Re: [理工] [電磁]平面波

看板Grad-ProbAsk作者 (biglp)時間14年前 (2011/11/19 11:15), 編輯推噓3(3010)
留言13則, 4人參與, 最新討論串2/3 (看更多)
※ 引述《big1p (biglp)》之銘言: : 1. : _ _ : A uniform plane wave with E = axEx propagates in a lossless simple medium : (εr=4, μr=1, σ=0) in the +z direction. Assume that Ex is sinusoidal with a : -4 : frequency 100 MHz and has a maximum value of 10 (V/m) at t=0 and z=1/8(m). : -8 : Determine the locations where Ex is a positive maximum when t=10 (s). : 2. : The instantaneous expression for the magnetic field intensity of a uniform : plane wave propagating in the +y direction in air is given by : _ _ -6 7 : H = az4*10 cos(10πt-ky+π/4) (A/m) : Determine the locations where Hz vanishes at t=3(ms). : 對這種題目沒什麼概念.. : 煩請各位高手指點一下 第一題會了 第二題的詳解是 π π π -π k= ---- (rad/s), t=0 時, H=0 => -ky + --- = + --- => ky = --- 30 4 2 4 -λ n 得 y= --- ±---λ, n=0,1,2,3... 之後把λ算出來代進去 8 2 y=-7.5±30n (m), n=1,2,3... 有點看不懂 麻煩指點一下謝謝 -- ※ 發信站: 批踢踢實業坊(ptt.cc) ◆ From: 111.249.155.142

11/20 00:05, , 1F
我試著代了時間項 為2pi的整數倍 用cos和角公式可以去掉
11/20 00:05, 1F

11/20 00:06, , 2F
剩後面-ky+π/4項 再來就令cos為0找出答案就ok
11/20 00:06, 2F

11/20 09:14, , 3F
詳解過程似乎跟時間無關...而且你的方法也有點不懂耶..
11/20 09:14, 3F

11/20 21:44, , 4F
磁場是振幅項跟cos項相乘 因為振幅為常數不為零 所以只
11/20 21:44, 4F

11/20 21:44, , 5F
能令後面cos項為0 所以就看使cos為零的條件為哪些
11/20 21:44, 5F

11/20 21:46, , 6F
[(10^7π*3*10^-3) - ky+π/4] 在cos項裡=0 用和角展開
11/20 21:46, 6F

11/20 21:48, , 7F
變成 cos(ky+π/4)=0 所以ky+π/4 = (2n-1)π/2 n=自然數
11/20 21:48, 7F

11/20 22:12, , 8F
會想不到用和角展開耶..可以直接令整個=(n-1/2)π嗎?
11/20 22:12, 8F

11/20 22:12, , 9F
這樣的話前面那一項是3*10^4次方耶..怎麼算阿 ?
11/20 22:12, 9F

11/20 22:43, , 10F
就是直接令(n-1/2)π,解出來沒錯
11/20 22:43, 10F

11/21 00:31, , 11F
其實是一樣的意思 只是10^7π*3*10^-3這項不覺得很大嗎XD
11/21 00:31, 11F

11/21 00:32, , 12F
只是把他消去而已 主要觀念還是就磁場消失 所以讓他=0
11/21 00:32, 12F

09/11 14:36, , 13F
剩後面-ky+π/4項 https://daxiv.com
09/11 14:36, 13F
文章代碼(AID): #1Ennz9kE (Grad-ProbAsk)
討論串 (同標題文章)
文章代碼(AID): #1Ennz9kE (Grad-ProbAsk)