JEE Physics 2025 Question with Solution

Given: A coil of area $A$ and $N$ turns rotates with angular velocity $\omega$ in a uniform magnetic field $B$. The magnetic field is parallel to the plane of the coil at the required instant.

Find: Magnetic flux $\phi$ and induced emf $\varepsilon$ at that instant.

Magnetic flux through the coil is

$$\phi = BA\cos\theta$$

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

When $\vec{B}$ is parallel to the plane of the coil, it is perpendicular to the normal. Hence,

$$\theta = 90^\circ$$

and therefore

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

For a rotating coil, Faraday's law gives

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

If

$$\phi = BA\cos(\omega t)$$

then

$$\varepsilon = -N\frac{d}{dt}\left(BA\cos(\omega t)\right) = NAB\omega\sin(\omega t)$$

At the instant when $\theta = 90^\circ$, the rate of change of flux is maximum, so

$$\varepsilon = NAB\omega$$

Therefore, the magnetic flux is $0$ and the induced emf is $NAB\omega$. Hence, the correct option is C.

Discrepancy noted: The answer key and the solution mark option D, but the worked solution clearly concludes $\phi = 0$ and $\varepsilon = NAB\omega$, which matches option C.

Given: $\vec{B}$ is parallel to the plane of the coil.

Find: The values of $\phi$ and $\varepsilon$.

The important orientation fact is that magnetic flux depends on the component of $\vec{B}$ along the area vector. When the field lies in the plane of the coil, the field has zero component along the normal.

So,

$$\phi = 0$$

However, the coil is rotating, so this zero flux is changing instantaneously at the maximum rate. Therefore the induced emf is maximum:

$$\varepsilon_{\max} = NAB\omega$$

Thus the required pair is $\phi = 0$ and $\varepsilon = NAB\omega$, so the correct option is C.