306 Electromagnetic WavesAdditional ProblemsP34.47(a)P=SA:=×LNMOQP1 3404 1 496 103 77 1011226WmmW2ejejπ..(b)ScB=max202µsoBScmax..==××=−224101340300 10335078NAmsT22ejSEc=max202soEcSmax××=−22 4103 00 101 3401 01078µπejbgkV mP34.48Suppose you cover a 1.7 m-by-0.3 m section of beach blanket. Suppose the elevation angle of the Sunis 60°. Then the target area you fill in the Sun’s field of view is17033004cos.m mm2af°=.Now IAUAtUIAt06 05 0436006msJ...~.P34.49(a)εθ=−ddtddtBABΦcosafεωθω=AddtBtABtmaxmaxcoscossincosbgεπtfBAft=maxsincostrfBft=maxcos sinThus,maxmaxcos=2rfBwhere is the angle between the magnetic field and the normal to the loop.(b)If Eis vertical, Bis horizontal, so the plane of the loop should be verticaland the plane should contain the line of sight of the transmitter .P34.50(a)FGM mRGMRrSSgrav
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This note was uploaded on 12/14/2011 for the course PHY 203 taught by Professor Staff during the Fall '11 term at Indiana State University .