从题设条件无法判定粒子从a和b哪个边界射出,应讨论如下:
(1)若带电粒子从边界a射出,轨迹恰好与边界b相切时粒子轨迹如图所示:
由几何知识可得:L=R+Rcosθ…①
又Bvq=
…②
解得R=
,L=
(1+coθ),可见若L≥
(1+cosθ)时,粒子将从a边界射出,由图可知粒子轨迹对应的圆心角为(2π-2θ),
所以粒子所用时间为:t=
?T=
![](https://video.ask-data.xyz/img.php?b=https://iknow-pic.cdn.bcebos.com/b999a9014c086e06e74b28cf01087bf40bd1cb57?x-bce-process=image%2Fresize%2Cm_lfit%2Cw_600%2Ch_800%2Climit_1%2Fquality%2Cq_85%2Fformat%2Cf_auto)
(2)若粒子从边界b射出,当L<
(1+cosθ)时,粒子将从边界b射出,又存在两种情况,
①轨迹如图所示:
![](https://video.ask-data.xyz/img.php?b=https://iknow-pic.cdn.bcebos.com/4b90f603738da977dbcd91afb351f8198718e350?x-bce-process=image%2Fresize%2Cm_lfit%2Cw_600%2Ch_800%2Climit_1%2Fquality%2Cq_85%2Fformat%2Cf_auto)
由图可知:L=Rsinα+Rsinβ=R(sinα+sinβ),其中α=
-θ,解得β=arc(
)
轨迹圆弧对应的圆心角为φ=α+β=
?θ+β所以射出经历的时间为t=
?T=
②如图所示:
![](https://video.ask-data.xyz/img.php?b=https://iknow-pic.cdn.bcebos.com/e4dde71190ef76c6c2ecedaf9e16fdfaaf516773?x-bce-process=image%2Fresize%2Cm_lfit%2Cw_600%2Ch_800%2Climit_1%2Fquality%2Cq_85%2Fformat%2Cf_auto)
由几何关系可知L=Rsinα-Rsinβ=Rcoθ-Rsinβ=R(cosθ-sinβ)
解得β=arcsin
()所以圆心角为φ
α?β=?θ?β所经历的时间为t=
答:粒子再次从磁场里射出经历的时间可能为间为:t=
或t=
或t=