Electromagnetic FieldsThis revised edition provides patient guidance in its clear and organized presentation of problems. It is rich in variety, large in number and provides very careful treatment of relativity. One outstanding feature is the inclusion of simple, standard examples demonstrated in different methods that will allow students to enhance and understand their calculating abilities. There are over 145 worked examples; virtually all of the standard problems are included. |
From inside the book
Results 1-3 of 35
Page 429
... propagation is given by ± 2 , corresponding to the sign of k . Suppose that k = 0 ; then we see that it is possible to have E20 and B2 # 0 . But we see from ( 24-17 ) that , in this case , w = 0 as well , so that we actually have a ...
... propagation is given by ± 2 , corresponding to the sign of k . Suppose that k = 0 ; then we see that it is possible to have E20 and B2 # 0 . But we see from ( 24-17 ) that , in this case , w = 0 as well , so that we actually have a ...
Page 439
... propagation constant k , we have 4 = 4。ei ( k5 - wr ) = 4。e1 ( kŵ · r - wt ) ( 24-84 ) Now if we always choose the normal în to be in the direction of propagation of the wave , we can take k to be positive and define a propagation ...
... propagation constant k , we have 4 = 4。ei ( k5 - wr ) = 4。e1 ( kŵ · r - wt ) ( 24-84 ) Now if we always choose the normal în to be in the direction of propagation of the wave , we can take k to be positive and define a propagation ...
Page 443
... propagation . Similarly , the average total energy density becomes ( u ) = a2 2μως Ze - 285 | Eo | 2 = e -285 = 2μv2 Eol2 ( 24-113 ) with the additional use of ( 24-49 ) and ( 24-40 ) . We see that in this case ( S ) = ( u ) vk just as ...
... propagation . Similarly , the average total energy density becomes ( u ) = a2 2μως Ze - 285 | Eo | 2 = e -285 = 2μv2 Eol2 ( 24-113 ) with the additional use of ( 24-49 ) and ( 24-40 ) . We see that in this case ( S ) = ( u ) vk just as ...
Other editions - View all
Common terms and phrases
Ampère's law angle assume axis becomes bound charge boundary conditions bounding surface calculate capacitance capacitor charge density charge distribution charge q circuit conductor consider constant coordinates corresponding Coulomb's law current density curve cylinder defined dielectric dipole direction displacement distance E₁ electric field electromagnetic electrostatic energy equal evaluate example Exercise expression field point flux force free charge free currents frequency function given induction infinitely long integral integrand k₂ Laplace's equation located Lorentz transformation magnetic magnitude material Maxwell's equations normal components obtained origin parallel particle perpendicular plane wave plates point charge polarized position vector potential difference quadrupole quantities radiation radius rectangular region result satisfy scalar scalar potential shown in Figure solenoid sphere spherical tangential components unit vacuum vector potential velocity volume write written xy plane zero Απερ дх Мо