Q.(a) Obtain the expression for the induced emf by changing relative orientation of the coil with the magnetic field (Graph not necessary). OR

(a) Differentiating the flux $\Phi=NBA\cos\omega t$ through a rotating coil gives a sinusoidal emf $e=e_0\sin\omega t$; (b) similar-triangle geometry for a spherical mirror gives $1/v+1/u=1/f$ and magnification $m=-v/u$. Both alternatives answered below.

(a) Induced emf from a rotating coil

Consider a rectangular coil of $N$ turns and area $A$, rotating with constant angular velocity $\omega$ about an axis perpendicular to a uniform magnetic field $B$. Let $\theta=\omega t$ be the angle between the normal to the coil and $B$ at time $t$ (starting with the coil's plane along $B$, i.e. $\theta=0$ at $t=0$).

The flux linked with the coil at time $t$:

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

By Faraday's law, the induced emf is the negative rate of change of flux:

$e=-\dfrac{d\Phi}{dt}=-NBA\dfrac{d}{dt}(\cos\omega t)=NBA\omega\sin\omega t$

Writing $e_0=NBA\omega$ (the peak emf):

$e=e_0\sin\omega t$

So, as the relative orientation between the coil and the field changes with rotation, the induced emf varies sinusoidally with time — this is the principle of the AC generator.

(b) Mirror equation and lateral magnification

Setup. Consider a concave mirror forming a real image. Let the object $O$ be at distance $u$ from the pole $P$, and its real, inverted image $I$ form at distance $v$, both measured along the principal axis.

Derivation (using similar triangles). Let the object height be $h$ (object $AB$, with $B$ on the axis) and image height $h'$ (image $A'B'$). Triangles $ABP$ and $A'B'P$ (formed by the ray through the pole, which reflects making equal angles with the axis) are similar:

$\dfrac{A'B'}{AB}=\dfrac{PB'}{PB}=\dfrac{v}{u}$

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