PX157 - D2 - induced electric fields

\vec{F} = q \vec{v} \times \vec{B} = \vec{0}

ϵ = - \frac{d ϕ_{B}}{d t}

eg: solenoid with current ( $I_{s} (t)$ ), cross-sectional area ( $A$ ), with $n$ turns pre unit length, and magnetic field:

\vec{B} = {\begin{cases} \vec{0} & o u t s i d e \\ μ_{0} n I_{s} \hat{z} & i n s i d e \end{cases}

ϕ_{B} = \iint_{S} \vec{B} \cdot d \vec{S} = μ_{0} n I_{s} A

ϵ = - μ_{0} \frac{N}{L} A \frac{d I_{s}}{d t}

I_{w} = \frac{ϵ}{R} = \frac{μ_{0} N A}{L R} \frac{d I_{s}}{d t}

\vec{F} = q (\vec{E} + \vec{v} \times \vec{B})

ϵ = \oint_{l o o p} \frac{\vec{F}}{q} \cdot d \vec{l}

since the wire is not moving, $\vec{v} = 0$ , and there is no external magnetic field, the only contribution can come from the electric field:

ϵ = \oint_{l o o p} \vec{E} \cdot d \vec{l}

\oint_{C} \vec{E} \cdot d \vec{l} = - \frac{d ϕ_{B}}{d t}

this poses a problem
remember: in electrostatics, an electric field starts at a positive charge and ends at a negative charge
$\int_{a}^{b} \vec{E} \cdot d \vec{l}$ represents the potential difference between the start and end points
for a closed path, the start and the end points are the same, so the potential difference is zero, ie: $\oint_{C} \vec{E} \cdot d \vec{l} = 0$
this is only valid in time-independent magnetic problems
$\oint_{C} \vec{E} \cdot d \vec{l} \neq 0 :$ if there is a changing magnetic field
$ϕ_{B}$ generates closed loops of electric fields, called the induced electric field
around a loop of induced electric field, since $\vec{F} = q \vec{E}$ , the electric field does net work around a closed loop, and a potential can no longer be defined
the concept of electrostatic potential no longer works
solenoid's power output over the whole solenoid:

P = | ϵ | I = (μ_{0} \frac{N A}{L} \frac{d I_{s}}{d t}) (I_{s} N)

\begin{aligned} B & = μ_{0} \frac{N}{L} I_{s} \\ ⟹ P & = B A \frac{d I_{s}}{d t} N \\ = \frac{L A}{μ_{0}} B \frac{d B}{d t} \\ ∴ P & = \frac{L A}{2 μ_{0}} \frac{d B^{2}}{d t} \\ e n e r g y, U & = \int P d t = L A \frac{B^{2}}{2 μ_{0}} \\ e n e r g y d e n s i t y, u & = \frac{U}{V o l .} = \frac{B^{2}}{2 μ_{0}} \end{aligned}

- for electric field, energy density, $u = \frac{1}{2} ϵ_{0} E^{2}$

comparing with $E = m c^{2} :$
- hydrogen atom:
  $V o l . = 6.2 \times 10^{- 31} m^{3}$
  $B \approx 12.5 T$

U = \frac{B^{2}}{2 μ_{0}} V o l . = m c^{2} ⟹ m \approx 4.3 \times 10^{- 40} k g

conclusion:
- resistor: dissipates energy
- capacitor: stores energy in $\vec{E}$
- coil (solenoid, inductor): stores energy in $\vec{B}$