Slide 1: Faraday’s Law of Induction - Mutual and Self-Inductance - Example 2

• In the last lecture, we discussed Faraday’s Law of Induction and self-inductance of a solenoid.
• Today, we will continue our discussion on Faraday’s Law of Induction and move on to mutual inductance.
• We will also solve an example problem to understand the concept better.
• Let’s begin!

Slide 2: Mutual Inductance

• Mutual inductance refers to the induction of an electromotive force (emf) in one coil due to a changing magnetic field produced by another coil.
• It is the property of a pair of coils that determines the amount of voltage induced in one coil when the current through the other coil changes.
• Symbolically, mutual inductance is represented by the letter M.
• The unit of mutual inductance is henry (H).

Slide 3: Calculation of Mutual Inductance

• The mutual inductance between two coils can be calculated using the formula: M = (N2 * Φ1) / I1 where N2 is the number of turns in the second coil, Φ1 is the magnetic flux produced by the first coil, and I1 is the current flowing through the first coil.

Slide 4: Example Problem - Coaxial Solenoids

• Consider two coaxial solenoids S1 and S2.
• S1 has N1 turns and carries a current I1.
• S2 has N2 turns and is placed inside S1 with a distance of separation ’d’ between them.
• We want to find the mutual inductance between these two coils.

Slide 5: Step 1 - Magnetic Field Produced by S1

• We start by calculating the magnetic field produced by the solenoid S1.
• The magnetic field at any point inside a solenoid is given by the equation: B1 = (μ0 * N1 * I1) / l1 where μ0 is the permeability of free space, N1 is the number of turns in S1, I1 is the current flowing through S1, and l1 is the length of S1.

Slide 6: Step 2 - Magnetic Flux through S2

• Next, we need to calculate the magnetic flux (Φ1) through the second solenoid S2.
• The magnetic flux through the second solenoid can be calculated using the equation: Φ1 = B1 * A2 where B1 is the magnetic field produced by S1 and A2 is the cross-sectional area of S2.

Slide 7: Step 3 - Calculating Mutual Inductance

• Now that we have calculated the magnetic flux, we can substitute the values into the formula for mutual inductance: M = (N2 * Φ1) / I1
• Substituting the values, we get: M = (N2 * B1 * A2) / I1

Slide 8: Mutual Inductance - Example problem

• Let’s consider a specific example to understand the calculation of mutual inductance involving coaxial solenoids.
• Solenoid S1 has 200 turns, carries a current of 2 A, and has a length of 0.4 m.
• Solenoid S2 has 400 turns and is placed inside S1 with a separation of 0.2 m.
• We want to find the mutual inductance between these two solenoids.

Slide 9: Step 1 - Magnetic Field Produced by S1

• Using the formula, we can calculate the magnetic field produced by S1: B1 = (μ0 * N1 * I1) / l1 Substituting the given values, we get: B1 = (4π * 10^-7 Tm/A * 200 turns * 2 A) / 0.4 m

Slide 10: Step 2 - Magnetic Flux through S2

• Next, let’s calculate the magnetic flux through S2: Φ1 = B1 * A2 Substituting the known values, we get: Φ1 = B1 * (π * (0.1 m)^2)

11. Step 3 - Calculating Mutual Inductance (Continued)

• Now that we have calculated the magnetic flux, we can substitute the values into the formula for mutual inductance: M = (N2 * Φ1) / I1
• Substituting the values from the previous step, we get: M = (400 turns * B1 * A2) / 2 A

12. Step 3 - Calculating Mutual Inductance (Continued)

• Simplifying the expression, we have: M = (200 turns * B1 * π * (0.1 m)^2) / A

13. Step 3 - Calculating Mutual Inductance (Continued)

• Substituting the value of B1 we calculated previously, we get: M = (200 turns * (4π * 10^-7 Tm/A * 200 turns * 2 A) / 0.4 m) * π * (0.1 m)^2 / A

14. Step 3 - Calculating Mutual Inductance (Continued)

• Simplifying further, we find: M = (4 * 10^-4 π Tm) / A

15. Step 3 - Calculating Mutual Inductance (Continued)

• Since the unit of mutual inductance is henry (H), we convert the above expression to the desired unit: M = 4 * 10^-4 H

16. Summary of Mutual Inductance Example Problem:

• We have found that the mutual inductance between the given coaxial solenoids is 4 * 10^-4 H.
• This indicates that a changing current in one coil will induce an emf of 4 * 10^-4 volts in the other coil.

17. Applications of Mutual Inductance:

• Mutual inductance plays a crucial role in the functioning of transformers and various other electrical devices.
• It allows for efficient transmission of electrical energy and voltage transformation between different circuits.
• Mutual inductance is also utilized in inductive proximity sensors and electromagnetic compatibility testing.

18. Self-Inductance: Recap

• Before moving on, let’s recap self-inductance briefly.
• Self-inductance refers to the property of a circuit element, typically an inductor, to resist changes in the current flowing through it.
• It is represented by the symbol L and measured in henry (H).

19. Calculating Self-Inductance:

• The self-inductance of an inductor can be calculated using the formula: L = (μ0 * N^2 * A) / l where μ0 is the permeability of free space, N is the number of turns, A is the cross-sectional area of the coil, and l is the length of the coil.

20. Summary:

• In this lecture, we discussed mutual inductance and learned how to calculate it using the appropriate formulas.
• We also solved an example problem involving mutual inductance between two coaxial solenoids.
• Finally, we recapped self-inductance and the formula to calculate it.
• In the next lecture, we will dive deeper into the applications of mutual inductance and explore more related topics.

21. Applications of Mutual Inductance (Continued):

• Mutual inductance is the basis for the functioning of transformers, which are essential in electrical power distribution systems.
• Transformers use the concept of mutual inductance to efficiently transfer electrical energy between different voltage levels.
• Mutual inductance allows for step-up or step-down voltage transformations, making it possible to transmit electricity over long distances with minimal losses.
• Inductive charging systems for electric vehicles also utilize mutual inductance to wirelessly transfer energy from a charging pad to the vehicle’s battery.

22. Mutual Inductance in Electric Circuits:

• Mutual inductance affects the behavior of interconnected circuits.
• When two coils are inductively coupled, a change in current in one coil induces an emf in the other coil, leading to a redistribution of current in the system.
• This concept is employed in circuits with mutual inductance to achieve desired characteristics, such as oscillations in LC circuits or signal coupling in amplifiers.

23. Magnetic Field Energy in Coils:

• Mutual inductance is closely related to magnetic field energy stored in coils.
• When current flows through a coil, it generates a magnetic field and stores energy in that field.
• If there is a change in current, the magnetic field changes, leading to a change in magnetic energy.
• Mutual inductance determines how much of the magnetic energy is transferred between coils and how it influences the behavior of the circuit.

24. Coupled Inductors:

• Coupled inductors are two or more inductors that are magnetically linked or physically positioned close to each other.
• By adjusting the arrangement and values of the inductors, engineers can control the amount of mutual inductance and influence circuit behavior.
• Coupled inductors are extensively used in power converters, where they play a vital role in regulating voltage levels, reducing noise, and improving efficiency.

25. Calculation of Mutual Inductance for Other Geometries:

• The formula we discussed earlier assumes coils with ideal geometric properties (solenoids).
• For other geometries, such as circular or rectangular loops, the calculation of mutual inductance becomes more complex.
• However, similar principles can be applied, considering the overlapping area, distance between coils, and the direction of current flow.

26. Factors Affecting Mutual Inductance:

• The value of mutual inductance depends on various factors:
  • Number of turns in each coil: Increasing the number of turns increases the mutual inductance.
  • Area of overlap between coils: A larger overlap area leads to a higher mutual inductance.
  • Separation distance between coils: A smaller distance between coils increases the mutual inductance.
  • Relative orientations of coils: The alignment and orientation of coils affect the mutual inductance.

27. Induced EMF in Mutual Inductance:

• The emf induced in a coil due to mutual inductance can be calculated using Faraday’s Law of Induction.
• The induced emf (ε) is given by the equation: ε = -M(dI1/dt)
• Here, dI1/dt represents the rate of change of current in the first coil, and M is the mutual inductance between the coils.

28. Limitations of Mutual Inductance:

• Mutual inductance assumes ideal conditions and neglects factors like magnetic leakage and resistance.
• Magnetic leakage refers to the magnetic field lines that do not intersect the other coil, reducing the effective mutual inductance.
• Resistance in the coils can cause energy losses and affect the behavior of the system.
• These factors need to be considered for more accurate predictions and practical applications.

29. Summary:

• In this lecture, we explored the concept of mutual inductance and learned how to calculate it for coaxial solenoids.
• We discussed its applications in transformers, electric circuits, and energy transfer systems.
• We also touched upon coupled inductors, factors affecting mutual inductance, induced EMF, and the limitations of the concept.
• Understanding mutual inductance is essential for analyzing electromagnetic systems and designing efficient electrical devices.

30. Next Lecture:

• In the next lecture, we will delve into the topic of electromagnetic waves and their properties.
• We will explore the nature of electromagnetic waves, their speed, behavior in different media, and various applications.
• Be prepared for a comprehensive discussion on one of the fundamental aspects of physics.