Physics Class 12 Part I…
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Eddy Currents Inductance

🎓 Class 12 Physics CBSE Theory Ch 6 – Electromagnetic Induction ⏱ ~14 min
🌐 ભાષા:

આ MCQ મોડ્યુલ આના પર આધારિત છે: Eddy Currents Inductance

આ મૂલ્યાંકન આના પર આધારિત હશે: Eddy Currents Inductance

મૂલ્યાંકન બનાવવામાં તેમની સામગ્રી સામેલ કરવા ચિત્રો, PDF અથવા Word દસ્તાવેજ અપલોડ કરો.

Eddy Currents Inductance

6.8 Eddy Currents

If a bulk piece of conductor (not a wire loop) experiences a changing magnetic flux, induced currents flow in closed loops within the volume of the conductor itself. These swirling currents - they look like the eddies of a stream - are called eddy currents (or Foucault currents).

Two faces of eddy currents:
  • Useful: induction heating (cooktops, melting metals), magnetic braking (trains, dynamometers), induction motors, metal detectors, electricity meters.
  • Wasteful: heat losses in transformer cores, motor stators, choke coils. Mitigated by laminating the iron - thin sheets break the eddy-current loops.
Solid core - large eddy loops Laminated - small loops →
Fig 6.5 Lamination of a transformer core breaks large eddy loops into small high-resistance ones, sharply reducing I²R losses.

6.9 Inductance

The flux through any circuit due to a current passing through it - or through a neighbouring circuit - is proportional to that current (in the linear, non-saturating regime). The proportionality constant is called inductance.

6.9.1 Mutual inductance

Two coils 1 and 2 are placed near each other. A current I1 in coil 1 produces a flux Φ2 through coil 2:

\[N_2\Phi_2 = M_{21}\,I_1\]

M21 is the mutual inductance of coil 2 with respect to coil 1. By symmetry M12 = M21 = M.

Induced EMF in coil 2 due to changing I1:

\[\varepsilon_2 = -M\dfrac{dI_1}{dt}\]

For two coaxial solenoids of length ℓ, areas A1 < A2, with N1 and N2 turns:

\[M = \mu_0 \dfrac{N_1 N_2 A_1}{\ell}\]

(if a soft-iron core fills both, replace μ0 by μ).

Outer (N₂, A₂) Inner (N₁, A₁)
Fig 6.6 Coaxial solenoids: mutual inductance M = μ₀N₁N₂A₁/ℓ, set by the geometry.

6.9.2 Self-inductance

Even a single coil has flux through itself due to its own current:

\[N\Phi = L\,I\]

L is the self-inductance. By Faraday's law:

\[\varepsilon = -L\dfrac{dI}{dt}\]

This "back EMF" opposes any change in current. For a long solenoid of length ℓ, area A, n = N/ℓ turns per metre:

\[L = \mu_0 n^2 A\,\ell = \mu_0\dfrac{N^2 A}{\ell}\]

6.9.3 Energy stored in an inductor

To establish a current I in an inductor against the back EMF, work is done by the source. The work goes into the magnetic field around the inductor:

\[U_B = \dfrac{1}{2}LI^2\]

This is to a magnetic circuit what ½CV² is to a capacitor. Equivalently in terms of B (for a long solenoid): u = B²/(2μ0) is the energy density of the magnetic field in vacuum.

Worked Example 6.7 - Self-inductance of a solenoid

Example 6.7 L3 Apply

An air-core solenoid is 0.50 m long, 4.0 cm² cross section, with 1500 turns. Find its self-inductance and the back EMF if the current changes from 0 to 3.0 A in 0.10 s.

L = μ0N²A/ℓ = (4π×10⁻⁷)(1500)²(4×10⁻⁴)/0.50 = (4π×10⁻⁷)(2.25×10⁶)(4×10⁻⁴)/0.50 = 2.26 × 10⁻³ H ≈ 2.3 mH.

|ε| = L|dI/dt| = 2.26 × 10⁻³ × (3.0/0.10) = 6.78 × 10⁻² V ≈ 68 mV.

Worked Example 6.8 - Mutual inductance of coaxial solenoids

Example 6.8 L3 Apply

Inner solenoid: ℓ = 0.30 m, A1 = 2.0 × 10⁻⁴ m², N1 = 600 turns. Outer (over the same length): N2 = 1500. Find M.

M = μ0N1N2A1/ℓ = (4π×10⁻⁷)(600)(1500)(2.0×10⁻⁴)/0.30 ≈ 7.54 × 10⁻⁴ H ≈ 0.75 mH.

Worked Example 6.9 - Energy stored

Example 6.9 L3 Apply

An inductor of L = 0.50 H carries 4.0 A. Find the energy stored.

U = ½ L I² = 0.5 × 0.50 × 16 = 4.0 J.

Interactive: Solenoid Self-Inductance Builder L3 Apply

Adjust geometry of an air-core solenoid; see L, back-EMF for a given dI/dt, and stored energy.

L = 2.26 mH  |  U = ½LI² = 10.18 mJ
Activity 6.3 - Spark when you open a switchL4 Analyse
  1. Connect a coil (or relay) in series with a battery and a switch.
  2. Close the switch, then quickly open it. (Stand back!)
Predict: Will the spark be larger when closing or opening the switch? Why?

The spark occurs when opening the switch. The current is interrupted very rapidly (small Δt), so |ε| = L|dI/dt| can be very large - hundreds of volts even from a 1.5 V cell. This high voltage ionises the air in the gap, producing the spark. (This is exactly how an automobile ignition coil generates ~30 kV from 12 V.)

Competency-Based Questions L1-L6

A 200-mH inductor carries a current that rises from 0 to 4.0 A in 50 ms.
1. The induced back-EMF is: L3 Apply
  • (a) 8 V
  • (b) 16 V
  • (c) 0.8 V
  • (d) 80 V
(b) ε = L dI/dt = 0.20 × (4.0/0.050) = 0.20 × 80 = 16 V.
2. The energy stored at I = 4.0 A is: L3 Apply
  • (a) 0.4 J
  • (b) 1.6 J
  • (c) 3.2 J
  • (d) 8.0 J
(b) U = ½ × 0.20 × 16 = 1.6 J.
3. Define self-inductance and state its SI unit. L1 Remember
L = NΦ/I, the flux linkage per unit current. SI unit: henry (H) = V·s/A = Wb/A.
4. Why are transformer cores laminated rather than solid? L4 Analyse
Laminations break the long eddy-current paths into many short, high-resistance ones. The same B variation produces a much smaller eddy current per lamination ⇒ I²R loss falls dramatically. Lamination thickness is chosen so that skin depth at the operating frequency is greater than the lamination thickness.
5. Design and dimension an inductor that stores 1 J of energy at 5 A current using an air-core solenoid. List your assumed geometry. L6 Create
U = ½ L I² ⇒ L = 2 × 1/(5²) = 0.08 H. With L = μ₀N²A/ℓ choose ℓ = 0.50 m, A = 5 × 10⁻³ m² (radius 4 cm) ⇒ N² = Lℓ/(μ₀A) = 0.08 × 0.5/(4π × 10⁻⁷ × 5 × 10⁻³) ≈ 6.4 × 10⁶ ⇒ N ≈ 2530 turns. Adding an iron core with μr = 100 reduces N by factor 10 to ~250 turns - much more practical.

Assertion-Reason Pairs L4 Analyse

Options: (A) Both true, R correct explanation. (B) Both true, R not the explanation. (C) A true, R false. (D) A false, R true.

Assertion: A solid metal disc rotated near a magnet experiences a retarding force.
Reason: Eddy currents in the disc set up forces opposing the relative motion (Lenz's law).
(A). Reason directly explains assertion. This is the principle of magnetic braking.
Assertion: The self-inductance of a solenoid increases when an iron core is inserted.
Reason: L ∝ μ; iron has μr ≫ 1 ⇒ much higher μ than vacuum.
(A). Reason directly explains assertion.
Assertion: The mutual inductance between two coils depends on the current in either coil.
Reason: M is defined as N2Φ2/I1.
(D). Assertion is false: M depends only on geometry, number of turns and the medium - not on current. Reason is the correct definition.

Frequently Asked Questions - Eddy Currents Inductance

What is the main concept covered in Eddy Currents Inductance?
In NCERT Class 12 Physics Chapter 6 (Electromagnetic Induction), "Eddy Currents Inductance" covers the core principles and equations students need for board exam success. The MyAiSchool lesson explains the topic with definitions, derivations, worked examples, and interactive simulations. Key formulas and dimensional analysis are included to build conceptual depth and problem-solving skills aligned with the CBSE 2025-26 syllabus.
How is Eddy Currents Inductance useful in real-life applications?
Real-life applications of "Eddy Currents Inductance" from NCERT Class 12 Physics Chapter 6 include electronics, communication systems, medical imaging, solar energy, semiconductor devices, and modern technology. The MyAiSchool lesson links every concept to a tangible example so students see physics as a problem-solving framework for the physical world, not as abstract formulas.
What are the key formulas in Eddy Currents Inductance?
Key formulas in "Eddy Currents Inductance" (NCERT Class 12 Physics Chapter 6 Electromagnetic Induction) are derived step-by-step in the MyAiSchool lesson. Students should memorize the final formula AND understand its derivation for full board marks. Each formula is listed with its dimensional formula, SI unit, applicability range, and common pitfalls. The Summary section at the end of each part includes a quick-reference formula card.
How does this part connect to other parts of Chapter 6?
NCERT Class 12 Physics Chapter 6 (Electromagnetic Induction) is structured so each part builds on the previous one. "Eddy Currents Inductance" connects directly to neighbouring parts via shared definitions, units, and methodology. The MyAiSchool lesson cross-references related concepts with internal links so students can navigate the whole chapter as one connected story rather than disconnected fragments.
What types of CBSE board questions come from Eddy Currents Inductance?
CBSE board questions from "Eddy Currents Inductance" typically include: (1) 1-mark MCQs on definitions and formulas, (2) 2-mark short-answer derivations or applications, (3) 3-mark numerical problems with units, (4) 5-mark long-answer derivations followed by application. The MyAiSchool lesson tags each Competency-Based Question (CBQ) with Bloom level (L1-L6) so students know how to study for each weight.
How can students use the interactive simulation effectively?
The interactive simulation in the "Eddy Currents Inductance" lesson allows students to adjust input parameters (sliders or selectors) and see physical quantities update in real time. To use it effectively: (1) try extreme values to understand limiting cases, (2) compare with the analytical formula, (3) check unit consistency, (4) test special configurations from worked examples. The simulation reinforces conceptual intuition that pure formula manipulation cannot.
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Physics Class 12 Part I – NCERT (2025-26)
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