A coil of area A and N turns is rotating with angular velocity $\mathrm{\omega }$ in a uniform magnetic field $\overrightarrow{\mathrm{B}}$ about an axis perpendicular to $\overrightarrow{\mathrm{B}}$. Magnetic flux $\mathrm{\phi }$ and induced emf $\mathrm{\epsilon }$ across it, at an instant when $\overrightarrow{\mathrm{B}}$ is parallel to the plane of coil, are :

Magnetic flux through the coil is given by:

$$\phi = N B A \cos\theta$$

Since the coil is rotating with angular velocity $\omega$, the angle at time $t$ is:

$$\theta = \omega t$$

Thus, the flux varies as:

$$\phi = N B A \cos(\omega t)$$

At the instant when the magnetic field is parallel to the plane of the coil, the normal to the coil is perpendicular to the field. This means:

$$\theta = 90^\circ \quad \Rightarrow \quad \cos(90^\circ) = 0$$

 $$\phi = N B A \times 0 = 0$$

According to Faraday's Law, the induced EMF is:

$$\varepsilon = - \frac{d\phi}{dt}$$

Differentiating $\phi = N B A \cos(\omega t)$:

$$\frac{d\phi}{dt} = - N B A \omega \sin(\omega t)$$

 $$\varepsilon = N B A \omega \sin(\omega t)$$

At the instant when $B$ is parallel to the plane of the coil, we already determined that $\theta = 90^\circ$ (i.e., $\omega t = 90^\circ$ or $\pi/2$).

Since $\sin 90^\circ = 1$, we get:

$$\varepsilon = N B A \omega$$