1 / 31

ENE 325 Electromagnetic Fields and Waves

ENE 325 Electromagnetic Fields and Waves. Lecture 6 Capacitance and Magnetostatics. Review (1). Conductor and boundary conditions tangential electric field, E t , = 0 for equipotential surface. normal electric flux density D n =  s . Dielectric

jersey
Télécharger la présentation

ENE 325 Electromagnetic Fields and Waves

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript


  1. ENE 325Electromagnetic Fields and Waves Lecture 6 Capacitance and Magnetostatics

  2. Review (1) • Conductor and boundary conditions • tangential electric field, Et, = 0 for equipotential surface. • normal electric flux density Dn = s. • Dielectric • macroscopic electric dipoles and bound charges • Polarization is dipole moment per unit volume, • electric flux density in dielectric medium,

  3. Review (2) • Dielectric and boundary conditions • tangential electric field, Et1= Et2 • normal electric flux density .

  4. Outline • Capacitance • Static magnetic fields • Bio-Savart ‘s law • Magnetic field in different current configurations

  5. Capacitance • Capacitance depends on the shape of conductor and the permittivity of the medium. Capacitance has a unit of Farad or F. From then

  6. Capacitance for parallel plate configuration At lower plate, then The potential difference Let A = the plate area then Q = sA then

  7. Total energy stored in the capacitance

  8. Ex1 Determine the relative permittivity of the dielectric material inserted between a parallel plate capacitor if • C = 40 nF, d = 0.1 mm, and A = 0.15 m2 • d = 0.2 mm, E = 500 kV/m, and s = 10 C/m2

  9. Capacitance in various charge distribution configurations(1) • Coaxial cable Use Gauss’s law,

  10. Capacitance in various charge distribution configurations(2) • Sphere Use Gauss’s law,

  11. Capacitance in various charge distribution configurations(3) • A parallel plate capacitor with horizontal dielectric layers

  12. Capacitance in various charge distribution configurations(4) • A parallel plate capacitor with vertical dielectric layers

  13. Ex5 From the parallel capacitor shown, Find the total capacitance.

  14. Static magnetic fields (Magnetostatics)

  15. Introduction(1) • source of the steady magnetic field may be a permanent magnet, and electric field changing linearly with time or a direct current. • a schematic view of a bar magnet showing the magnetic field. Magnetic flux lines begin and terminate at the same location, more like circulation.

  16. Introduction(2) • Magnetic north and south poles are always together.

  17. Introduction(3) • Oersted’s experiment shows that current produces magnetic fields that loop around the conductor. The field grows weaker as one compass moves away from the source of the current.

  18. Bio-Savart law (1) • The law of Bio-Savart states that at any point P the magnitude of the magnetic field intensity produced by the differential element is proportional to the product of the current, the magnitude of the differential length, and the sine of the angle lying between the filament and a line connecting the filament to the point P at which the filed is desired. • The magnitude of the magnetic field intensity is inversely proportional to the square of the distance from the differential element to the point P.

  19. Bio-Savart law (2) • The direction of the magnetic field intensity is normal to the plane containing the differential filament and the line drawn from the filament to the point P. • Bio-Savart law is a method to determine the magnetic field intensity. It is an analogy to Coulomb’s law of Electrostatics.

  20. Bio-Savart law (3) from this picture: Total fieldA/m

  21. Magnetic field intensity resulting from an infinite length line of current (1) Pick an observation point P located on  axis. The current The vector from the source to the test point is a unit vector

  22. Magnetic field intensity resulting from an infinite length line of current(2) then From a table of Integral, then

  23. Magnetic field intensity resulting from a ring of current (1) • A ring is located on z = 0 plane with the radius a. The observation point is at z = h.

  24. Magnetic field intensity resulting from a ring of current (2) A unit vector

  25. Magnetic field intensity resulting from a ring of current (3) Consider a symmetry • components are cancelled out due to symmetry of two segments on the opposite sides of the ring. Therefore from we have

  26. Magnetic field intensity resulting from a ring of current (4) then We finally get

  27. Magnetic field intensity resulting from a rectangular loop of current(1) Find the magnetic field intensity at the origin. By symmetry, there will be equal magnetic field intensity at each half width (w/2).

  28. Magnetic field intensity resulting from a rectangular loop of current(2) Consider 0  x  w/2, y = -w/2 A unit vector We have

  29. Magnetic field intensity resulting from a rectangular loop of current(3) Then the total magnetic field at the origin is Look up the table of integral, we find then A/m.

  30. Bio-Savart law in different forms • We can express Bio-Savart law in terms of surface and volume current densities by replacing with and : where K = surface current density (A/m) I = K x width of the current sheet and

  31. Right hand rule • The method to determine the result of cross product (x).

More Related