them could be concluded by Faraday’s Law (i.e. changing magnetic flux)

Why $\vec E$ created by charges guarantees $\oint_L\vec E \cdot d\vec l= 0$

path A to B: $\vec E \parallel d\vec l$   so, we have $\int^B_A kq/r^2dl$ on path A to B
with the same logic, we have $\int^B_A kq/r^2(-dl)$ on path C to D
On path B to C and path D to A, since $\vec E \bot d\vec l$ , integrations on those two path would be zero.
So, $\oint_L\vec E \cdot d\vec l= \int^B_A kq/r^2dl + \int^B_A kq/r^2(-dl) = 0$

By contrast, for $\vec E$ created by changing $\vec B$,
since $\vec E \parallel d\vec l$,
$\oint_L\vec E \cdot d\vec l \neq 0$