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

ByteNotes

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

Application Question

Medium difficulty • Concept in a practical situation

Medium

Question 1

Applied Concept

A square loop of side 10 cm and resistance 0.5 Ω is placed in the east-west plane. A uniform magnetic field of 0.10 T is set up across the plane in the north-east direction. The magnetic field is decreased to zero in 0.70 s at a steady rate. Find the magnitude of the induced emf and current during this interval.

The angle between B (north-east) and the area vector A (horizontal, eastward) is θ = 45°. Initial flux: ΦB = BA cosθ = 0.10 × (0.1)² × cos45° = 0.10 × 10⁻² / √2 = 10⁻³/√2 Wb. Final flux = 0.
Induced emf: ε = |ΔΦB/Δt| = (10⁻³/√2) / 0.70 = 10⁻³/(√2 × 0.7) ≈ 1.0 mV.
Induced current: I = ε/R = 10⁻³ V / 0.5 Ω = 2 mA. Note: Earth's steady magnetic field does not contribute because it does not change during the experiment, so dΦB/dt = 0 for that component.

ByteNotes

Application Question

Medium difficulty • Concept in a practical situation

Medium

Question 2

Applied Concept

A circular coil of radius 10 cm, 500 turns, and resistance 2 Ω is placed with its plane perpendicular to the horizontal component of Earth's magnetic field (3.0 × 10⁻⁵ T). It is rotated about its vertical diameter through 180° in 0.25 s. Estimate the induced emf and current.

Initial flux: ΦB(initial) = BA cos0° = 3.0 × 10⁻⁵ × π × (0.1)² = 3π × 10⁻⁷ Wb. Final flux after 180° rotation: ΦB(final) = BA cos180° = −3π × 10⁻⁷ Wb. Change in flux = 6π × 10⁻⁷ Wb.
Induced emf: ε = N × ΔΦB/Δt = 500 × 6π × 10⁻⁷ / 0.25 = 500 × 6 × 3.14 × 10⁻⁷ / 0.25 ≈ 3.8 × 10⁻³ V.
Induced current: I = ε/R = 3.8 × 10⁻³ / 2 = 1.9 × 10⁻³ A. These are estimated (average) values; instantaneous values depend on the speed of rotation at each instant.

ByteNotes

Application Question

Medium difficulty • Concept in a practical situation

Medium

Question 3

Applied Concept

A metallic rod of 1 m length is rotated with a frequency of 50 rev/s, with one end hinged at the centre of a circular metallic ring of radius 1 m. A uniform magnetic field of 1 T parallel to the axis exists everywhere. Find the emf between the centre and the ring.

The rod rotates with angular velocity ω = 2πν = 2π × 50 = 100π rad/s. Using the formula for the emf of a rotating rod: ε = (1/2)BωR² = (1/2) × 1.0 × 100π × (1)².
ε = (1/2) × 1.0 × 2π × 50 × 1 = 50π ≈ 157 V. The emf is generated because free electrons in the rotating rod experience a Lorentz force and accumulate at the rim, creating a potential difference.
This can also be confirmed using Method II: treating the rotating rod and a radius of the ring as a loop, the rate of change of area of the sector swept gives the same result ε = (1/2)BωR² = 157 V.

ByteNotes

Application Question

Medium difficulty • Concept in a practical situation

Medium

Question 4

Applied Concept

A long solenoid with 15 turns per cm has a small loop of area 2.0 cm² placed inside the solenoid normal to its axis. If the current through the solenoid changes steadily from 2.0 A to 4.0 A in 0.1 s, find the induced emf in the loop.

The number of turns per unit length: n = 15 turns/cm = 1500 turns/m. The rate of change of current in the solenoid: dI/dt = (4.0 − 2.0)/0.1 = 20 A/s.
The magnetic field inside the solenoid: B = μ0nI. The rate of change of flux through the small loop (area A = 2.0 × 10⁻⁴ m²): dΦB/dt = μ0n(dI/dt)A = 4π × 10⁻⁷ × 1500 × 20 × 2.0 × 10⁻⁴.
Induced emf: ε = dΦB/dt = 4π × 10⁻⁷ × 1500 × 20 × 2.0 × 10⁻⁴ ≈ 7.54 × 10⁻⁶ V ≈ 7.54 μV. The small loop acts as a secondary coil (1 turn) in a mutual inductance arrangement with the solenoid.

ByteNotes

Application Question

Medium difficulty • Concept in a practical situation

Medium

Question 5

Applied Concept

A rectangular wire loop of sides 8 cm and 2 cm with a small cut is moving out of a uniform magnetic field of 0.3 T directed normal to the loop with velocity 1 cm/s normal to the longer side. Calculate the emf developed across the cut and the duration for which it lasts.

When moving normal to the longer side (8 cm), the effective length cutting field lines is the shorter side: l = 2 cm = 0.02 m. Motional emf: ε = Blv = 0.3 × 0.02 × 0.01 = 6 × 10⁻⁵ V.
Duration: The loop exits completely when the shorter side (2 cm) has fully crossed the boundary; time = width/velocity = 0.02 m / 0.01 m/s = 2 s.
The induced voltage of 6 × 10⁻⁵ V lasts for 2 s while the loop is crossing the boundary. Once the loop is completely outside the field region, dΦB/dt = 0 and the emf drops to zero.

ByteNotes

Application Question

Medium difficulty • Concept in a practical situation

Medium

Question 6

Applied Concept

A 1.0 m long metallic rod is rotated with an angular frequency of 400 rad/s about an axis normal to the rod through one end. A uniform magnetic field of 0.5 T parallel to the axis exists. Calculate the emf developed between the centre and the ring.

Using the rotating rod emf formula: ε = (1/2)BωR², where B = 0.5 T, ω = 400 rad/s, and R = 1.0 m.
ε = (1/2) × 0.5 × 400 × (1.0)² = (1/2) × 200 = 100 V.
The emf arises because each element dr of the rod at distance r moves with velocity v = ωr, generating a small emf dε = Bωr dr; integrating from 0 to R gives ε = (1/2)BωR² = 100 V.

ByteNotes

Application Question

Medium difficulty • Concept in a practical situation

Medium

Question 7

Applied Concept

A horizontal wire 10 m long extending east to west falls with a speed of 5.0 m/s at right angles to the horizontal component of Earth's magnetic field (0.30 × 10⁻⁴ T). Find (a) the induced emf and (b) which end of the wire is at higher potential.

The wire falls vertically with velocity v = 5.0 m/s perpendicular to the horizontal component of Earth's field BH = 0.30 × 10⁻⁴ T. The wire (east-west) is perpendicular to both v and BH (which points north). Motional emf: ε = BHlv = 0.30 × 10⁻⁴ × 10 × 5.0 = 1.5 × 10⁻³ V = 1.5 mV.
Using Fleming's right-hand rule (or Lenz's law): The wire moves downward, BH points north. The force on positive charges (qv × B) is directed from west to east, so conventional current flows from west to east inside the wire.
Since current flows from west to east inside the wire, the west end is at a lower potential (analogous to negative terminal) and the east end is at a higher potential (analogous to positive terminal) — the east end is at higher potential.

ByteNotes

Application Question

Easy difficulty • Concept in a practical situation

Easy

Question 8

Applied Concept

Current in a circuit falls from 5.0 A to 0.0 A in 0.1 s. If an average emf of 200 V is induced, estimate the self-inductance of the circuit.

The average rate of change of current: |dI/dt| = ΔI/Δt = (5.0 − 0.0)/0.1 = 50 A/s.
Using the self-inductance formula for back emf: |ε| = L|dI/dt|, so L = |ε|/|dI/dt| = 200/50 = 4 H.
This large self-inductance (4 H) implies that the circuit strongly opposes the change in current; such inductors are used in power systems to limit sudden changes in current and protect circuit components from large transient voltages.

ByteNotes

Application Question

Easy difficulty • Concept in a practical situation

Easy

Question 9

Applied Concept

A pair of adjacent coils has a mutual inductance of 1.5 H. If the current in one coil changes from 0 to 20 A in 0.5 s, what is the change in flux linkage with the other coil?

The induced emf in coil 1 due to changing current in coil 2 is ε1 = −M(dI2/dt), so the average emf is |ε| = M × ΔI/Δt = 1.5 × 20/0.5 = 60 V.
The change in flux linkage ΔN1Φ1 = M × ΔI2 = 1.5 × (20 − 0) = 30 Wb (weber-turns).
This can be verified using the definition M = N1ΔΦ1/ΔI2: ΔN1Φ1 = M × ΔI2 = 1.5 × 20 = 30 Wb-turns. The change in flux linkage of 30 Wb is the total flux linked with all turns of coil 1.

ByteNotes

Application Question

Easy difficulty • Concept in a practical situation

Easy

Question 10

Applied Concept

Kamla pedals a stationary bicycle attached to a 100-turn coil of area 0.10 m². The coil rotates at half a revolution per second in a magnetic field of 0.01 T perpendicular to the rotation axis. What is the maximum voltage generated?

Given: N = 100 turns, A = 0.10 m², v = 0.5 Hz, B = 0.01 T. Using the AC generator peak emf formula: ε0 = NBAω = NBA(2πv).
ε0 = 100 × 0.01 × 0.10 × 2 × 3.14 × 0.5 = 100 × 0.01 × 0.10 × 3.14 = 0.314 V.
The maximum voltage generated is 0.314 V. While small, this example demonstrates the principle of an AC generator and the feasibility of generating electricity from human-powered mechanical energy — an alternative possibility for small-scale power generation.

ByteNotes

Premium Access

Unlock all Application Questions

You are viewing 10 free application questions from this chapter. Upgrade to Premium to unlock the remaining 12 questions.

Upgrade to Premium12 questions locked

Chapter sections

Chapter 6: Electromagnetic Induction | Application Questions

A circular coil of radius 10 cm, 500 turns, and resistance 2 Ω is placed with its plane perpendicular to the horizontal component of Earth's magnetic field (3.0 × 10⁻⁵ T). It is rotated about its vertical diameter through 180° in 0.25 s. Estimate the induced emf and current.

A metallic rod of 1 m length is rotated with a frequency of 50 rev/s, with one end hinged at the centre of a circular metallic ring of radius 1 m. A uniform magnetic field of 1 T parallel to the axis exists everywhere. Find the emf between the centre and the ring.

A long solenoid with 15 turns per cm has a small loop of area 2.0 cm² placed inside the solenoid normal to its axis. If the current through the solenoid changes steadily from 2.0 A to 4.0 A in 0.1 s, find the induced emf in the loop.

A rectangular wire loop of sides 8 cm and 2 cm with a small cut is moving out of a uniform magnetic field of 0.3 T directed normal to the loop with velocity 1 cm/s normal to the longer side. Calculate the emf developed across the cut and the duration for which it lasts.

A 1.0 m long metallic rod is rotated with an angular frequency of 400 rad/s about an axis normal to the rod through one end. A uniform magnetic field of 0.5 T parallel to the axis exists. Calculate the emf developed between the centre and the ring.

A horizontal wire 10 m long extending east to west falls with a speed of 5.0 m/s at right angles to the horizontal component of Earth's magnetic field (0.30 × 10⁻⁴ T). Find (a) the induced emf and (b) which end of the wire is at higher potential.

Current in a circuit falls from 5.0 A to 0.0 A in 0.1 s. If an average emf of 200 V is induced, estimate the self-inductance of the circuit.

A pair of adjacent coils has a mutual inductance of 1.5 H. If the current in one coil changes from 0 to 20 A in 0.5 s, what is the change in flux linkage with the other coil?

Kamla pedals a stationary bicycle attached to a 100-turn coil of area 0.10 m². The coil rotates at half a revolution per second in a magnetic field of 0.01 T perpendicular to the rotation axis. What is the maximum voltage generated?

Unlock all Application Questions