Q 12-06-044JEE MainJEE Main 2025 (29 Jan, Shift 1)Easy
A coil of area $A$ and $N$ turns is rotating with angular velocity $\omega$ in a uniform magnetic field $\vec B$ about an axis perpendicular to $\vec B$. Magnetic flux $\varphi$ and induced emf $\varepsilon$ across it, at an instant when $\vec B$ is parallel to the plane of the coil, are
Answer: (C) $\varphi = 0,\ \varepsilon = NAB\omega$
Let $\theta = \omega t$ be the angle between $\vec B$ and the normal to the coil:
$$\varphi = NAB\cos\omega t, \qquad \varepsilon = -\frac{d\varphi}{dt} = NAB\omega\sin\omega t$$
When $\vec B$ lies in the plane of the coil, the normal is perpendicular to $\vec B$ ($\theta = 90^\circ$): the flux is zero and the emf is at its maximum, $NAB\omega$.
Solution by Sreeraj P, M.Sc Physics