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Lecture 3-6 Self Inductance and Mutual Inductance (pg. 36 – 42)

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Presentation on theme: "Lecture 3-6 Self Inductance and Mutual Inductance (pg. 36 – 42)"— Presentation transcript:

1 Lecture 3-6 Self Inductance and Mutual Inductance (pg. 36 – 42)
ECT1026 Field Theory Lecture 3-6 Self Inductance and Mutual Inductance (pg. 36 – 42)

2 3.6 Self Inductance & Mutual Inductance
ECT1026 Field Theory 2007/2008 3.6 Self Inductance & Mutual Inductance Self Inductance inductor When the switch S is closed, the current in the circuit will rise slowly to its final value. Because of this finite rate of rise of current, the magnetic field produced by it also changes with time and, as a result of that, an induced emf is associated with the circuit. A circuit with an inductor and resistor

3 3.6 Self Inductance & Mutual Inductance
ECT1026 Field Theory 2007/2008 3.6 Self Inductance & Mutual Inductance This phenomenon, the appearance of an induced emf in a circuit associated with the changes in its own magnetic field is called self - induction. The corresponding property is called self - inductance. A circuit element, such as a coil, that is designed specifically to have self-inductance is called an inductor. The induced emf appears across the inductor in the circuit. Moreover, an inductor is also a circuit element that stores energy in a magnetic field.

4 3.6 Self Inductance & Mutual Inductance
ECT1026 Field Theory 2007/2008 3.6 Self Inductance & Mutual Inductance Inductor: Similar to electrical capacitor – can store magnetic energy Store energy in the volume comprising the inductors Typical examples: Helical geometry Cylindrical Core

5 3.6 Self Inductance & Mutual Inductance
ECT1026 Field Theory 2007/2008 3.6 Self Inductance & Mutual Inductance Magnetic Flux Linkage,  (pronouns as lambda) Magnetic flux linkage  is defined as the total magnetic flux linking a given circuit:  = N

6 3.6 Self Inductance & Mutual Inductance
ECT1026 Field Theory 2007/2008 3.6 Self Inductance & Mutual Inductance The self-inductance is defined as: Total Flux Linkage Current that produce flux The unit of inductance is the henry (H), equivalent to one Weber-turn per ampere, i.e., 1 H =1 Wb/A

7 3.6 Self Inductance & Mutual Inductance
ECT1026 Field Theory 2007/2008 3.6 Self Inductance & Mutual Inductance The induced emf in a circuit due to self – induction: If I changes, then there is an induced emf around the loop by Faraday’s Law: Self inductance of a circuit depends on its size and shape

8 3.6 Self Inductance & Mutual Inductance
ECT1026 Field Theory 2007/2008 3.6 Self Inductance & Mutual Inductance Parallel Wire Transmission Lines and Coaxial Transmission Lines Two separate conductors, the flux linkage refers to the flux  through the closed surface between the two conductors. To find L, need to determine magnetic flux through area S

9 3.6 Self Inductance & Mutual Inductance
ECT1026 Field Theory 2007/2008 3.6 Self Inductance & Mutual Inductance Mutual Inductance Two coils A and C are placed next to each other When S is closed, some of the flux produced by the current in A becomes linked with C and the emf induced in C circulates a momentary current through the galvanometer G.

10 3.6 Self Inductance & Mutual Inductance
ECT1026 Field Theory 2007/2008 3.6 Self Inductance & Mutual Inductance Mutual Inductance When S is opened, the collapse of the flux induces an emf in the reverse direction in C. (The magnitude and direction of the emf are given by Faraday’s Law and Lenz’s Law). Since a change of current in one coil is accompanied by change of flux linked with the other coil that result the emf induced in the later. The two coils are said to have mutual inductance. Mutual inductance is defined as the ability of one current loop to induce an emf in the adjacent loop when the current in first loop changes. It is measured by a parameter called mutual inductance, the unit is Henry.

11 3.6 Self Inductance & Mutual Inductance
ECT1026 Field Theory 2007/2008 3.6 Self Inductance & Mutual Inductance Expressions for Mutual Inductance (M) Flux 12 through C2 Two coils: Coil 1 (C1): N1 number of turns Coil 2 (C2): N2 number of turns Current of I1 flowing in the coil C1 produces a flux 12 through coil C2. Hence, the mutual inductance between the two coils is: (Henry, H)

12 3.6 Self Inductance & Mutual Inductance
ECT1026 Field Theory 2007/2008 3.6 Self Inductance & Mutual Inductance Expressions for Mutual Inductance (M) On the other hand, M12 is defined as: (Henry) Where N1 = number of loops in loop 1, and is the flux 21 generated by current I2 in loop 2 and linked to loop 1.

13 3.6 Self Inductance & Mutual Inductance
ECT1026 Field Theory 2007/2008 3.6 Self Inductance & Mutual Inductance Expression for mutual inductance M Mutual inductance M21 of coil 2 with respect to coil 1 Mutual inductance M12 of coil 1 with respect to coil 2 The mutual inductance depends upon the geometrical arrangements of the two coils and the permeability of the magnetic region. For a linear magnetic medium such as free space, and if the geometry of the relative arrangement of the coils is the same:

14 3.6 Self Inductance & Mutual Inductance
ECT1026 Field Theory 2007/2008 3.6 Self Inductance & Mutual Inductance Example 3.6-1 Show that the inductance of a long solenoid of N turns, cross-sectional area A, and length l is given by: Solution: Self-inductance: For a long solenoid of N-turns, and length l and carrying a current I, I B n: number of turns per unit length N: number of turns n = N/l

15 3.6 Self Inductance & Mutual Inductance
ECT1026 Field Theory 2007/2008 3.6 Self Inductance & Mutual Inductance Example 3.6-1 Thus, magnetic flux through each turn is: Thus, self –inductance is: (proved)

16 3.6 Self Inductance & Mutual Inductance
ECT1026 Field Theory 2007/2008 3.6 Self Inductance & Mutual Inductance Example 3.6-2 Show that the inductance of a toroid with N turns (of rectangular cross-section) of height h and having mean inner and outer radii of a and b is: h a b Calculate the inductance if N = 450; a = 10cm, b = 20cm, and h = 10cm. Toroid showing Rectangular cross-section

17 3.6 Self Inductance & Mutual Inductance
ECT1026 Field Theory 2007/2008 3.6 Self Inductance & Mutual Inductance Example 3.6-2: Solution Inside the core: B & ds ? Assume air core (o) Recall: Magnetic flux density B flow through the cross section area of the magnetic core is dr h B ? ds ? ds = hdr

18 3.6 Self Inductance & Mutual Inductance
ECT1026 Field Theory 2007/2008 3.6 Self Inductance & Mutual Inductance Example 3.6-2: Solution Calculate the inductance if N = 450; a = 10cm, b = 20cm, and h = 10cm. L = 2.08 mH

19   3.6 Self Inductance & Mutual Inductance Example 3.6-3
ECT1026 Field Theory 2007/2008 3.6 Self Inductance & Mutual Inductance Example 3.6-3 Two coils X and Y of 1000 turns each lie in parallel planes such that 80% of flux produced by one coil linked to another. If a current of 5A flowing in X produces a flux of 0.5mWb in it. Find mutual inductance between X and Y. Solution: Mutual inductance M21 of coil 2 with respect to coil 1 : Magnetic flux linked to Y generated by X Take X as 1 and Y as 2 IX = 5; NX = NY = 1000; YX = 0.80.510-3 Wb MYX = 0.08 H

20 3.6 Self Inductance & Mutual Inductance
ECT1026 Field Theory 2007/2008 3.6 Self Inductance & Mutual Inductance Example – Excise Problem For a long solenoid with N1 turns, length l, and cross section of A is concentric with a second coil with N2 turns, length l1, and cross-section A2. Find the mutual inductance assuming free-space condition.

21 3.6 Self Inductance & Mutual Inductance
ECT1026 Field Theory 2007/2008 3.6 Self Inductance & Mutual Inductance Example – Extra Example Develop an expression for the inductance per unit length of a coaxial transmission line. The conductors have radii a and b, insulating material has a linear permeability m.

22 Cross-sectional view of coaxial transmission line
ECT1026 Field Theory 2007/2008 3.6 Self Inductance & Mutual Inductance Example – Solution Cross-sectional view of coaxial transmission line

23 3.6 Self Inductance & Mutual Inductance
ECT1026 Field Theory 2007/2008 3.6 Self Inductance & Mutual Inductance Example – Solution Due to current I in the inner conductor, magnetic field generated in the region between conductors is: B = m I 2pr Assume long conductor:  B  ds a b  = Flux through the planar surface between conductors: = a b m I 2pr ( l dr ) ds = l dr = m I l 2p ln( ) b a

24 3.6 Self Inductance & Mutual Inductance
ECT1026 Field Theory 2007/2008 3.6 Self Inductance & Mutual Inductance Solutions The inductance per unit length L’ is L’ = L l Recall: L = I = I l = m 2p ln( ) b a Recall: m I l b  = ln( ) 2p a


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