With the CBSE Class 12 Physics Board Exam scheduled on 20 February 2026, students should now focus on revising important chapters like Electromagnetic Induction, which require both conceptual understanding and accurate application of formulas.
During the final phase of preparation, students often find this chapter difficult while solving numerical questions related to induced emf, magnetic flux, and Lenz’s Law, especially under exam time constraints. Revising these concepts at this stage is important because questions based on derivations, formula application, and real-life circuit situations are commonly included in the board examination.
NCERT Solutions for Class 12 Physics Chapter 6 help students understand key concepts through step-by-step explanations and solved numericals. Focusing on this chapter now supports better accuracy in calculations, improves confidence in solving application-based questions, and strengthens overall preparation for the upcoming CBSE Class 12 Physics Board Exam 2026.
CBSE Class 12 Physics Previous Year Question Papers
These solutions help you revise quickly, understand step-by-step problem-solving methods, and practice questions that are most likely to appear in the exam.
NCERT Solutions for Class 12 Physics Chapter 6 Electromagnetic Induction
NCERT Solutions for Class 12 Physics Chapter 6 provide clear and accurate answers to all textbook questions, explained in a simple and logical manner.
Each solution follows the NCERT guidelines, making it easier for students to understand important concepts like Faraday’s Laws, Lenz’s Law, induced EMF, and self & mutual induction.
EMI NCERT Solutions for Class 12 Physics help students build strong conceptual clarity, which is essential for board exams.
Class 12 Electromagnetic Induction NCERT Solutions (All Exercises)
EMI NCERT Solutions for Class 12 Physics are given here. Practicing all exercises regularly helps improve problem-solving speed, accuracy, and confidence, especially important during last-minute exam preparation.
Question 1. Predict the direction of induced current in the situations described by the following Figs. 6.18(a) to (f ).
Solution : Lenz’s law shows the direction of induced current in a closed loop. In the given two figures they shows the direction of induced current when the North pole of a bar magnet is moved towards and away from a closed loop respectively. 
Question 2. Use Lenz’s law to determine the direction of induced current in the situations described by Fig. 6.19: (a) A wire of irregular shape turning into a circular shape; (b) A circular loop being deformed into a narrow straight wire.
Question 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?
Solution : Number of turns on the solenoid = 15 turns/cm = 1500 turns/m Number of turns per unit length, n = 1500 turns The solenoid has a small loop of area, A = 2.0 cm2 = 2 × 10−4 m2 Current carried by the solenoid changes from 2 A to 4 A. Change in current in the solenoid, di = 4 − 2 = 2 A Change in time, dt = 0.1 s Induced emf in the solenoid is given by Faraday’s law as: e = dØ/dt
Where, Ø= Induced flux through the small loop = BA ...
(ii) B = Magnetic field = μ0ni μ0 = Permeability of free space = 4π×10−7 H/m Hence, equation (i) reduces to:
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Hence, the induced voltage in the loop is 7.5 x 10 -6 v Question 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?
Solution : Length of the rectangular wire, l = 8 cm = 0.08 m Width of the rectangular wire, b = 2 cm = 0.02 m Hence, area of the rectangular loop, A = lb = 0.08 × 0.02 = 16 × 10−4 m2 Magnetic field strength, B = 0.3 T Velocity of the loop, v = 1 cm/s = 0.01 m/s (a) Emf developed in the loop is given as: e = Blv = 0.3 × 0.08 × 0.01 = 2.4 × 10−4 V
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= 0.02/0.01 = 2s
Hence, the induced voltage is 2.4 × 10−4 V which lasts for 2 s.
(b) Emf developed, e = Bbv = 0.3 × 0.02 × 0.01 = 0.6 × 10−4 V
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= 0.08/0.01 = 8s Hence, the induced voltage is 0.6 × 10−4 V which lasts for 8 s.
Question 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.
Solution : Length of the rod, l = 1 m Angular frequency,ω = 400 rad/s Magnetic field strength, B = 0.5 T One end of the rod has zero linear velocity, while the other end has a linear velocity of lω. Average linear velocity of the rod,
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Emf developed between the centre and the ring,
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= 100V
Hence, the emf developed between the centre and the ring is 100 V.
Question 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?
Solution : Wire’s Length, l = 10 m Speed of the wire with which it is falling, v = 5.0 m/s Strength of magnetic field, B = 0.3×10−4Wbm −2
(a) EMF induced in the wire, e = Blv =0.3×10 −4 ×5×10=1.5×10 −3 V
(b) We can determine the direction of the induced current by using the Fleming’s right hand thumb rule, here the current is flowing in the direction from West to East.
(c) In this case the eastern end of the wire will be having higher potential
Question 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 induced, give an estimate of the self-inductance of the circuit.
Solution : Initial current, I1 = 5.0 A Final current, I2 = 0.0 A Change in current,dl = I 1 - I 2 = 5A Time taken for the change, t = 0.1 s Average emf, e = 200 V For self-inductance (L) of the coil, we have the relation for average emf as: e = L di/dt L = e/(di/dt) 200/(5/0.1) = 4H Hence, the self induction of the coil is 4 H.
Question 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?
Solution : Given, a pair of adjacent coils. Mutual inductance, M = 1.5 H Current in the coil, I = 20 A Time ,t = 0.5 s Using formula, Φ = MI we get, Φ = 1.5 x 20 = 30H Hence, the change in the flux linkage is 30 Wb.
How to Prepare Chapter 6 Physics Class 12 NCERT Exercise-Wise Solutions?
To learn Electromagnetic Induction effectively, follow these learning tips:
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Start by understanding the theory from the NCERT textbook before jumping to numericals.
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Learn and revise key formulas and laws daily, especially Faraday’s and Lenz’s laws.
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Solve NCERT questions exercise-wise to maintain a structured approach.
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Analyze solved examples carefully to understand how formulas are applied.
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Practice numericals regularly and try solving them without looking at solutions first.
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Revise important derivations and concepts a few days before the exam for better retention.