Electromagnetic Induction (SAQs)

Physics-2 | 9. Electromagnetic Induction – SAQs:
Welcome to SAQs in Chapter 9: Electromagnetic Induction. This page includes the most crucial FAQs for Short Answer Questions. Answers are given in simple English, with a Telugu explanation, and formatted according to the exam style. This will help you understand key concepts and aim for top marks in your final exams.

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SAQ-1 : Describe the ways in which Eddy currents are used to advantage.

Introduction

Eddy currents are a phenomenon resulting from the movement of a conductor in an external magnetic field, generating currents when the magnetic flux associated with the conductor changes. These currents, understood through Faraday’s laws of electromagnetic induction, are often considered a loss in many magnetic systems but have numerous beneficial applications.

Applications and Advantages of Eddy Currents

1. Non-Destructive Testing: Eddy currents are crucial in non-destructive testing, identifying defects as small as 0.5 mm. They are highly sensitive to surface defects and can detect anomalies even through surface coatings.
2. Conductivity Measurements: These currents enable precise conductivity measurements, essential in materials science and engineering.
3. Position Sensing: Eddy current sensors, or inductive sensors, are used for measuring the position or change in position of a conductive target, applicable in robotics and automation.
4. Induction Furnaces: In metallurgy, eddy currents are used in induction furnaces for melting metals and other materials.
5. Electric Meters: They play a role in the operation of electric meters, ensuring accurate functioning.
6. Magnetic Braking: Utilized in magnetic braking systems in trains, eddy currents created in the wheels by an electromagnet aid in slowing down the vehicle when brakes are applied.

Summary

Eddy currents, generated by changing magnetic flux, are incredibly versatile and serve various useful purposes across different fields. Harnessing these currents offers significant insights into electromagnetism and its practical implementations, showcasing their importance beyond being mere losses in magnetic systems.

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SAQ-2 : Obtain an expression for the emf induced across a conductor which is moved in a uniform magnetic field which is perpendicular to the plane of motion.

Introduction

Motional electromotive force (emf) is an emf induced in a conductor due to its motion in a magnetic field. This concept is a key aspect of electromagnetism, derived from Faraday’s law of electromagnetic induction. We will explore the calculation of motional emf using a straight wire moving in a magnetic field.

Step-by-Step Calculation of Motional EMF

1. Conductor in Magnetic Field: Consider a straight conducting wire of length AB = l placed in a uniform magnetic field.
2. Wire Movement and Area Change: The wire is moved by a distance x to a new position CD. The magnetic flux (Φ_B​) linked with the loop ABCD can be calculated using Φ_B ​= Blx, where B is the magnetic field strength and A is the area enclosed by the loop, given by lx.
3. Rate of Change in Magnetic Flux: Since the displacement x varies with time, there is a rate of change in the magnetic flux. According to Faraday’s law, this changing flux induces an emf in the conductor. The rate of change of flux is expressed as $$\varepsilon = -\frac{d\Phi_B}{dt}$$
4. Expression for Induced EMF: Substituting Φ_B​ = Blx into Faraday’s law, we get: $$\varepsilon = -\frac{d}{dt}(Blx)$$ Simplifying this, the motional emf (ε) is expressed as: $$\varepsilon = -Blv$$ In this equation, v is the velocity of the conductor, derived from $$\frac{dx}{dt}$$

Summary

The negative sign in the final equation for motional emf indicates that the induced emf opposes its cause, in accordance with Lenz’s law. Understanding motional emf is essential as it has significant applications in various electromechanical systems and devices. This concept plays a pivotal role in the generation of electrical energy in different contexts.

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SAQ-3 : Obtain an expression for the mutual inductance of two long co-axial solenoids.

Introduction

Mutual inductance is a key concept in electromagnetism that describes the ability of one coil to induce an electromotive force (emf) in another coil due to a change in current in the first coil. We will explore the calculation of mutual inductance using two coaxial solenoids as an example.

Step-by-Step Calculation of Mutual Inductance

1. Solenoid Specifications: Consider two long coaxial solenoids, denoted as S1 (inner solenoid) and S2 (outer solenoid). The area of S1 is A₁ with turns per unit length n₁, and S₂ has corresponding values A₂ and n₂.
2. Total Number of Turns: Let N₁ and N₂ be the total number of turns in coils S₁ and S₂, respectively.
3. Magnetic Flux Induction: Assume a current I₂ flows through S₂, inducing a magnetic flux (Φ2Φ2) through S₁. The magnetic field (B₂) due to I₂ in S₂ is given by B₂ = μ₀​n2I2, where �0μ0​ is the permeability of free space.
4. Flux Linkage Calculation: The flux linked with each turn of S1 (Φ2) is B2A1. Therefore, the total flux linked with N1 turns is: Φ2 = B2A1N1 Φ2 = μ₀​n2I2A1N1 This is denoted as equation (i).
5. Mutual Inductance Definition: According to mutual inductance, Φ2 = M12I2, where M12 is the mutual inductance of solenoid 1 with respect to solenoid 2. This is denoted as equation (ii).
6. Deriving Mutual Inductance: By comparing equations (i) and (ii), we obtain: M12 = μ₀​n2N1A1 Substituting N1 = n1l, the expression for M12 becomes: M12 = μ₀​n1n2lA1
7. Mutual Inductance Symmetry: The mutual inductance of solenoid 2 with respect to solenoid 1 (M21) can be similarly calculated and is equal to M12.

Summary

This calculation demonstrates that mutual inductance depends on the physical characteristics of the coils, such as the number of turns and the area, as well as the permeability of the medium. Understanding mutual inductance is vital for the operation of various electrical and electronic devices, including transformers and induction coils.

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SAQ-4 : Obtain an expression for the magnetic energy stored in a solenoid in terms of the magnetic field, area and length of the solenoid.

Introduction

The magnetic energy stored in a solenoid is a key aspect of electromagnetic theory with applications in devices like inductors and transformers. We will derive an expression for the magnetic energy stored in a solenoid, considering the magnetic field (B), cross-sectional area (A), and length (l) of the solenoid.

Step-by-Step Calculation of Magnetic Energy

1. Permittivity Equation: Start with the equation for the permittivity of free space $$\varepsilon_0 = \frac{1}{c^2 \mu_0}$$​ Here, c is the speed of light, and μ₀​ is the permeability of free space. This is equation (1).
2. Electric Field from Magnetic Field: The electric field (E) generated by a magnetic field of strength B is given by: $$E = \frac{B}{c}​$$ This is equation (2).
3. Energy Density in Magnetic Field: The energy stored per unit volume in the magnetic field is given by: $$\frac{U}{Al} = \frac{1}{2} \times \varepsilon_0 \times E^2$$ Substituting values from equations (1) and (2) into this equation: $$\frac{U}{Al} = \frac{1}{2} \times \frac{1}{c^2 \mu_0} \times \left(\frac{B}{c}\right)^2$$
4. Total Magnetic Energy: After simplification, we get the expression for total magnetic energy stored in the solenoid: $$U = \frac{1}{2} \times \frac{1}{\mu_0} \times B^2 \times Al$$ This simplifies to: $$U = \frac{1}{2} \times B^2 \times \mu_0 \times Al$$

Summary

The magnetic energy stored in a solenoid is given by:

$$U = \frac{1}{2} B^2 \mu_0 Al$$

Understanding this equation is crucial for designing electrical components and systems such as transformers and inductors, where the storage of energy in the magnetic field is essential.