Re: [問題] 高中物理
QvB/2 = Mv^2 /R
v = QBR/2M
K = 0.5 Mv^2 = (QBR)^2 /8M
※ 引述《shmily000 (愛雪兒)》之銘言:
: 電荷為Q,質量為M的質點,以30的夾角進入磁場B中,並作半徑為R的圓周運動時,
: 此帶動質點的動能為
: 2 2 2
: 請問一下這一題答案給 Q B R 有沒有錯
: -------
: 8m
: 我的想法是垂直磁場的速度 = V/2
: 就知道QBR = M(V/2)
: 所以V = 2QBR/M
: 2 2 2
: 2 Q B R
: 因此1/2MV = 2 -----------
: m
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