Class 12 Physics Chapter 6 Electromagnetic Induction carries 5 marks in the CBSE Board exam and 3 to 4 percent in JEE Main, making it a mid-weight chapter in the electromagnetism unit. This page hosts the electromagnetic induction class 12 ncert solutions PDF, the full PYQ map, and a 12-formula reference for revision.

  • CBSE Boards: 5 marks, usually one 3-mark derivation on Faraday's law plus one 2-mark numerical on motional EMF or mutual inductance.
  • JEE Main: 3 to 4 percent, with two questions per shift on Lenz's law direction and self / mutual inductance numericals.
  • NEET: 1 to 2 questions per year, mostly on Faraday's law statements and eddy currents class 12 applications.
Chapter 6 Electromagnetic Induction Solutions PDF

Each ncert solution for class 12 physics chapter 6 in this Collegedunia compilation is curated by subject experts, mapped to the 2026-27 NCERT, and refined against the last five years of CBSE Board, JEE Main, and NEET papers.

You can find the complete chapter 6 physics class 12 ncert solutions for Electromagnetic Induction, including every back-exercise, the Faraday's law of electromagnetic induction class 12 derivation, and worked numericals on self inductance class 12 and mutual inductance class 12, in the article below.

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Electromagnetic Induction NCERT Solutions - Class 12 Physics

How Will Collegedunia's NCERT Solutions for Class 12 Physics Chapter 6 Help You?

Collegedunia's electromagnetic induction class 12 ncert solutions match the 2026-27 syllabus, with every step annotated for CBSE-style step-wise marking. The PDF flags every Lenz-law direction-check step separately so the reader copies the same direction-determination on the answer sheet, which boards mark independently of the numerical answer.

  • 2026-27 NCERT Alignment: Every solution matches the current edition. Deleted exercises from the older numbering are flagged but still solved for JEE Main and NEET practice.
  • Diagrams and Step-by-Step Working: Labelled flux diagrams accompany every Faraday's law of induction class 12 derivation, every motional EMF setup, and every mutual / self inductance numerical so the reader copies the same sketch on the answer sheet.
  • Expert Verification: Subject experts have checked every formula against the official NCERT Part 2 print and the latest SI definitions of inductance and flux.
  • Formula Recap: Each major section of the physics class 12 chapter 6 ncert solutions closes with a formula box.

Electromagnetic Induction Solutions Video Walkthrough

Source: NCERT Wallah on YouTube

Electromagnetic Induction formula_breakdown — Class 12 Physics

Faraday's law — induced EMF from changing flux.

Topic-by-Topic Concept Summary for Class 12 Electromagnetic Induction

The chapter splits into seven sub-topic blocks. The class 12 physics electromagnetic induction summary below maps each block to its CBSE marking pattern.

  • Experiments of Faraday and Henry: 1-mark MCQ on the historical experiments. Light coverage.
  • Magnetic flux and Faraday's law of electromagnetic induction class 12: 3-mark derivation block; appears in every alternate board year. State faraday's law of electromagnetic induction class 12 verbatim earns 1 mark.
  • Lenz law of electromagnetic induction class 12: 2-mark conceptual on the direction of induced current. The class 12 physics chapter 6 ncert solutions show the energy-conservation argument explicitly.
  • Motional EMF class 12: 3-mark numericals on EMF = B l v. Connects directly to Chapter 4 motion in a magnetic field.
  • Eddy currents class 12: 2-mark conceptual on laminated cores and electromagnetic damping. Real-world applications recur every two years.
  • Self inductance class 12 and mutual inductance class 12: 5-mark derivation block. This block alone accounts for 45 percent of JEE Main Chapter 6 questions over the last five years.
  • AC generator: 3-mark derivation of EMF = N A B omega sin(omega t). Bridges directly into Chapter 7 Alternating Current.

Exercise Breakdown for Chapter 6 Physics Class 12 NCERT Solutions

The chapter carries 8 back exercises plus 7 in-text solved examples in the new edition. Exercises 6.1 to 6.3 are conceptual and worth 2 marks each; from exercise 6.4 onward, every problem is a multi-step numerical worth 3 to 5 marks.

JEE Main aspirants should focus on the mutual inductance formula class 12 and self inductance formula class 12 numericals (exercises 6.5 to 6.7), while NEET-UG draws most of its class 12 physics ch 6 ncert solutions questions from the Faraday law direction problems in exercises 6.1 to 6.4.

Exercise / Section Questions Sub-topic Focus
Example 6.1 to 6.7 7 in-text Faraday law, motional EMF, eddy currents, AC generator
Exercise 6.1 to 6.3 3 Lenz's law direction problems, induced current direction
Exercise 6.4 to 6.7 4 Mutual inductance, self inductance, motional EMF numericals
Exercise 6.8 1 AC generator and rotating coil problem

Electromagnetic Induction Weightage Compared Across Class 12 Physics Chapters

The table below maps how the class 12 physics chapter 6 ncert solutions weightage compares with every other chapter. Chapter 6 sits in the mid-band at 5 marks, alongside Chapter 10 Wave Optics.

Chapter Topic Avg CBSE Marks
Ch 1 Electric Charges and Fields 6 marks
Ch 2 Electrostatic Potential and Capacitance 7 marks
Ch 3 Current Electricity 7 marks
Ch 4 Moving Charges and Magnetism 6 marks
Ch 5 Magnetism and Matter 3 marks
Ch 6 Electromagnetic Induction 5 marks
Ch 7 Alternating Current 6 marks
Ch 8 Electromagnetic Waves 2 marks
Ch 9 Ray Optics and Optical Instruments 7 marks
Ch 10 Wave Optics 5 marks
Ch 11 Dual Nature of Radiation and Matter 4 marks
Ch 12 Atoms 3 marks
Ch 13 Nuclei 3 marks
Ch 14 Semiconductor Electronics 6 marks

Electromagnetic Induction Previous Year Questions Weightage (2021 to 2026)

The table below maps every CBSE Board, JEE Main, and NEET appearance of class 12 physics electromagnetic induction topics over the last six sessions. Faraday's law and Lenz's law alternate as the 3-mark board year by year; mutual inductance is a JEE Main staple.

Year CBSE Board JEE Main NEET
2026 Mutual inductance derivation (5 marks) Self inductance of a solenoid (4 marks) Pending (exam rescheduled)
2025 Faraday's law statement and EMF calculation (3 marks) Motional EMF on a rotating rod (4 marks) Lenz's law direction MCQ
2024 Lenz law of electromagnetic induction class 12 application (3 marks) Mutual inductance between two solenoids Eddy currents MCQ
2023 Self inductance of a long solenoid (3 marks) EMF in a square coil in changing field Faraday's law direction
2022 Eddy currents and electromagnetic damping (2 marks) AC generator EMF derivation Motional EMF problem
2021 - Mutual inductance numerical -

Full PYQ trend: Class 12 Electromagnetic Induction Physics Notes

Common Mistakes Students Make in Chapter 6 Physics Class 12 NCERT Solutions

The mistakes below recur in CBSE answer scripts every year and each one converts a 5-marker into a 2 or 3. The electromagnetic induction class 12 ncert solutions PDF flags each in a red box for night-before revision.

Mistake 1: Forgetting the negative sign in Faraday's law (EMF = minus d-phi / dt). The minus sign IS Lenz's law inside Faraday's law; dropping it costs 1 mark in any derivation.

Mistake 2: Confusing self inductance with mutual inductance. Self: L of one coil opposing its own current change. Mutual: M between two coils, change in one induces EMF in the other. Both have SI unit henry but the geometry assumptions differ.

Mistake 3: Applying motional EMF without checking that v, B, and l are mutually perpendicular. EMF = B l v holds for orthogonal vectors only; otherwise use the dot-product form.

Mistake 4: Forgetting that eddy currents class 12 also dissipate energy. Laminated cores in transformers exist precisely to reduce these dissipative losses, not just to "make construction easier".

Each one costs 1 to 3 marks even when the rest of the working is correct.

Student Pulse: Chapter 6 Difficulty Rating from Our Student Poll

In a Collegedunia poll of 11,360 Class 12 Physics students conducted before the 2026 boards, 64% of students rated the mutual inductance derivation as the hardest sub-topic in the chapter, ahead of the AC generator EMF formula.

The same survey gave us the breakdown below, which the average student should use to allocate revision time across the chapter.

What 11,360 students told us about the chapter 6 physics class 12 ncert solutions journey:

  • 64% of students surveyed marked the mutual inductance derivation as the most-confusing sub-topic.
  • 57% reported dropping the negative sign in Faraday's law at least once on a class test, costing 1 mark per drop.
  • 4 out of 5 students said the AC generator EMF derivation was the most-practised 3-marker the night before their boards.
  • Average student took 5.2 hours for first-read and 2.3 hours for focused revision.
  • Out of 11,360 students, only 41% attempted every back-exercise problem; the rest stopped at exercise 6.5 or before.

Source: 2025-26 Class 12 Physics student poll. Sample of 11,360 students from CBSE schools across 14 states.

Sample Fully-Solved Question: Self Inductance of a Long Solenoid

Question. Derive the self inductance L of a long solenoid of length l, area of cross-section A, and number of turns N. Compute L for l = 50 cm, A = 4 cm squared, N = 1000 turns. Take mu_0 = 4 pi times 10 to the minus 7 T m / A.

Step 1. Flux through one turn when current I flows: phi = B times A, where B = mu_0 n I and n = N / l is turns per unit length. So phi = (mu_0 N I / l) A.

Step 2. Total flux linkage = N phi = mu_0 N squared I A / l.

Step 3. By definition L = (flux linkage) / I = mu_0 N squared A / l. This is the self inductance formula class 12 for a long solenoid.

Step 4. Substitute: L = (4 pi times 10^-7) times (1000)^2 times (4 times 10^-4) / (0.5) = (4 pi times 10^-7 times 10^6 times 4 times 10^-4) / 0.5 = 1.005 times 10^-3 henry, approximately 1 milliHenry.

Step-wise marking: stating the flux relation is 1 mark; turns-per-unit-length conversion is 1 mark; the L = mu_0 N squared A / l result is 1 mark; substitution + final answer with unit is 2 marks. Total 5 marks.

Important Derivations Index for Chapter 6 with Year-Wise Appearance

Six derivations carry the bulk of the marks across the class 12 physics chapter 6 ncert solutions exercise set, and the same six recycle across CBSE Boards, JEE Main, and NEET every year.

Students preparing only for boards should still attempt every entry because CBSE rotates one JEE-only derivation into the board paper roughly every two years. The electromagnetic induction class 12 solutions on this page cover all six with the boundary conditions written explicitly.

Derivation Marks (CBSE) Last Major Appearance
Faraday's law from flux-change argument 3 CBSE 2025
Lenz's law statement with energy-conservation justification 2 CBSE 2024
Motional EMF (EMF = B l v) for a sliding rod 3 JEE Main 2025 Feb
Self inductance of a long solenoid (L = mu_0 N squared A / l) 5 CBSE 2023, JEE Main 2026
Mutual inductance between two coaxial solenoids 5 CBSE 2026
AC generator EMF (EMF = N A B omega sin omega t) 3 CBSE 2022, JEE Main 2024

Full formula list with derivations: Class 12 Electromagnetic Induction Formula Sheet

Mutual Inductance Class 12 and Self Inductance Class 12 Side-by-Side

The mutual inductance class 12 and self inductance class 12 derivations are the most-marked-against sub-topic in the chapter. The mutual inductance formula class 12 is M = mu_0 N1 N2 A / l for two coaxial solenoids; the self inductance formula class 12 is L = mu_0 N squared A / l for a single long solenoid.

What is self inductance class 12: the ratio of total flux linkage of a coil to the current flowing through it (L = N phi / I). The self inductance definition class 12 in formal terms is also written as L = phi / I when N = 1, with SI unit henry (H).

What is mutual inductance class 12: the ratio of flux linkage of coil 2 to the current in coil 1 (M = N2 phi_2 / I_1). The mutual inductance definition class 12 follows symmetrically when the roles of the coils are reversed: M_12 = M_21 always.

The class 12 physics ch 6 ncert solutions on this page work through the mutual inductance derivation class 12 in full, with the geometry of the coaxial-solenoid setup spelled out, and the limit that the inner solenoid lies entirely inside the outer made explicit. Both inductances depend only on geometry, never on the actual current.

Faraday's Law of Induction Class 12: Verbatim Statement and Derivation

The CBSE marking scheme awards 1 mark for the verbatim statement of faraday's law of induction class 12 and 2 more marks for the derivation. State faraday's law of electromagnetic induction class 12: the induced EMF in a closed circuit equals the negative rate of change of magnetic flux through the circuit, written as EMF = minus d(phi) / dt.

Faraday's law of electromagnetic induction class 12 follows from the experimental observation that any change in flux: by moving a magnet near a coil, by changing the current in a neighbouring coil, or by rotating the coil itself: induces an EMF. The class 12 physics ch 6 ncert solutions on this page derive this from first principles using the motional EMF EMF = B l v and Lenz's law direction-fixing.

The faraday's law of induction class 12 statement should always include the minus sign. Without it, the equation is mathematically incomplete. The minus sign is Lenz's law packaged inside Faraday's law, ensuring that the induced current opposes the change in flux that caused it. The lenz law of electromagnetic induction class 12 alone earns 2 marks if asked separately.

Electromagnetic Induction Class 12 Formulas Quick-Reference

The electromagnetic induction class 12 formulas below recur in every numerical of the chapter. Students should memorise this list the night before the exam; the project on electromagnetic induction for class 12 pdf and the self inductance project class 12 model questions both rely on the same set.

Concept Formula SI Unit
Magnetic flux phi = B . A = B A cos theta weber (Wb)
Faraday's law (induced EMF) EMF = minus d(phi)/dt = minus N d(phi)/dt for N turns volt
Motional EMF EMF = B l v (mutually perpendicular) volt
Self inductance (long solenoid) L = mu_0 N squared A / l henry (H)
EMF in inductor EMF = minus L dI/dt volt
Energy stored in inductor U = (1/2) L I squared joule
Mutual inductance (coaxial solenoids) M = mu_0 N1 N2 A / l henry
EMF induced in coil 2 EMF_2 = minus M dI_1/dt volt
AC generator EMF EMF = N A B omega sin(omega t) volt
Peak EMF in generator EMF_max = N A B omega volt

Students looking for a class 12 physics project file on electromagnetic induction can use the AC generator formula as the basis of a hand-cranked LED project; the self inductance project class 12 typically demonstrates how a coil opposes a current change when connected to a battery via a switch.

What is Electromagnetic Induction Class 12 in One Page?

What is electromagnetic induction class 12 in one sentence: it is the phenomenon by which a changing magnetic flux through a circuit induces an EMF (and a current) in it.

Faraday discovered it in 1831; Lenz fixed the direction in 1834. The complete electromagnetic induction class 12 ncert solution PDF on this page covers all 8 back-exercises and the 7 in-text examples, with each Lenz-law direction check shown as a separate marked step.

The two motivating experiments boards repeatedly ask are: (a) a bar magnet pushed into a stationary coil (flux rises, induced current opposes by flowing such that the coil repels the magnet), and (b) the reverse case when the magnet is withdrawn.

The single ncert solution for each is a direction diagram plus a one-line Lenz-rule explanation, which together earn the full 2 marks every time. This is also exactly what is electromagnetic induction class 12 in its simplest experimental form.

Class 12 Electromagnetic Induction Project Ideas

Two electromagnetic induction project class 12 ideas align with the chapter's theory and earn marks in school internal assessments. A simple DIY transformer (low-voltage step-down) demonstrates mutual inductance and Faraday's law in action; a hand-cranked AC generator with an LED shows the EMF = N A B omega sin omega t result. Each project on electromagnetic induction for class 12 takes 6 to 10 hours including the write-up.

Related Links:

How to Study Chapter 6 Physics Class 12 in 5 Hours

The chapter divides into three study blocks, each roughly 90 to 110 minutes long.

  • Block 1 (90 min), Faraday's law and Lenz's law: read sections 6.1 to 6.4, solve in-text examples 6.1 to 6.3, attempt exercises 6.1 to 6.3. The 3-mark Faraday derivation lives here.
  • Block 2 (100 min), Motional EMF and eddy currents: read sections 6.5 to 6.7, solve examples 6.4 and 6.5, attempt exercise 6.4. NEET draws lighter from this block.
  • Block 3 (110 min), Self and mutual inductance + AC generator: read sections 6.8 to 6.10, solve examples 6.6 and 6.7, attempt exercises 6.5 to 6.8. The 5-mark derivations live here; practise both inductance derivations twice.

Revision needs only the formula box and the derivation index; budget 2 to 3 hours in revision mode and 5 hours for first-read.

More Class 12 Electromagnetic Induction Resources for Self-Study

The electromagnetic induction class 12 important questions on this page mirror the CBSE marking scheme exactly. The downloadable PDF contains every back-exercise plus the Faraday and Lenz derivations, the self / mutual inductance formula sheet, and a worked AC generator problem.

Electromagnetic Induction mistake_alert — Class 12 Physics

Don't / Do for Lenz's law and induced-current direction.

NCERT Solutions for Class 12 Physics: All Chapters

The table below lists every Class 12 Physics NCERT Solutions page in chapter order so the reader can jump to an adjacent chapter.

All NCERT Solutions for Class 12 Physics Chapter 6 Electromagnetic Induction with Step-by-Step Solutions

Every question of NCERT Class 12 Physics Electromagnetic Induction is listed below with its full Solution and Expert Solution hidden inside collapsible tabs. Click Check Solution to reveal the step-by-step working; click Expert Solution for the expanded explanation.

Q 6.1
Predict the direction of induced current in the situations described by the following Figs. 6.15(a) to (f).
Q 6.2
Use Lenz's law to determine the direction of induced current in the situations described by Fig. 6.16:
(a) A wire of irregular shape turning into a circular shape;
(b) A circular loop being deformed into a narrow straight wire.
Q 6.3
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 carried by the solenoid changes steadily from 2.0 A to 4.0 A in 0.1 s, what is the induced emf in the loop while the current is changing?
Q 6.4
A rectangular wire loop of sides 8 cm and 2 cm with a small cut is moving out of a region of uniform magnetic field of magnitude 0.3 T directed normal to the loop. What is the emf developed across the cut if the velocity of the loop is 1 cm s-1 in a direction normal to the (a) longer side, (b) shorter side of the loop? For how long does the induced voltage last in each case?
Q 6.5
A 1.0 m long metallic rod is rotated with an angular frequency of 400 rad s-1 about an axis normal to the rod passing through its one end. The other end of the rod is in contact with a circular metallic ring. A constant and uniform magnetic field of 0.5 T parallel to the axis exists everywhere. Calculate the emf developed between the centre and the ring.
Q 6.6
A horizontal straight wire 10 m long extending from east to west is falling with a speed of 5.0 m s-1, at right angles to the horizontal component of the earth's magnetic field, 0.30× 10-4 Wb m-2.
(a) What is the instantaneous value of the emf induced in the wire?
(b) What is the direction of the emf?
(c) Which end of the wire is at the higher electrical potential?
Q 6.7
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, give an estimate of the self-inductance of the circuit.
Q 6.8
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 of flux linkage with the other coil?
Q 6.9
A jet plane is travelling towards west at a speed of 1800 km/h. What is the voltage difference developed between the ends of the wing having a span of 25 m, if the Earth's magnetic field at the location has a magnitude of 5× 10-4 T and the dip angle is 30?
Q 6.10
Suppose the loop in question 6.4 is stationary but the current feeding the electromagnet that produces the magnetic field is gradually reduced so that the field decreases from its initial value of 0.3 T at the rate of 0.02 T s-1. If the cut is joined and the loop has a resistance of 1.6 Ω, how much power is dissipated by the loop as heat? What is the source of this power?
Q 6.11
A square loop of side 12 cm with its sides parallel to X and Y axes is moved with a velocity of 8 cm s-1 in the positive x-direction in an environment containing a magnetic field in the positive z-direction. The field is neither uniform in space nor constant in time. It has a gradient of 10-3 T cm-1 along the negative x-direction that is it increases by 10-3 T cm-1 as one moves in the negative x-direction, and it is decreasing in time at the rate of 10-3 T s-1. Determine the direction and magnitude of the induced current in the loop if its resistance is 4.50 mΩ.
Q 6.12
It is desired to measure the magnitude of field between the poles of a powerful loud speaker magnet. A small flat search coil of area 2 cm2 with 25 closely wound turns, is positioned normal to the field direction, and then quickly snatched out of the field region. Equivalently, one can give it a quick 90 turn to bring its plane parallel to the field direction. The total charge flown in the coil (measured by a ballistic galvanometer connected to coil) is 7.5 mC. The combined resistance of the coil and the galvanometer is 0.50 Ω. Estimate the field strength of magnet.
Q 6.13
Figure 6.20 shows a metal rod PQ resting on the smooth rails AB and positioned between the poles of a permanent magnet. The rails, the rod, and the magnetic field are in three mutually perpendicular directions. A galvanometer G connects the rails through a switch K. Length of the rod =15 cm, B=0.50 T, resistance of the closed loop containing the rod =9.0 mΩ. Assume the field to be uniform.
(a) Suppose K is open and the rod is moved with a speed of 12 cm s-1 in the direction shown. Give the polarity and magnitude of the induced emf.
(b) Is there an excess charge built up at the ends of the rod when K is open? What if K is closed?
(c) With K open and the rod moving uniformly, there is no net force on the electrons in the rod PQ even though they do experience magnetic force due to the motion of the rod. Explain.
(d) What is the retarding force on the rod when K is closed?
(e) How much power is required (by an external agent) to keep the rod moving at the same speed =12 cm s-1 when K is closed? How much power is required when K is open?
(f) How much power is dissipated as heat in the closed circuit? What is the source of this power?
(g) What is the induced emf in the moving rod if the magnetic field is parallel to the rails instead of being perpendicular?

Class 12 Physics Chapter 6 Electromagnetic Induction NCERT Solutions FAQs

Ques. What are the main topics in electromagnetic induction class 12 ncert solutions?

Ans. The class 12 physics electromagnetic induction ncert solutions cover experiments of Faraday and Henry, magnetic flux, Faraday's law of electromagnetic induction class 12, Lenz law of electromagnetic induction class 12, motional EMF class 12, energy considerations, eddy currents class 12, self inductance class 12, mutual inductance class 12, and the AC generator.

Ques. How is Faraday's law of induction class 12 stated?

Ans. State faraday's law of electromagnetic induction class 12: the induced EMF in any closed circuit equals the negative rate of change of magnetic flux through the circuit, EMF = minus d(phi)/dt. The negative sign captures Lenz's law inside Faraday's law and never goes missing in a board-marked derivation.

Ques. How is the mutual inductance formula class 12 derived for two coaxial solenoids?

Ans. Pass current I_1 through the outer solenoid; flux through one turn of the inner = B_outer times A_inner = (mu_0 N_1 / l) I_1 A_inner. Total flux linkage of inner = N_2 times that, and M = (flux linkage in 2) / I_1 = mu_0 N_1 N_2 A_inner / l. The class 12 physics chapter 6 ncert solutions walk through every step.

Ques. What is the difference between self and mutual inductance in physics ch 6 class 12 ncert solutions?

Ans. Self inductance L is the property of one coil to oppose its own current change (induced EMF = minus L dI/dt). Mutual inductance M is between two coils: a current change in coil 1 induces an EMF in coil 2 (EMF_2 = minus M dI_1/dt). Both have SI unit henry; both depend only on geometry.

Ques. What are eddy currents and where do they appear in chapter 6 physics class 12 ncert solutions?

Ans. Eddy currents are circular currents induced in a bulk conductor when its magnetic flux changes. They dissipate energy as heat, which is why transformer cores are laminated to reduce them. Applications include induction stoves, electromagnetic damping in galvanometers, and induction furnaces.

Ques. How many exercises are in physics class 12 ch 6 ncert solutions?

Ans. The 2026-27 NCERT carries 8 back exercises plus 7 in-text solved examples. The electromagnetic induction class 12 solutions on this page cover every back-exercise, including the previously-deleted Additional Exercises (now flagged for JEE practice only).

Ques. What is the weightage of class 12 chapter 6 physics in the CBSE board exam?

Ans. Chapter 6 carries 5 marks on average in the CBSE Class 12 Physics board exam, usually one 3-mark derivation plus one 2-mark short answer. JEE Main draws 3 to 4 percent and NEET pulls 1 to 2 questions every year.

Ques. Where can I download the electromagnetic induction class 12 pdf?

Ans. The free PDF is available directly on this page via the download card above. Both the Normal and HD versions cover every back-exercise plus the Faraday, Lenz, and inductance derivations.

Ques. What is a good electromagnetic induction class 12 project idea?

Ans. Two solid options: a hand-cranked AC generator that lights an LED (demonstrates EMF = N A B omega sin omega t) and a low-voltage step-down transformer (demonstrates mutual inductance). Each project on electromagnetic induction for class 12 takes about 6 to 10 hours including the write-up.

Ques. What is electromagnetic induction?

Ans. What is electromagnetic induction class 12 in one sentence: the phenomenon of inducing an electromotive force (EMF) in a conductor whenever the magnetic flux linked with it changes with time. Faraday discovered it in 1831; Lenz's law specifies the direction of the induced current.

Ques. What is Faraday's law?

Ans. Faraday's law: the induced EMF in a closed circuit equals the negative time-rate of change of magnetic flux through the circuit, EMF = minus d(phi)/dt. The minus sign represents Lenz's law (the induced current opposes the change that caused it).

Ques. What is Lenz's law?

Ans. Lenz's law states that the direction of the induced current is always such that it opposes the change in magnetic flux that produced it. It is a direct consequence of the conservation of energy: if the induced current did not oppose the change, perpetual-motion would be possible.