Advertisement Space
Electromagnetic Induction NCERT Solutions & Important Questions
Welcome back to examspark.in! I'm Lucky, and today we're going to master CBSE Class 12 Physics Chapter 6: Electromagnetic Induction (EMI). In this chapter, you'll discover how changing magnetic fields can actually create electric currents—the exact science behind generators and transformers. Mastering EMI is crucial for your 2026 board exams and competitive exams like JEE and NEET. Bina kisi tension ke, let's clear all your doubts and boost your exam prep!
Join Telegram
PDF download locked
Use 20 Spark coins to unlock this chapter PDF. Unlocked PDFs stay in your vault.
Chapter NameElectromagnetic Induction
SubjectPhysics
Class12
BoardCBSE / State Boards
DifficultyModerate
Exam WeightageAround 7-8 Marks (Combined with Alternating Current)
Learning Objectives
After completing this chapter, students will be able to:
- Understand the concept of Magnetic Flux and how it changes.
- Apply Faraday's Laws of Electromagnetic Induction to calculate induced EMF.
- Use Lenz's Law to find the exact direction of induced currents.
- Calculate Motional EMF across a conductor moving in a magnetic field.
- Differentiate between Self-Inductance and Mutual Inductance and solve related numericals.
- Understand the basic working principle of an AC Generator.
Key Concepts, Definitions and Formulas
- Magnetic Flux ($\Phi_B$): The total number of magnetic field lines passing normally through a given area.
Formula: $$\Phi_B = \mathbf{B} \cdot \mathbf{A} = BA \cos\theta$$
SI Unit: Weber (Wb) - Faraday's Law of Induction: The magnitude of the induced EMF is directly proportional to the rate of change of magnetic flux linked with the circuit.
Formula: $$\varepsilon = -N \frac{d\Phi_B}{dt}$$ - Lenz's Law: The direction of the induced current is such that it opposes the change in magnetic flux that produced it. (This gives the negative sign in Faraday's law).
- Motional EMF: The EMF induced across the ends of a conductor of length $l$ moving with velocity $v$ perpendicular to a magnetic field $B$.
Formula: $$\varepsilon = Blv$$ - Self-Inductance ($L$): The property of a coil by virtue of which it opposes any change in the strength of current flowing through it.
Formula: $$\varepsilon = -L \frac{dI}{dt}$$
Energy Stored: $$U = \frac{1}{2} L I^2$$
SI Unit: Henry (H) - Mutual Inductance ($M$): The property of two neighboring coils where a changing current in one induces an EMF in the other.
Formula: $$\varepsilon_2 = -M \frac{dI_1}{dt}$$
Advertisement Space
Full NCERT Solutions
(Note: These solutions strictly follow the latest rationalized NCERT textbook for 2026-27 Board Exams)
NCERT Q1: 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 $1 \text{ cm s}^{-1}$ 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?
Answer:
Step 1: Note down given values.
Length, $l = 8 \text{ cm} = 0.08 \text{ m}$
Breadth, $b = 2 \text{ cm} = 0.02 \text{ m}$
Velocity, $v = 1 \text{ cm s}^{-1} = 0.01 \text{ m s}^{-1}$
Magnetic Field, $B = 0.3 \text{ T}$
(a) Moving normal to the longer side ($l$):
Here, the EMF is induced across the longer side, so we use $l$ in the formula.
$$\varepsilon = B l v$$
$$\varepsilon = 0.3 \times 0.08 \times 0.01 = 2.4 \times 10^{-4} \text{ V}$$
Time it lasts: The voltage lasts until the shorter side ($b$) completely exits the field.
$$t = \frac{\text{distance}}{\text{velocity}} = \frac{0.02}{0.01} = 2 \text{ seconds}$$
(b) Moving normal to the shorter side ($b$):
Here, the EMF is induced across the shorter side.
$$\varepsilon = B b v$$
$$\varepsilon = 0.3 \times 0.02 \times 0.01 = 0.6 \times 10^{-4} \text{ V}$$
Time it lasts: The voltage lasts until the longer side ($l$) completely exits the field.
$$t = \frac{0.08}{0.01} = 8 \text{ seconds}$$
Step 1: Note down given values.
Length, $l = 8 \text{ cm} = 0.08 \text{ m}$
Breadth, $b = 2 \text{ cm} = 0.02 \text{ m}$
Velocity, $v = 1 \text{ cm s}^{-1} = 0.01 \text{ m s}^{-1}$
Magnetic Field, $B = 0.3 \text{ T}$
(a) Moving normal to the longer side ($l$):
Here, the EMF is induced across the longer side, so we use $l$ in the formula.
$$\varepsilon = B l v$$
$$\varepsilon = 0.3 \times 0.08 \times 0.01 = 2.4 \times 10^{-4} \text{ V}$$
Time it lasts: The voltage lasts until the shorter side ($b$) completely exits the field.
$$t = \frac{\text{distance}}{\text{velocity}} = \frac{0.02}{0.01} = 2 \text{ seconds}$$
(b) Moving normal to the shorter side ($b$):
Here, the EMF is induced across the shorter side.
$$\varepsilon = B b v$$
$$\varepsilon = 0.3 \times 0.02 \times 0.01 = 0.6 \times 10^{-4} \text{ V}$$
Time it lasts: The voltage lasts until the longer side ($l$) completely exits the field.
$$t = \frac{0.08}{0.01} = 8 \text{ seconds}$$
NCERT Q2: A 1.0 m long metallic rod is rotated with an angular frequency of $400 \text{ rad s}^{-1}$ 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.
Answer:
Step 1: Identify the given values.
Length of rod, $l = 1.0 \text{ m}$
Angular frequency, $\omega = 400 \text{ rad s}^{-1}$
Magnetic field, $B = 0.5 \text{ T}$
Step 2: Apply the rotational EMF formula.
The EMF induced across a rotating rod is given by:
$$\varepsilon = \frac{1}{2} B l^2 \omega$$
Step 3: Substitute and calculate.
$$\varepsilon = \frac{1}{2} \times 0.5 \times (1.0)^2 \times 400$$
$$\varepsilon = 0.25 \times 400 = 100 \text{ V}$$
Final Answer: The EMF developed is 100 V.
Step 1: Identify the given values.
Length of rod, $l = 1.0 \text{ m}$
Angular frequency, $\omega = 400 \text{ rad s}^{-1}$
Magnetic field, $B = 0.5 \text{ T}$
Step 2: Apply the rotational EMF formula.
The EMF induced across a rotating rod is given by:
$$\varepsilon = \frac{1}{2} B l^2 \omega$$
Step 3: Substitute and calculate.
$$\varepsilon = \frac{1}{2} \times 0.5 \times (1.0)^2 \times 400$$
$$\varepsilon = 0.25 \times 400 = 100 \text{ V}$$
Final Answer: The EMF developed is 100 V.
NCERT Q3: A circular coil of radius 8.0 cm and 20 turns is rotated about its vertical diameter with an angular speed of $50 \text{ rad s}^{-1}$ in a uniform horizontal magnetic field of $3.0 \times 10^{-2} \text{ T}$. Obtain the maximum and average emf induced in the coil. If the coil forms a closed loop of resistance $10 \Omega$, calculate the maximum value of current in the coil.
Answer:
Step 1: List the given values.
Radius, $r = 8.0 \text{ cm} = 0.08 \text{ m}$
Area, $A = \pi r^2 = 3.14 \times (0.08)^2 \approx 0.0201 \text{ m}^2$
Turns, $N = 20$
Angular speed, $\omega = 50 \text{ rad s}^{-1}$
Magnetic field, $B = 3.0 \times 10^{-2} \text{ T}$
Resistance, $R = 10 \Omega$
Step 2: Calculate Maximum EMF ($\varepsilon_0$).
$$\varepsilon_0 = N A B \omega$$
$$\varepsilon_0 = 20 \times 0.0201 \times (3.0 \times 10^{-2}) \times 50$$
$$\varepsilon_0 = 1000 \times 0.0201 \times 3.0 \times 10^{-2} = 0.603 \text{ V}$$
Step 3: Calculate Average EMF.
Over a full cycle, the average induced EMF in a rotating coil is always Zero.
Step 4: Calculate Maximum Current ($I_0$).
$$I_0 = \frac{\varepsilon_0}{R} = \frac{0.603}{10} = 0.0603 \text{ A}$$
Final Answer: Maximum EMF is 0.603 V, Average EMF is 0 V, and Maximum current is 0.0603 A.
Step 1: List the given values.
Radius, $r = 8.0 \text{ cm} = 0.08 \text{ m}$
Area, $A = \pi r^2 = 3.14 \times (0.08)^2 \approx 0.0201 \text{ m}^2$
Turns, $N = 20$
Angular speed, $\omega = 50 \text{ rad s}^{-1}$
Magnetic field, $B = 3.0 \times 10^{-2} \text{ T}$
Resistance, $R = 10 \Omega$
Step 2: Calculate Maximum EMF ($\varepsilon_0$).
$$\varepsilon_0 = N A B \omega$$
$$\varepsilon_0 = 20 \times 0.0201 \times (3.0 \times 10^{-2}) \times 50$$
$$\varepsilon_0 = 1000 \times 0.0201 \times 3.0 \times 10^{-2} = 0.603 \text{ V}$$
Step 3: Calculate Average EMF.
Over a full cycle, the average induced EMF in a rotating coil is always Zero.
Step 4: Calculate Maximum Current ($I_0$).
$$I_0 = \frac{\varepsilon_0}{R} = \frac{0.603}{10} = 0.0603 \text{ A}$$
Final Answer: Maximum EMF is 0.603 V, Average EMF is 0 V, and Maximum current is 0.0603 A.
NCERT Q4: A horizontal straight wire 10 m long extending from east to west is falling with a speed of $5.0 \text{ m s}^{-1}$, at right angles to the horizontal component of the earth's magnetic field, $0.30 \times 10^{-4} \text{ Wb m}^{-2}$. (a) What is the instantaneous value of the emf induced in the wire? (b) What is the direction of the emf?
Answer:
Step 1: Note the values.
Length, $l = 10 \text{ m}$
Velocity, $v = 5.0 \text{ m s}^{-1}$
Magnetic field, $B = 0.30 \times 10^{-4} \text{ T}$
(a) Calculate induced EMF:
$$\varepsilon = B l v$$
$$\varepsilon = (0.30 \times 10^{-4}) \times 10 \times 5.0$$
$$\varepsilon = 1.5 \times 10^{-3} \text{ V} = 1.5 \text{ mV}$$
(b) Direction of EMF:
Using Fleming's Right-Hand Rule:
Magnetic Field (Forefinger): South to North
Motion (Thumb): Downwards
Induced Current (Middle finger): West to East
Final Answer: Instantaneous EMF is 1.5 mV and the direction is from West to East.
Step 1: Note the values.
Length, $l = 10 \text{ m}$
Velocity, $v = 5.0 \text{ m s}^{-1}$
Magnetic field, $B = 0.30 \times 10^{-4} \text{ T}$
(a) Calculate induced EMF:
$$\varepsilon = B l v$$
$$\varepsilon = (0.30 \times 10^{-4}) \times 10 \times 5.0$$
$$\varepsilon = 1.5 \times 10^{-3} \text{ V} = 1.5 \text{ mV}$$
(b) Direction of EMF:
Using Fleming's Right-Hand Rule:
Magnetic Field (Forefinger): South to North
Motion (Thumb): Downwards
Induced Current (Middle finger): West to East
Final Answer: Instantaneous EMF is 1.5 mV and the direction is from West to East.
NCERT Q5: 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.
Answer:
Step 1: Identify variables.
Change in current, $dI = 0.0 - 5.0 = -5.0 \text{ A}$
Time taken, $dt = 0.1 \text{ s}$
Induced EMF, $\varepsilon = 200 \text{ V}$
Step 2: Use the formula for Self-Inductance.
$$\varepsilon = -L \frac{dI}{dt}$$
Step 3: Calculate $L$.
$$200 = -L \left(\frac{-5.0}{0.1}\right)$$
$$200 = L \times 50$$
$$L = \frac{200}{50} = 4 \text{ H}$$
Final Answer: The self-inductance of the circuit is 4 Henry (H).
Step 1: Identify variables.
Change in current, $dI = 0.0 - 5.0 = -5.0 \text{ A}$
Time taken, $dt = 0.1 \text{ s}$
Induced EMF, $\varepsilon = 200 \text{ V}$
Step 2: Use the formula for Self-Inductance.
$$\varepsilon = -L \frac{dI}{dt}$$
Step 3: Calculate $L$.
$$200 = -L \left(\frac{-5.0}{0.1}\right)$$
$$200 = L \times 50$$
$$L = \frac{200}{50} = 4 \text{ H}$$
Final Answer: The self-inductance of the circuit is 4 Henry (H).
NCERT Q6: 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?
Answer:
Step 1: Note the given data.
Mutual Inductance, $M = 1.5 \text{ H}$
Change in current, $\Delta I = 20 - 0 = 20 \text{ A}$
Step 2: Apply the Flux-Mutual Inductance relation.
The magnetic flux linked with the second coil is directly proportional to the current in the first coil:
$$\Phi = M I$$
Therefore, change in flux $\Delta \Phi = M \Delta I$
Step 3: Calculate.
$$\Delta \Phi = 1.5 \times 20 = 30 \text{ Wb}$$
Final Answer: The change of flux linkage is 30 Weber (Wb).
Step 1: Note the given data.
Mutual Inductance, $M = 1.5 \text{ H}$
Change in current, $\Delta I = 20 - 0 = 20 \text{ A}$
Step 2: Apply the Flux-Mutual Inductance relation.
The magnetic flux linked with the second coil is directly proportional to the current in the first coil:
$$\Phi = M I$$
Therefore, change in flux $\Delta \Phi = M \Delta I$
Step 3: Calculate.
$$\Delta \Phi = 1.5 \times 20 = 30 \text{ Wb}$$
Final Answer: The change of flux linkage is 30 Weber (Wb).
Extra Important Questions (Board Style)
Practicing these mix questions is the secret to board exam phodna. We've handpicked these for your 2026 prep!
Multiple Choice Questions (MCQs)
1. The polarity of induced EMF is found by using:
a) Ampere's law
b) Biot-Savart Law
c) Lenz's Law
d) Fleming's Left Hand Rule
Answer: (c) Lenz's Law.
2. A solid metal block is placed in an alternating magnetic field. The block gets heated up due to:
a) High resistance
b) Eddy currents
c) Mutual induction
d) Hysteresis loss
Answer: (b) Eddy currents.
Explanation: Changing magnetic flux induces circular circulating currents inside bulk metals called eddy currents, which generate heat.
Explanation: Changing magnetic flux induces circular circulating currents inside bulk metals called eddy currents, which generate heat.
3. The unit of self-inductance is equivalent to:
a) Volt-second/Ampere
b) Volt-Ampere/second
c) Ampere-second/Volt
d) Weber/meter
Answer: (a) Volt-second/Ampere.
Explanation: $L = \frac{\varepsilon \cdot dt}{dI}$
Explanation: $L = \frac{\varepsilon \cdot dt}{dI}$
4. Two pure inductors each of inductance $L$ are connected in series. Their equivalent inductance is:
a) $L/2$
b) $L$
c) $2L$
d) $L^2$
Answer: (c) $2L$.
5. Magnetic flux is a:
a) Vector quantity
b) Scalar quantity
c) Phasor
d) Tensor quantity
Answer: (b) Scalar quantity.
Explanation: It is the dot product of two vectors ($\mathbf{B}$ and $\mathbf{A}$).
Explanation: It is the dot product of two vectors ($\mathbf{B}$ and $\mathbf{A}$).
Short Answer Questions (2-3 Marks)
6. State Faraday's laws of electromagnetic induction.
Answer:
* First Law: Whenever there is a change in the magnetic flux linked with a circuit, an EMF is induced in it.
* Second Law: The magnitude of the induced EMF is directly proportional to the rate of change of magnetic flux linked with the circuit ($\varepsilon = -N \frac{d\Phi_B}{dt}$).
Difficulty: Easy
* First Law: Whenever there is a change in the magnetic flux linked with a circuit, an EMF is induced in it.
* Second Law: The magnitude of the induced EMF is directly proportional to the rate of change of magnetic flux linked with the circuit ($\varepsilon = -N \frac{d\Phi_B}{dt}$).
Difficulty: Easy
7. Why does a metallic piece become very hot when it is surrounded by a coil carrying high-frequency alternating current?
Answer: The high-frequency alternating current produces a rapidly changing magnetic field. This changing magnetic flux induces strong eddy currents within the bulk of the metallic piece. Due to the metal's resistance, these eddy currents dissipate a large amount of electrical energy as heat ($I^2R$ loss), making the metal hot. This is the principle of induction heating.
Difficulty: Moderate
Difficulty: Moderate
8. Define Mutual Inductance and state its SI unit.
Answer: Mutual inductance is the property of two coils by virtue of which a change of current in one coil induces an EMF in the neighboring coil. Its SI unit is the Henry (H).
Difficulty: Easy
Difficulty: Easy
Long Answer Questions (5 Marks)
9. Derive an expression for the motional EMF induced across a conductor moving in a uniform magnetic field using the concept of Lorentz force.
Answer Hint:
* Consider a rod of length $l$ moving with velocity $v$ perpendicular to field $B$.
* An electron (charge $e$) inside the rod experiences Lorentz force $F_m = evB$. This force pushes electrons to one end, creating an electric field $E$.
* At equilibrium, electric force $F_e$ balances magnetic force $F_m$. So, $eE = evB \Rightarrow E = vB$.
* The potential difference (EMF) $\varepsilon = E \times l = Blv$.
Difficulty: High
* Consider a rod of length $l$ moving with velocity $v$ perpendicular to field $B$.
* An electron (charge $e$) inside the rod experiences Lorentz force $F_m = evB$. This force pushes electrons to one end, creating an electric field $E$.
* At equilibrium, electric force $F_e$ balances magnetic force $F_m$. So, $eE = evB \Rightarrow E = vB$.
* The potential difference (EMF) $\varepsilon = E \times l = Blv$.
Difficulty: High
10. (a) Define Self-Inductance. (b) Derive an expression for the self-inductance of a long solenoid of cross-sectional area A, length l, and having N turns.
Answer Hint:
* (a) Give definition.
* (b) Magnetic field inside a solenoid $B = \mu_0 n I = \mu_0 \left(\frac{N}{l}\right) I$.
* Flux through one turn $\phi = B \cdot A = \mu_0 \left(\frac{N}{l}\right) I \cdot A$.
* Total flux $\Phi = N \phi = \frac{\mu_0 N^2 A I}{l}$.
* Comparing with $\Phi = L I$ we get $L = \frac{\mu_0 N^2 A}{l}$.
Difficulty: High
* (a) Give definition.
* (b) Magnetic field inside a solenoid $B = \mu_0 n I = \mu_0 \left(\frac{N}{l}\right) I$.
* Flux through one turn $\phi = B \cdot A = \mu_0 \left(\frac{N}{l}\right) I \cdot A$.
* Total flux $\Phi = N \phi = \frac{\mu_0 N^2 A I}{l}$.
* Comparing with $\Phi = L I$ we get $L = \frac{\mu_0 N^2 A}{l}$.
Difficulty: High
11. What are Eddy Currents? Explain any two practical applications of eddy currents.
Answer Hint:
* Eddy currents are loops of electrical current induced within bulk conductors by a changing magnetic field.
* Application 1: Magnetic Braking in Trains: Strong electromagnets are situated above the rails. When activated, the induced eddy currents in the rails oppose the motion of the train, stopping it smoothly.
* Application 2: Induction Furnaces: Used to melt metals. High-frequency AC is passed through a coil surrounding the metal, producing huge eddy currents that generate enough heat to melt the metal.
Difficulty: Moderate
* Eddy currents are loops of electrical current induced within bulk conductors by a changing magnetic field.
* Application 1: Magnetic Braking in Trains: Strong electromagnets are situated above the rails. When activated, the induced eddy currents in the rails oppose the motion of the train, stopping it smoothly.
* Application 2: Induction Furnaces: Used to melt metals. High-frequency AC is passed through a coil surrounding the metal, producing huge eddy currents that generate enough heat to melt the metal.
Difficulty: Moderate
Assertion-Reason Questions
(A = Both true, R is correct explanation. B = Both true, R is not correct explanation. C = A true, R false. D = A false, R true)
12. Assertion: Lenz's law violates the principle of conservation of energy.
Reason: The induced EMF always opposes the change in magnetic flux.
Answer: (D) A is false, R is true.
Explanation: Lenz's law actually supports the conservation of energy. The mechanical work done against the opposing force is converted into electrical energy.
Explanation: Lenz's law actually supports the conservation of energy. The mechanical work done against the opposing force is converted into electrical energy.
13. Assertion: An artificial satellite with a metal surface moving in the equatorial plane of the earth will not experience any induced EMF.
Reason: The satellite's motion is parallel to the earth's magnetic field lines at the equator.
Answer: (A) Both are true, and R is the correct explanation.
Explanation: No flux is cut if motion is parallel to the field lines.
Explanation: No flux is cut if motion is parallel to the field lines.
Case-Based Questions
14. Faraday's Coil and Magnet Experiment:
Faraday performed a classic experiment where he moved a bar magnet towards and away from a coil connected to a galvanometer. He observed deflections only when the magnet was in motion.
Q1. What happens to the galvanometer deflection if the magnet is moved faster?
(Answer: The deflection increases because the rate of change of flux $\frac{d\Phi}{dt}$ is higher, leading to a higher induced EMF).
Q2. If the North pole is pushed into the coil, what polarity does the face of the coil acquire?
(Answer: North polarity, to oppose the incoming North pole according to Lenz's law).
(Answer: The deflection increases because the rate of change of flux $\frac{d\Phi}{dt}$ is higher, leading to a higher induced EMF).
Q2. If the North pole is pushed into the coil, what polarity does the face of the coil acquire?
(Answer: North polarity, to oppose the incoming North pole according to Lenz's law).
15. Transformers and Mutual Induction:
A transformer works on the principle of mutual induction. The primary coil carries an alternating current, which creates a changing magnetic flux that is linked to the secondary coil via a soft iron core.
Q1. Can a transformer operate on a steady Direct Current (DC)?
(Answer: No, steady DC produces a constant magnetic flux. Without a changing flux, no EMF is induced in the secondary coil).
(Answer: No, steady DC produces a constant magnetic flux. Without a changing flux, no EMF is induced in the secondary coil).
Common Mistakes Students Make
- Ignoring the Negative Sign: In Faraday's law ($\varepsilon = -N \frac{d\Phi_B}{dt}$), forgetting the negative sign is a huge mistake. That sign represents Lenz's law!
- Flux Angle Confusion: In $\Phi = BA \cos\theta$, remember that $\theta$ is the angle between the magnetic field vector and the normal (perpendicular) to the area, NOT the surface of the area itself. Concept ekdam clear rakhna!
- Fleming's Rules: Mixing up Left Hand (for Force/Motors) and Right Hand (for Induced Current/Generators). Memory Trick: "Right hand generates, left hand forces."
- Units of Inductance: Writing 'Weber' instead of 'Henry' for Self or Mutual Inductance.
Exam Preparation Tips
- Focus on Lenz's Law Applications: Board exams always feature a 1 or 2-mark conceptual question asking you to determine the direction of induced current in a loop. Practice the Right-Hand Grip rule thoroughly!
- Master the Derivations: The derivation for Motional EMF ($\varepsilon = Blv$) and Self-Inductance of a Solenoid ($L = \mu_0 n^2 A / l$) are incredibly high-yield for 3-mark or 5-mark sections.
- Write Step-by-Step: During the exam, if a numerical feels tough, write down the given values, the formula, and substitute the values. CBSE marks are step-by-step; you'll grab partial marks even if the final calculation goes wrong.
- Revision Strategy: Make a single A4 sheet with all formulas (Flux, Faraday, Motional EMF, $L$, $M$) and stick it on your wall. Look at it every morning.
FAQ Section
Is Electromagnetic Induction important for Class 12 Boards?
Absolutely! Chapter 6 (EMI) combined with Chapter 7 (Alternating Current) forms a major high-scoring unit in the CBSE Class 12 Physics syllabus.
Where can I download NCERT PDF and chapter notes?
You can easily download complete NCERT textbook PDFs, handwritten notes, and solution worksheets directly from examspark.in.
Which questions are most important from EMI?
Derivations of motional EMF, numericals calculating self-inductance, and reasoning questions based on Lenz's Law and eddy currents are the most repeatedly asked questions.
What is the difference between self and mutual inductance?
Self-inductance is the induction of EMF in a single coil due to changing current within itself. Mutual inductance involves two coils, where a changing current in the primary coil induces an EMF in the secondary coil.
How do I figure out the direction of induced current?
Use Lenz's Law combined with the Right-Hand Grip rule. Ask yourself: "What is causing the flux to change?" Then, direct your current to create a magnetic field that fights back against that change!
Chapter 6, Electromagnetic Induction, is the bridge that connects magnetism with electricity in the real world. By understanding these concepts, you aren't just memorizing for an exam—you are learning how the electricity powering your home is actually generated! Keep your concepts clear, practice the Right-Hand rules, and you'll find this chapter incredibly scoring. Stay focused, revise these Updated NCERT Solutions regularly, and definitely practice the PYQs (Previous Year Questions). Remember, bina tension ke board exams phodenge! Download more free notes and mock tests from examspark.in. Keep learning and stay awesome!
More Class 12 Physics Chapters
Ch 1: Electric Charges and Fields
Ch 2: Electrostatic Potential and Capacitance
Ch 3: Current Electricity
Ch 4: Moving Charges and Magnetism
Ch 5: Magnetism and Matter
Ch 6: Electromagnetic Induction
Ch 7: Alternating Current
Ch 8: Electromagnetic Waves
Ch 9: Ray Optics
Ch 10: Wave Optics
Ch 11: Dual Nature of Radiation and Matter
Ch 12: Atoms
Ch 13: Nuclei
Class 12 Mock QUIZs
Ch 14: Semiconductor Electronics