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EMF2016 ELECTROMAGNETIC THEORY

EMF2016 ELECTROMAGNETIC THEORY. MAXWELL EQUATIONS. Maxwell equations in differential form. (Gauss’s law for electric field). where E = electric field intensity D = electric flux density H = magnetic field intensity B = magnetic flux density

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EMF2016 ELECTROMAGNETIC THEORY

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  1. EMF2016ELECTROMAGNETIC THEORY MAXWELL EQUATIONS

  2. Maxwell equations in differential form (Gauss’s law for electric field) where E = electric field intensityD = electric flux density H = magnetic field intensityB = magnetic flux density v = free electric charge volume density J = free conduction current density (Gauss’s law for magnetic field) (Faraday’s law) (Ampere’s law)

  3. Constitutive equations where  is medium permittivity  is medium permeability

  4. Static fields Electrostatic  .D= v E = 0 Magnetostatic .B = 0 H = J

  5. Gauss’s law for electric field Integrate over a volume V gives Apply divergence theorem to the left integral, It says “The total outflow of electric flux from a closed surface S equals the total free charge enclosed”, which is commonly known as Gauss’s law for electric field.

  6. Gauss’s law for electric field – cont. The Gauss’s law integral can also be written in terms of E, where

  7. Gauss’s law for magnetic field Integrate over a volume V gives Apply divergence theorem to the left integral, It says “There is no net outflow of magnetic flux from any closed surface S”, which implies that the net magnetic charge enclosed by any closed surface is always zero, i.e. there is no isolated magnetic charge, magnetic charges always appear in pair.

  8. Maxwell-Faraday’s law Integrate over a surface area S gives Apply Stoke’s theorem to the left integral,

  9. Maxwell-Faraday’s law – cont. If the loop is stationary, the partial differentiation can be changed to ordinary differentiation, which is generally known as Faraday’s law.

  10. Maxwell-Faraday’s law – cont. • The above formulation has assumed that the loop C is stationary, which is necessary for changing the partial differentiation to ordinary differentiation. • Hence, the resulting electromotive force (emf) is solely due to the time varying magnetic field and is generally known as transformer emf. • This is different from the original Faraday’s law which includes motional emfdue to moving conductor.

  11. Maxwell-Ampere’s law Integrate over a surface area S gives Apply Stoke’s theorem to the left integral, which is generally known as Ampere’s law.

  12. Maxwell equations in integral form (Gauss’s law for electric field) (Gauss’s law for magnetic field) (Faraday’s law) (Ampere’s law)

  13. Kirchhoff’s voltage law The Maxwell-Faraday equation, implies that the sum of voltages around a loop is not zero but equals the emf due to electromagnetic induction, If the voltage sum includes the emf, we still have the Kirchhoff’s voltage law

  14. Kirchhoff’s current law Let The Maxwell-Ampere equation becomes It follows that

  15. Kirchhoff’s current law – cont. Applying divergence theorem, It says “The total outflow current from any volume (including circuit node) is always zero”, which is Kirchhoff’s current law,

  16. Faraday’s law • An electromotive force (emf) will be induced in a loop only if the magnetic flux linking the loop surface CHANGES with time. • The magnitude of the emf induced is equal to the negative time rate of change of the magnetic flux linkage.

  17. Lenz’s law The induced current is always in such a direction as to APPOSE the CHANGE of magnetic flux that produced it.

  18. Faraday’s law: Example I [Ulaby, pp. 232]

  19. Stationary loop in time-varying magnetic field [Ulaby, Figure 6-2, pp. 234]

  20. Transformer emf [Ulaby, pp. 234] The current I flowing through the circuit is given by

  21. Transformer emf: Example 1 An inductor is formed by winding N turns of a thin conducting wire into a circular loop of radius a. The inductor loop is in the x-y plane with its center at the origin, and it is connected to a resistor R, as shown in Fig. 6-3. In the presence of a magnetic field given by B =B0 (y2 + z3) sin(t), where  is the angular frequency. Find (a) the magnetic flux linking a single turn of the inductor, (b) the emf, given that N = 10, B0 = 0.2 T, a = 10 cm, and  =103 rad/s, (c) the polarity of Vemftr at t = 0, and (d) the induced current in the circuit for R = 1 k (assume the wire resistance to be negligibly small).

  22. Transformer emf: Example 1 – cont. [Ulaby, pp. 235]

  23. Transformer emf: Example 1 – cont. Solution: (a) The magnetic flux linking each turn of the inductor is  = sBds = s [B0 (y2 + z3) sin t] zds= 3a2B0sint. (Because ds = z ds, the resulting Vemf will be counter clockwise) (b) From Faraday’s law, Vemf = N dF/dt= d(3Na2B0sint)/dt = 3Na2B0cost. For N = 10, a = 0.1 m,  = 103 rad/s, and B0 = 0.2 T, Vemf = 188.5 cos(103t) (V).

  24. Transformer emf: Example 1 – cont. c) Vemf= V1 – V2 = 188.5 (V) Hence, point 2 is at higher potential then point 1. Alternatively, at t = 0, d/dt > 0 and Vemf= 188.5 V. Because the flux is increasing, the current I must be in the direction as shown in the Figure to satisfy Lenz’s law. Consequently, point 2 is at a higher potential than point 1. d) Because I is defined clockwise, I= Vemf/R= (188.5/103) cos(103t) = 0.19 cos(103t) (A).

  25. Transformer emf: Example 2 Determine the voltage V1 and V2 across the 2- and 4- resistors shown in figure below. The loop is located in the x-y plane, its area is 4m2, the magnetic flux density is B= -z0.3t (T), and the internal resistance of the wire may be ignored. [Ulaby, pp. 236]

  26. Transformer emf: Example 2 – cont. Solution:  = sBds = s(-z0.3t)  zds = -0.3t x 4 = -1.2t Vemftr= -d/dt = 1.2 (counter-clockwise) Because I is defined counter-clockwise, I= Vemftr / (R1 + R2) = 1.2 / (2 + 4) = 0.2 A The voltages across R1 and R2 are V1= I R1= 0.2 2 = 0.4 V V2= I R2= 0.2 4 = 0.8 V

  27. Transformer emf: Example 3 • A coil consists of 100 turns of wire wrapped around a square frame of sides 0.25 m. The coil is centred at the origin with each of its sides parallel to the x- or y- axis. Find the induced emf across the open circuited ends of the coil if the magnetic field is given by • i) • ii) (T) (T)

  28. Transformer emf: Example 3 – cont. (a)

  29. Transformer emf: Example 3 – cont. (b)

  30. Lorentz force equation (electric force) (magnetic force) (electromagnetic force on moving charge)

  31. Current carrying conductor in static magnetic field [Ulaby, Figure 5.2, pp. 190 ]

  32. Moving conductor in static magnetic field [Ulaby, pp. 238]

  33. Motional emf

  34. The electric motor and generator (Source: The Ulaby textbook, p. 307)

  35. Motional emf: Example 1 The rectangular loop shown in Fig. 6-8 has a constant width l, but its length xincreases with time as a conducting bar slides at a uniform velocity u in a static magnetic field B= zB0x. Note that B increases linearly with x. The bar starts from x = 0 at t = 0. Find the motional emf between terminals 1 and 2 and the current I flowing through the resistor R. Assume that the loop resistance Ri<< R.

  36. Motional emf: Example 1 – cont. [Ulaby, pp. 239]

  37. Motional emf: Example 1 – cont. Solution: The sliding bar, being the only part of the circuit that crosses the lines of the field B , is the only part of contour (2341) that contributes to motional emf. (in terms of sliding bar position x) (in terms of time t)

  38. Moving conductor in time-varying magnetic field For the general case of a single-turn conducting loop moving in a time-varying magnetic field, the total induced emf is the sum of a transformer emf component and a motional emfcomponent. It can be shown that the sum is equivalent to the emf given by the original Faraday’s law.

  39. Charge-current continuity By the principle of charge conservation, the time rate of decrease of charge within a given volume must be equal to the net outward current flow through the closed surface of the volume, In differential form,

  40. Incompleteness of original Ampere’s law Original Ampere’s law in differential form, The divergence of the curl of any vector field is identically equal to zero, This implies It follows that, by conservation of charges, But, observations have shown that v/t is not necessarily zero.

  41. Displacement current To solve the previous Ampere’s law problem, Maxwell added a term so that It follows that, By conservation of charges, By Gauss’s law, Therefore, Jdis termed displacement current density.

  42. Displacement current: Example A parallel plate capacitor with plate area of 5 cm2and plate separation of 3 mm has a voltage 50 sin(103t) V applied to its plates. Calculate the displacement current assuming  = 2o . A y  Id d  x V ~ + Solution:

  43. Displacement current: Example – cont. Solution: (continued)

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