Q.Prove theoretically the relation between e.m.f. induced in a coil and rate of change of magnetic flux in electromagnetic induction. A parallel plate air condenser has a capacity of 20 $\mu$F. What will be the new capacity if:

Faraday's law of EMI is derived from the motional emf of a conductor sliding on rails in a magnetic field, then generalised to any changing flux; the capacitor problem applies $C=\varepsilon_0A/d$ and $C=K\varepsilon_0A/d$ directly. The OR alternative places hydrogen's spectral series on the energy-level diagram, then compares incident and threshold frequency for the photoelectric problem.

Part 1 — Relation between induced emf and rate of change of flux:

Consider a conducting rod $PQ$ of length $l$, free to slide with velocity $v$ along two parallel rails, in a region of uniform magnetic field $B$ perpendicular to the plane of the rails. Each free electron in the rod, moving with the rod at velocity $v$, experiences a magnetic force $F = qvB$ (from $\vec F = q\vec v\times\vec B$), which pushes charges along the rod and sets up a potential difference (motional emf) between its ends. The work done per unit charge in moving the charge along the length $l$ of the rod is

$e = \int_0^l (vB)\,dl = Blv \qquad (\text{for }v, B, l\text{ mutually perpendicular}).$

Now, as the rod slides a distance $dx$ in time $dt$ (so $v=dx/dt$), the circuit's enclosed area changes by $dA = l\,dx$, and the magnetic flux through the circuit changes by $d\Phi = B\,dA = Bl\,dx$. So

$e = Blv = Bl\frac{dx}{dt} = \frac{d(Blx)}{dt} = \frac{d\Phi}{dt}.$

Lenz's law fixes the sign: the induced emf always opposes the change producing it, so

$e = -\frac{d\Phi}{dt}.$

For a coil of $N$ turns, each turn contributes equally, so the total induced emf is

$e = -N\frac{d\Phi}{dt},$

which is Faraday's law of electromagnetic induction in its general form — valid for any cause of flux change (relative motion, changing current, changing area, etc.), not just the sliding-rod case used to derive it.

Part 2 — Parallel-plate capacitor: For a parallel-plate capacitor with plate area $A$, plate separation $d$, and (if present) a dielectric of constant $K$ filling the gap, $C = \dfrac{K\varepsilon_0 A}{d}$ (with $K=1$ for air/vacuum). Given $C_0 = 20\ \mu\text{F}$ initially (air-filled):

1. If the separation $d$ is doubled (area and medium unchanged), $C \propto 1/d$, so the new capacitance is half: $C' = 20/2 = 10\ \mu\text{F}$.
2. If instead a dielectric slab of $K=8$ completely fills the original gap ($d$, $A$ unchanged), $C \propto K$, so the new capacitance is $C'' = K\times C_0 = 8\times 20 = 160\ \mu\text{F}$. …