The [[EMF]] around a closed path is equal to the negative of the time rate of change of the magnetic [[Flux]] enclosed by the path.
$$\mathcal{E} = -\frac{d \Phi_B}{dt}$$
Accordingly, a [[Solenoid]] with $N$ total turns will have an EMF $N$ times as large as that induced by just one of its loops:
$$\mathcal{E} = -N\frac{d \Phi_B}{dt}$$
Of a rotating loop with $N$ total turns and angular velocity $\omega$:
$$\mathcal{E} = -N\frac{d \Phi_{B}}{dt} = NBA \omega sin(\omega t)$$
If the path is stationary:
$$\oint \vec{E} \cdot d \vec{l} = -\frac{d \Phi_B}{dt}$$
Inside a [[Solenoid]]:
$$\oint \vec{E} \cdot d \vec{l} = -\frac{dB}{dt}A = -\mu_{0}n \frac{dI}{dt} A$$