From Faraday’s Law
$$ V_{\text{emf}} = -\frac{d\Phi}{dt} = \oint_C \mathbf{E} \cdot d\mathbf{l} $$
Motional EMF using magnetic force $F_m = qu \times B$
$$ V_{\text{emf}} = \oint_C \frac{\mathbf{F}_m}{q} \cdot d\mathbf{l} = \oint_C (\mathbf{u} \times \mathbf{B}) \cdot d\mathbf{l}
$$
Rotating bar in a magnetic field:
$$ V_{\text{emf}} = \int_{\text{Bar}} (\mathbf{u} \times \mathbf{B}) \cdot d\mathbf{l} = \int_0^l (r \omega \hat{\phi} \times B_0 \hat{z}) \cdot dr \hat{r} = \frac{\omega B_0 l^2}{2} $$
Total EMF:
$$ V_{\text{emf}} = -\frac{d\Phi}{dt} \newline = -\iint_S \frac{\partial \mathbf{B}}{\partial t} \cdot d\mathbf{s} + \oint_C (\mathbf{u} \times \mathbf{B}) \cdot d\mathbf{l} \newline = V_{\text{emf}}^{\text{tr}} + V_{\text{emf}}^{\text{m}} $$
Moving loop near time-varying current
$$ V_{\text{emf}} = V_{\text{emf}}^{\text{tr}} + V_{\text{emf}}^{\text{m}} \newline V_{\text{emf}}^{\text{tr}} = -\int_C^b \frac{\partial}{\partial t} \left( \frac{\mu_0 i(t)}{2 \pi r} \right) \cdot dr \hat{\phi} = -\frac{\mu_0 b}{2\pi} \ln\left(\frac{c+a}{c}\right) \frac{d i(t)}{dt} \newline V_{\text{emf}}^{\text{m}} = \frac{\mu_0 \omega I_0 b}{2 \pi} \ln \left( \frac{c+a}{c} \right) \sin \omega t
$$
Transformer equation (without motional EMF)
$$ V_2 = -N_2 \frac{d\Phi}{dt} $$
Ideal transformer relationships
$$ \frac{V_2}{V_1} = \frac{N_2}{N_1}, \quad \frac{I_2}{I_1} = \frac{N_1}{N_2}, \quad Z_{\text{in}} = \left( \frac{N_1}{N_2} \right)^2 Z_L $$
Generator EMF
$$ V_{\text{emf}} = -\frac{d\Phi}{dt} = \oint (\mathbf{u} \times \mathbf{B}) \cdot d\mathbf{l} = A \omega B_0 \sin \omega t $$
Ampere’s Law with displacement current
$$ \nabla \times \mathbf{H} = \mathbf{J} + \frac{\partial \mathbf{D}}{\partial t} $$
Displacement current density
$$ \mathbf{J}_d = \frac{\partial \mathbf{D}}{\partial t} $$
Charge density $\rho _v = \nabla \cdot D$
$$ \nabla \cdot J = \frac{\delta \rho_v}{\delta t} $$
Integral form with conduction and displacement currents
$$ \oint_C \mathbf{H} \cdot d\mathbf{l} = \int_S \mathbf{J} \cdot d\mathbf{s} + \int_S \frac{\partial \mathbf{D}}{\partial t} \cdot d\mathbf{s} = I_c + I_d = I_{\text{enc}} $$
Electric field in a lossy capacitor
$$ \mathbf{E}(t) = \frac{v_s(t)}{d} \hat{z} = \frac{V_0 \cos \omega t}{d} \hat{z} $$