Question

What do you understand by electromagnetic induction? State the Faraday's laws of electromagnetic induction.
OR
Obtain the formula for the capacitance of a parallel plate capacitor when a dielectric medium is partially filled between its plates.

Hint:

Option 1: the induced emf equals the rate of change of magnetic flux, \(e = -N\,d\Phi/dt\), and exists only while the flux changes. Option 2: add the potential drops across the air gap \((d-t)\) and the dielectric slab of thickness \(t\), then use \(C = Q/V\).

Answer 1

Option 1: Electromagnetic Induction and Faraday's Laws

Step 1: Meaning of electromagnetic induction. Electromagnetic induction is the phenomenon in which an emf (called induced emf), and hence an electric current (induced current, if the circuit is closed), is produced in a coil whenever the magnetic flux linked with the coil changes. It was discovered by Michael Faraday and Joseph Henry. The induced current flows only as long as the flux keeps changing.

Step 2: Magnetic flux. The magnetic flux linked with a coil of area \(A\) placed in a magnetic field \(B\) is \[\Phi = BA\cos\theta\] where \(\theta\) is the angle between the field \(\vec{B}\) and the area vector. The flux changes if \(B\), \(A\) or \(\theta\) changes, or if there is relative motion between the coil and a magnet.

Step 3: Faraday's first law. Whenever the magnetic flux linked with a closed circuit changes, an induced emf is set up in the circuit. This induced emf lasts only as long as the flux continues to change; the moment the flux becomes steady, the emf disappears.

Step 4: Faraday's second law. The magnitude of the induced emf is equal to the rate of change of magnetic flux linked with the circuit. For a coil of \(N\) turns, \[e = -N\frac{d\Phi}{dt}\] The negative sign represents Lenz's law: the induced emf opposes the very change of flux that produces it, which follows from the law of conservation of energy.

Step 5: Induced current. If the circuit has total resistance \(R\), the induced current is \[I = \frac{e}{R} = -\frac{N}{R}\frac{d\Phi}{dt}\] Thus a faster change of flux or a larger number of turns gives a larger induced emf and current.
\[\boxed{e = -N\dfrac{d\Phi}{dt}}\]

Option 2: Capacitance of a Parallel Plate Capacitor Partially Filled with Dielectric

Step 1: The arrangement. Consider a parallel plate capacitor whose plates each have area \(A\) and are separated by a distance \(d\). Let the plates carry charges \(+Q\) and \(-Q\), so the surface charge density is \(\sigma = Q/A\). A dielectric slab of dielectric constant \(K\) and thickness \(t\) (with \(t < d\)) is introduced between the plates; the remaining thickness \((d - t)\) contains air or vacuum.

Step 2: Electric field in the air region. In the air-filled part between the plates, the electric field due to the charged plates is \[E_0 = \frac{\sigma}{\varepsilon_0} = \frac{Q}{\varepsilon_0 A}\]
Step 3: Electric field inside the dielectric. Inside a dielectric of constant \(K\), the field is reduced by the factor \(K\): \[E = \frac{E_0}{K} = \frac{Q}{K\varepsilon_0 A}\]
Step 4: Total potential difference. The potential difference between the plates equals field times distance, added over the two regions: \[V = E_0(d - t) + E\,t = E_0(d-t) + \frac{E_0}{K}\,t = E_0\left[(d-t) + \frac{t}{K}\right]\] Substituting \(E_0 = Q/(\varepsilon_0 A)\): \[V = \frac{Q}{\varepsilon_0 A}\left[(d-t) + \frac{t}{K}\right]\]
Step 5: Capacitance. By definition \(C = Q/V\), therefore \[C = \frac{Q}{V} = \frac{\varepsilon_0 A}{(d-t) + \dfrac{t}{K}}\] Consistency check: If \(t = 0\) (no dielectric), \(C = \varepsilon_0 A/d\); if \(t = d\) (dielectric fills the whole gap), \(C = K\varepsilon_0 A/d\). Both are the standard results, so the formula is correct.
\[\boxed{C = \dfrac{\varepsilon_0 A}{(d-t) + \dfrac{t}{K}}}\]