Getting a top score in the Rajasthan Board of Secondary Education (RBSE) Class 12 Physics exam requires a deep understanding of every chapter. Chapter 6, Electromagnetic Induction (EMI), is one of the most crucial and scoring units. This comprehensive guide covers the essential concepts, formulas, and step-by-step solutions needed to secure full marks.
Section 1: Core Concepts of Class 12 Physics Chapter 6 Electromagnetic Induction (EMI)
Electromagnetic Induction is the phenomenon where an electric current is generated in a conductor when it is exposed to a changing magnetic field. Mastering the following foundational concepts is key:
1. Magnetic Flux (ϕB)
- Definition: The measure of the total number of magnetic field lines passing through a given area.
- Formula: ϕB=B
⋅A
=BAcosθ
- B: Magnitude of the magnetic field.
- A: Area of the loop.
- θ: Angle between the magnetic field vector (B
) and the area vector (A
).
- SI Unit: Weber (Wb) or T⋅m2.
2. Faraday’s Laws of Electromagnetic Induction
- First Law: Whenever the magnetic flux linked with a closed circuit changes, an electromotive force (EMF) is induced in the circuit.
- Second Law: The magnitude of the induced EMF (ϵ) is directly proportional to the rate of change of magnetic flux linked with the circuit.
- Formula: ϵ=−NdtdϕB
- N: Number of turns in the coil.
- dtdϕB: Rate of change of magnetic flux.
3. Lenz’s Law and Conservation of Energy
- Lenz’s Law: The polarity of the induced EMF is such that it tends to produce a current which opposes the change in magnetic flux that produced it.
- The negative sign in Faraday’s law represents Lenz’s Law.
- Significance: Lenz’s law is a direct consequence of the law of conservation of energy. The work done in moving the magnet against the repulsive/attractive force is converted into electrical energy.
4. Motional Electromotive Force (EMF)
- Definition: The EMF induced across the ends of a conductor due to its motion in a magnetic field.
- Formula (for a rod of length l moving with velocity v perpendicular to a uniform magnetic field B): ϵ=Blv
5. Eddy Currents
- Definition: Induced circulating currents developed in the body of a bulk metallic conductor when the magnetic flux linked with it changes. They are sometimes called Foucault currents.
- Applications (Uses): Magnetic braking in trains, electromagnetic damping, induction furnace.
- Disadvantages: Cause wastage of energy in the form of heat (Joule heating).
6. Self-Induction and Self-Inductance (L)
- Self-Induction: The phenomenon of the production of an induced EMF in a coil due to a change in the current flowing through the same coil.
- Self-Inductance (L): The ratio of the magnetic flux linked with a coil to the current flowing through it.
- Formula: L=INϕB * Induced EMF:ϵ=−LdtdI
- SI Unit: Henry (H).
7. Mutual Induction and Mutual Inductance (M)
- Mutual Induction: The phenomenon of the production of an induced EMF in a secondary coil due to a change in current in a nearby primary coil.
- Mutual Inductance (M): The ratio of the magnetic flux linked with the secondary coil to the current in the primary coil.
- Formula: M=I1N2ϕB2 * Induced EMF (in coil 2):ϵ2=−MdtdI1
- SI Unit: Henry (H).
Section 2: RBSE Class 12 Physics Chapter 6 Solved Numerical Problems
The theory is best cemented by practicing numerical problems. These solutions cover common patterns from the RBSE syllabus.
Problem 1: Calculation of Induced EMF (Faraday’s Law)
Question: 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 voltage in the loop?
Solution:
- Given:
- Turns per unit length, n=15 turns/cm=1500 turns/m
- Area of loop, A=2.0 cm2=2.0×10−4 m2
- Change in current, dI=(4.0−2.0) A=2.0 A
- Time taken, dt=0.1 s
- Permeability of free space, μ0=4π×10−7 H/m
- Step 1: Magnetic Field in the Solenoid: B=μ0nI
- Step 2: Magnetic Flux (ϕB) linked with the loop: ϕB=BA=(μ0nI)A (Since the loop is normal to the axis, cosθ=1)
- Step 3: Induced EMF (ϵ): ϵ=dtdϕB=dtd(μ0nAI)=μ0nAdtdI
- Step 4: Substitute Values: ϵ=(4π×10−7)×(1500)×(2.0×10−4)×(0.12.0)ϵ=4π×10−7×1500×2.0×10−4×20ϵ≈7.54×10−6 V
Answer: The induced voltage in the loop is 7.54×10−6 V.
Problem 2: Motional EMF
Question: A horizontal straight wire 10 m long extending from East to West is falling with a speed of 5.0 m/s at right angles to the horizontal component of the Earth’s magnetic field, 0.30×10−4 Wb/m2. What is the instantaneous value of the EMF induced in the wire?
Solution:
- Given:
- Length of the wire, l=10 m
- Speed, v=5.0 m/s
- Horizontal component of Earth’s magnetic field, BH=0.30×10−4 T (Note: Wb/m2=T)
- Concept: Since the wire is moving perpendicular to the magnetic field, the motional EMF formula is applicable. The vertical component of the Earth’s field is not relevant here as the wire is falling horizontally (velocity is horizontal), but the horizontal component of the magnetic field acts perpendicular to the wire’s length (E-W) and its velocity (vertical), generating the maximum possible motional EMF.
- Formula: ϵ=Blv
- Substitute Values: ϵ=(0.30×10−4 T)×(10 m)×(5.0 m/s)ϵ=1.5×10−3 V
Answer: The instantaneous EMF induced is 1.5×10−3 V.
Section 3: Important RBSE Board Exam Questions
For a first-rank preparation, focus on these frequently asked theoretical and numerical questions from previous RBSE papers:
| Topic | Type of Question | Marks (Expected) |
| Faraday’s & Lenz’s Law | State and explain Faraday’s laws of EMI. Prove that Lenz’s law is in accordance with the law of conservation of energy. | 2-3 Marks |
| Motional EMF | Derive the expression for motional EMF (ϵ=Blv) for a conducting rod moving in a uniform magnetic field. | 3 Marks |
| Self & Mutual Induction | Define Self-Inductance and Mutual Inductance. Write their S.I. units. Derive an expression for the self-inductance of a long solenoid. | 3-4 Marks |
| AC Generator | State the principle of an AC Generator. Obtain the expression for the induced EMF: ϵ=NBAωsin(ωt). | 4-5 Marks |
| Eddy Currents | What are Eddy Currents? Write any two applications and two disadvantages of Eddy Currents. | 2 Marks |
| Numerical Problems | Problems based on the calculation of induced EMF, Self-Inductance, or Mutual Inductance from given parameters. (e.g., Solenoid/Coil problems, Motional EMF calculations). | 2-3 Marks |
| Chapter 6 | (Open) |
Exam Strategy for Chapter 6
- Formula Sheet: Create a single-page formula sheet covering ϕB, Faraday’s Law, Motional EMF, Self-Inductance (L), and Mutual Inductance (M).
- Lenz’s Law Diagram: Practice the diagrams for Lenz’s law, showing the direction of induced current when a magnet approaches and recedes from a coil. This is a common 2-mark question.
- Derivation Practice: Practice the derivations for Motional EMF and the Self-Inductance of a Solenoid multiple times to ensure accuracy in final RBSE board exams.