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

Magnetic flux $$\varphi$$ through a coil with $$N$$ turns and area $$A$$ in a uniform magnetic field $$\overrightarrow{B}$$ is given by the formula $$\varphi = NAB\cos\theta\quad\text{-(1)}$$ where $$\theta$$ is the angle between the normal to the coil and the magnetic field.

Since the coil rotates with angular velocity $$\omega$$ about an axis perpendicular to $$\overrightarrow{B}$$, the angle varies as $$\theta = \omega t\,.$$ Substituting into $$(1)$$ gives $$\varphi = NAB\cos(\omega t)\quad\text{-(2)}$$

The induced emf $$\varepsilon$$ is given by Faraday’s law: $$\varepsilon = -\frac{d\varphi}{dt}\,. $$ Using $$(2)$$, we find $$\varepsilon = -\frac{d}{dt}\bigl(NAB\cos(\omega t)\bigr) = NAB\,\omega\sin(\omega t)\quad\text{-(3)}$$

When $$\overrightarrow{B}$$ is parallel to the plane of the coil, its normal is perpendicular to $$\overrightarrow{B}$$ so that $$\theta = 90^\circ = \frac{\pi}{2}\,.$$ Then from $$(2)$$, $$\cos\bigl(\tfrac{\pi}{2}\bigr)=0\implies \varphi = 0\,,$$ and from $$(3)$$, $$\sin\bigl(\tfrac{\pi}{2}\bigr)=1\implies \varepsilon = NAB\,\omega\,.$$ Hence at that instant $$\varphi = 0\,,\quad \varepsilon = NAB\,\omega\,,$$ which matches Option C.