我是這樣做 w = ω0 ; 之後的頻率ω2 = ω
F = ke^2/r^2 + evB = mv^2/r
其中 ke^2/r^2 = mrw^2 ; evB = eωrB
=> F = mrw^2 + erBω = mrω^2
解一元二次方程式
ω = (eB+sqrt[(eB)^2 +4(mw)^2])/2m = (eB/2m) + sqrt[(eB/2m)^2 + w^2]
考慮弱磁場 也就是 eB/2m << w 根號內的eB/2m 忽略
ω = eB/2m + w
end
※ 引述《pap641217 (Plain&AttractivePhysics)》之銘言：
: 題目在：
: http://tinyurl.com/ye6xzn3
: 怎麼樣就是得不到答案 A ，請網友幫忙提示一下 謝謝！！
--
I am the bone of my paper. Pen is my body, and ink is my blood.
I have finished over a thousand assignments.
Unknown to pass. Nor known to fail.
Have withstood pain to take many midterms.
Yet, those hands will never write anything.
So as I pray, Unlimited homeworks.
--
※ 發信站: 批踢踢實業坊(ptt.cc)
◆ From: 140.115.31.248
※ 編輯: nightkid 來自: 140.115.31.248 (01/27 15:38)