1. Magnetic flux through a rotating coil

For a coil of area $A$ in a magnetic field $\vec B$, the magnetic flux through one turn is

$\phi = BA\cos\theta$

where $\theta$ is the angle between $\vec B$ and the normal to the plane of the coil.

For a coil of $N$ turns, total flux linkage is

$\Phi = NBA\cos\theta$

2. Given instant: $\vec B$ is parallel to the plane of the coil

If $\vec B$ is parallel to the plane of the coil, then $\vec B$ is perpendicular to the normal of the coil. So,

$\theta = 90^\circ$

Hence flux through one turn is

$\phi = BA\cos 90^\circ = 0$

Therefore, magnetic flux is zero at that instant.

3. Induced emf

Induced emf is given by Faraday's law:

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

If the coil rotates with angular velocity $\omega$, then

$\Phi = NBA\cos(\omega t)$

So,

$\varepsilon = -\frac{d}{dt}\left(NBA\cos \omega t\right)$

$\varepsilon = NBA\omega \sin \omega t$

At the instant when $\vec B$ is parallel to the plane of the coil, we have

$\theta = 90^\circ \Rightarrow \sin\theta = 1$

Thus,

$\varepsilon = NBA\omega$

4. Match with options

• Flux $=0$
• Induced emf $= NAB\omega$

So the correct option is:

$\boxed{\text{D}}$

5. Comparison with stored answer

Stored correct answer: $\text{D}$

My derived answer also gives $\text{D}$, so they agree.