Chapter 6: Electromagnetic Induction Long Questions | physics Class 12 CBSE

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Chapter 6: Electromagnetic Induction Long Questions | physics Class 12 CBSE | ByteNotes.in

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Question 1

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Describe the three experiments of Faraday that established the phenomenon of electromagnetic induction. What common conclusion can be drawn from all three?

Experiment 6.1 (Magnet and coil): Moving the north pole of a bar magnet toward a coil connected to a galvanometer caused a deflection indicating induced current; the deflection reversed when the south pole was used or when the magnet was pulled away; the deflection was proportional to the speed of the magnet.
Experiment 6.2 (Two coils): The bar magnet was replaced by a second coil carrying steady current from a battery; moving this current-carrying coil toward or away from the first coil induced current in the first coil, showing that a current-carrying coil behaves like a magnet.
Experiment 6.3 (Stationary coils with switching current): Both coils were held stationary; pressing the tapping key caused a momentary deflection as current in C2 changed from zero to maximum; no deflection while key was held pressed (steady current); deflection reversed when key was released — proving relative motion is not essential.
In all three experiments, the galvanometer deflected only when the magnetic flux through the coil was changing — whether due to motion of a magnet, motion of a current-carrying coil, or switching on/off current in a stationary coil.
The common conclusion is: an emf is induced in a coil whenever the magnetic flux through it changes with time; the magnitude of the induced emf depends on the rate of change of flux, as summarised by Faraday's law: ε = −N(dΦB/dt).

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Question 2

Long Form

Derive the expression for motional emf using (a) Faraday's law and (b) the Lorentz force. Show that both approaches give the same result.

Setup: Consider a rectangular conductor PQRS in a uniform magnetic field B perpendicular to its plane. The arm PQ of length l is free to move with velocity v toward the left, with RQ = x at any instant.
Faraday's law approach: The magnetic flux enclosed by the loop is ΦB = Blx. Since x decreases as PQ moves, dΦB/dt = Bl(dx/dt) = −Blv. By Faraday's law, ε = −dΦB/dt = Blv.
Lorentz force approach: Any charge q in the moving rod PQ moves with velocity v in field B; the Lorentz force F = qvB acts on it toward Q. All charges experience the same force irrespective of position in the rod.
Work done in moving charge q from P to Q over length l is W = qvBl. Since emf = work done per unit charge, ε = W/q = Blv.
Both approaches yield ε = Blv, confirming that motional emf is the work done by the Lorentz force per unit charge, consistent with Faraday's law when the conductor moves in a time-independent magnetic field.

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Question 3

Long Form

Explain Lenz's law with a suitable example. Prove that Lenz's law is a consequence of the law of conservation of energy.

Lenz's law: The polarity of the induced emf is such that it tends to produce a current which opposes the change in magnetic flux that produced it. The negative sign in ε = −dΦB/dt represents this.
Example: When the north pole of a bar magnet is pushed toward a coil, the flux through the coil increases. The induced current flows counter-clockwise (viewed from the magnet's side) so that the coil's near face becomes a north pole — repelling the incoming magnet and opposing the increase in flux.
When the north pole is withdrawn, the flux decreases; the induced current reverses, making the near face a south pole — attracting the receding magnet and opposing the decrease in flux.
Conservation of energy proof: Suppose the induced current aided the incoming magnet instead of opposing it. The magnet would attract the coil and accelerate indefinitely without any energy input — a perpetual-motion machine, violating conservation of energy.
In reality, the induced current opposes the magnet's motion, so a person must do work to move the magnet. This mechanical work is converted into electrical energy, which is dissipated as Joule heat — fully consistent with conservation of energy.

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Question 4

Long Form

Derive the expression for mutual inductance of two long coaxial solenoids. Show that M12 = M21.

Setup: Two long coaxial solenoids S1 (inner, radius r1, n1 turns/length) and S2 (outer, radius r2, n2 turns/length), each of length l. Total turns: N1 = n1l and N2 = n2l.
Finding M12: Current I2 in S2 produces a uniform field B2 = μ0n2I2 inside S2 (and hence inside S1). Flux linkage with S1 is N1Φ1 = (n1l)(πr1²)(μ0n2I2) = μ0n1n2πr1²lI2. By definition N1Φ1 = M12I2, so M12 = μ0n1n2πr1²l.
Finding M21: Current I1 in S1 produces a field B1 = μ0n1I1 confined inside S1. The flux through S2 linked only via S1's cross-section gives N2Φ2 = (n2l)(πr1²)(μ0n1I1) = μ0n1n2πr1²lI1. By definition N2Φ2 = M21I1, so M21 = μ0n1n2πr1²l.
Comparing: M12 = M21 = M = μ0n1n2πr1²l. With a core of relative permeability μr, M = μrμ0n1n2πr1²l.
The equality M12 = M21 is a general result; it is especially useful when calculating one of them is geometrically difficult, as the easier one can be used as the value of M for the pair.

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Question 5

Long Form

Derive the expression for the instantaneous emf produced by an AC generator. Describe its variation with time.

Setup: A coil of N turns and area A rotates with constant angular speed ω in a uniform magnetic field B. At t = 0, the coil is perpendicular to B (θ = 0). At time t, the angle between A and B is θ = ωt.
Flux: The flux through the coil at time t is ΦB = BA cosθ = BA cos(ωt).
Applying Faraday's law: ε = −N(dΦB/dt) = −N·BA·d(cos ωt)/dt = NBAω sin(ωt) = ε0 sin(ωt), where ε0 = NBAω is the peak (maximum) emf.
The emf is zero at t = 0 (coil perpendicular to B, flux is maximum but rate of change is zero), rises to maximum ε0 at ωt = 90° (coil parallel to B), falls back through zero at ωt = 180°, reaches minimum −ε0 at ωt = 270°, and returns to zero at ωt = 360°.
Since the emf varies sinusoidally between +ε0 and −ε0, the current direction alternates periodically — hence the name alternating current (AC). The frequency of the AC equals the frequency of rotation v: ε = ε0 sin(2πvt).

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Question 6

Long Form

Compare the self-inductance of a solenoid with the capacitance of a parallel plate capacitor in terms of their dependence on geometry and material properties.

Self-inductance of a solenoid depends on geometry: L = μ0n²Al, where n is turns per unit length, A is cross-sectional area, and l is length. Capacitance of a parallel plate capacitor depends on geometry: C = ε0A/d, where A is plate area and d is plate separation.
Both depend on a material property of the medium: L depends on permeability (μ = μrμ0 for a magnetic core), and C depends on permittivity (ε = Kε0 for a dielectric). Inserting a high-μr material into a solenoid increases L; inserting a dielectric into a capacitor increases C.
Energy storage analogy: Energy stored in an inductor is UL = (1/2)LI² (magnetic energy), and energy stored in a capacitor is UC = (1/2)CV² (electrostatic energy); in both cases energy is stored in the field.
Energy density analogy: Magnetic energy density uB = B²/2μ0 is analogous to electrostatic energy density uE = (1/2)ε0E²; both are proportional to the square of the respective field strength.
Key difference: Inductance resists change in current (electromagnetic inertia), while capacitance stores charge and resists change in voltage; L is the analogue of mass (inertia) and C has no direct mechanical analogue in this sense.

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Question 7

Long Form

Analyse what happens in a circuit when a current-carrying inductor is connected to a resistor. How does self-inductance play the role of electrical inertia?

When a switch is closed in a circuit containing an inductor (L) and resistor (R), the current does not jump instantly to its final value but rises gradually; the self-induced emf (back emf) ε = −L(dI/dt) opposes the increase in current.
To establish a current I against this back emf, work must be done. The rate of work done is dW/dt = εI = LI(dI/dt). Integrating gives the total energy stored: W = (1/2)LI².
When the switch is opened, the current tends to fall suddenly to zero; the inductor now generates a large emf to oppose this decrease — similar to how a massive body resists deceleration.
This analogy is precise: L corresponds to mass m, current I corresponds to velocity v, and energy (1/2)LI² corresponds to kinetic energy (1/2)mv²; self-inductance is therefore the electromagnetic analogue of mass (inertia).
The self-induced emf is also called back emf because it always acts in the direction opposing the change in current — opposing growth when current increases and opposing decay when current decreases — making L the measure of electromagnetic inertia of the coil.

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Question 8

Long Form

Describe the principle of operation of an AC generator. Name the types of commercial AC generators and state their energy sources.

Principle: An AC generator works on Faraday's law of electromagnetic induction. When a coil rotates in a uniform magnetic field, the magnetic flux through it changes continuously, inducing an alternating emf ε = NBAω sin(ωt) = ε0 sin(ωt).
Key components: The armature (coil of N turns, area A) is mounted on a rotor shaft and rotated at angular speed ω in a magnetic field B. The ends of the coil connect to the external circuit via slip rings and carbon brushes.
Hydro-electric generators: Mechanical energy is provided by water falling from a height (from dams); the potential energy of water drives the rotation of the armature.
Thermal generators: Water is heated using coal or other fuels to produce high-pressure steam, which drives a turbine connected to the armature.
Nuclear power generators: Nuclear fuel (such as uranium) is used instead of coal to heat water and produce steam for rotating the armature. Modern generators produce up to 500 MW; in India, the rotation frequency is 50 Hz.

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Question 9

Long Form

Derive an expression for the magnetic energy stored in a solenoid and hence obtain the expression for magnetic energy density. Compare this with electrostatic energy density.

Energy stored in inductor: To establish current I against back emf ε = −L(dI/dt), the work done at any instant is dW = LI dI. Integrating from 0 to I gives the total energy W = (1/2)LI².
Expressing in terms of B: For a solenoid, B = μ0nI so I = B/(μ0n), and L = μ0n²Al. Substituting: W = (1/2)(μ0n²Al)(B/μ0n)² = B²Al/(2μ0).
Magnetic energy density: Dividing the energy W = B²Al/(2μ0) by the volume V = Al of the solenoid gives uB = B²/(2μ0).
Comparison with electrostatic energy density: The energy stored per unit volume in a parallel plate capacitor is uE = (1/2)ε0E². Both uB and uE are proportional to the square of the respective field strength.
Both expressions are general and valid for any region of space containing a magnetic or electric field respectively, not just for solenoids or capacitors. In both cases, energy resides in the field itself.

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Question 10

Long Form

Explain how the direction of induced current in different shaped loops (rectangular, triangular, irregular) moving into or out of a magnetic field region is determined using Lenz's law.

Lenz's law principle: The induced current always flows in a direction that opposes the change in magnetic flux through the loop. If flux is increasing, the induced current creates a field opposing the external field (into the page if B is out of page); if flux is decreasing, the induced current creates a field in the same direction as the external field.
Rectangular loop entering field (Example 6.4(i)): As the loop moves into the field, flux increases; the induced current flows in a direction (e.g., bcdab) that creates a magnetic field opposing the increasing flux — i.e., opposing the external field direction through the loop.
Triangular loop moving out (Example 6.4(ii)): As the triangular loop exits the field, flux decreases; the induced current flows along bacb to create a field in the same direction as the external field, opposing the decrease in flux.
Irregular loop moving out (Example 6.4(iii)): Similarly, as flux decreases, the induced current flows along cdabc to oppose the decrease — regardless of the shape, the direction is determined by Lenz's law.
Key observation from Example 6.4: No induced current flows when a loop is completely inside or completely outside the field region, because the flux is not changing in either case; induction occurs only during entry or exit.

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Chapter 6: Electromagnetic Induction | Long Questions

Describe the three experiments of Faraday that established the phenomenon of electromagnetic induction. What common conclusion can be drawn from all three?

Derive the expression for motional emf using (a) Faraday's law and (b) the Lorentz force. Show that both approaches give the same result.

Explain Lenz's law with a suitable example. Prove that Lenz's law is a consequence of the law of conservation of energy.

Derive the expression for mutual inductance of two long coaxial solenoids. Show that M12 = M21.

Derive the expression for the instantaneous emf produced by an AC generator. Describe its variation with time.

Compare the self-inductance of a solenoid with the capacitance of a parallel plate capacitor in terms of their dependence on geometry and material properties.

Analyse what happens in a circuit when a current-carrying inductor is connected to a resistor. How does self-inductance play the role of electrical inertia?

Describe the principle of operation of an AC generator. Name the types of commercial AC generators and state their energy sources.

Derive an expression for the magnetic energy stored in a solenoid and hence obtain the expression for magnetic energy density. Compare this with electrostatic energy density.

Explain how the direction of induced current in different shaped loops (rectangular, triangular, irregular) moving into or out of a magnetic field region is determined using Lenz's law.

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