Question

10.44 The electric field component of a uniform plane wave traveling in seawater (o = 4 S/m, & 81 8 M = Mo) is E = 8e - 0.12 cos (wt - 0.3z)a, V/m

Answer 1

0 Civen in this question Electric field -0.13 E = 8e cos (wt - 0.37 37) ax y/om where, 6= 45/m, 8160 and u = lo 7 Mo=4a *10 Mm fo= 8.854x10 pm a) NOW power densing calculating P = √ Real Ex H az using this formula 0:12 8e where, E ios (+-0.32) H = cals е lu where in te -0.12 8 te cost wt-0.32) ori HE MO 81 to -0.17 8*e OY ws (W+-0.32) H - 4** 107 81*8.854 *152 88€ 0.12 OY H cos (W+-032) 41.86 Now, H rame puting in equation 0 and henu

-0.12 P= Real 80012 se cos(WF-0.32) * ws(W7-0.32) 41.86 or, -0.22 p= 1 * 8²*e Tan 41.86 -022 az P = 0.76 power density of this enctrical : This is the field, the depth power is reduce b) when Now finding to 20dB. That means 2008 = 10²=100. density 1. NOW, Recut E2414) = 2008 2100 OY, -0.27 1.sze = 200 03 , -0.27 e = 65.41 -0.27 or, in 165:41) 11 4.18 or, k= -20.90 : the is depth 20.90 ( Because depth is not negerive vam) when the power de density is reduce to 2008. 57