vector which points in the direction of prop-agation and describes how energy propagatesin the wave,vectorS=1μ0vectorE×vectorB.Electromagnetic waves carry energy.Theenergy density of any electric field is given by,uE=12ǫ0E2.The energy density of any magnetic field isgiven by,uB=12B2μ0.So the total energy density of an electro-magnetic wave is given by the sum of thesetwo. For an electromagnetic wave, since the

kim (jk36442) – EM waves and particles – yeazell – (58005)6electric and magnetic fields oscillate bothuEanduBoscillate.Inthefollowingproblems,consideramonochromatic electromagnetic plane wavepropagating in thexdirection.At a par-ticular point in space, the magnitude of theelectric field has an instantaneous value of149 V/m in the positivey-direction.Thewave is traveling in the positivex-direction.xyzEwave propagationLet :μ0= 4π×10−7m·N/A.The Poynting vector isvectorS=1μ0vectorE×vectorB .Fora plane, electromagnetic wave,vectorEandvectorBarealways perpendicular to each other and to thedirection of propagation of the wave; in thiscase, it is in the direction of propagation andhas magnitudeS=E Bμ0=(149 V/m) (4.97011×10−7T)4π×10−7m·N/A=58.9308 W/m

Find the instantaneous magnitude of themagnetic field at the same point and time.Thespeedoflightis2.99792×108m/s,thepermeabilityoffreespaceis4π×2.015(part3of5)10.0pointsWhat are the directions of the instantaneousmagnetic field and the instantaneous Poynt-ing vector respectively?Explanation:Let :μ0= 4π×10−7m·N/A.The Poynting vector isvectorS=1μ0vectorE×vectorB .Fora plane, electromagnetic wave,vectorEandvectorBarealways perpendicular to each other and to thedirection of propagation of the wave; in thiscase, it is in the direction of propagation andhas magnitudeS=E Bμ0=(149 V/m) (4.97011×10−7T)4π×10−7m·N/A=58.9308 W/m

kim (jk36442) – EM waves and particles – yeazell – (58005)7in the ˆıdirection,vectorBmust be in theˆkdirec-tion. The Poynting vector is in the directionof propagation of the wave: ˆı.