$$ \mathcal E=-\frac{\text d \Phi_m}{\text dt}=-\frac{\text d}{\text dt}\int_S\vec B\cdot \text d\vec A $$
The EMF opposes the change
Consider a conducting disk of radius $a$ oriented $\perp$ to a field $B$ rotating at $\omega$
$$ \begin{aligned} \text{velocity}\quad \vec v&=r\omega \, \hat \theta \\ \text{Force per unit charge}&=\vec v\times\vec B = vB\, \hat r \\ \text d \mathcal E&=(\vec v \times \vec B)\cdot \text d\vec l \\ &=r\omega B\, \text{d} r \\ \text{Hence} \quad \mathcal E_\text{tot}&=\int^a_0r\omega B \, \text dr \\ &=\frac{1}{2}\omega B a^2 \end{aligned} $$
$$ \mathcal E=-\frac{\text d\Phi_m}{\text d t} \quad ;\quad \Phi_m=\int_S\vec B\cdot\text d\vec A \quad ; \quad \mathcal E=-\oint_C\vec E\cdot\text d\vec l $$
$$ \oint_C\vec E\cdot \text d \vec l=- \frac{\text d}{\text dt}\int_S \vec B\cdot \text d\vec A $$
💡 a changing $B$-field creates an $E$-field
Consider a metal plate pulled through a $B$-field
$$ \begin{aligned} \oint_C\vec B\cdot \text d \vec l &=\mu_0\int_S\left ( \vec j+\epsilon_0\frac{\partial \vec E}{\partial t} \right )\cdot \text d \vec A \\ \oint \vec E\cdot\text d\vec l&=-\frac{\text d}{\text dt}\int_S\vec B\cdot\text d\vec A
\end{aligned} $$