则:d=
1 |
2 |
由牛顿第二定律得:a=
qE |
m |
电场强度:E=
U |
2d |
由①②③解得时间为:t=
|
设平板的长度为L,则:L=v0t ⑤
撤去电场后,带电粒子在磁场中作半径为R的圆弧运动,
轨迹如下图所示:
由几何关系得:R2=(R-d)2+L2 ⑥
粒子作圆周运动,由牛顿第二定律得:qv0B=
m
| ||
R |
由④⑤⑥⑦解得磁感应强度:B=
2mv0U | ||
d(qU+4m
|
答:磁感应强度大小为:=
2mv0U | ||
d(qU+4m
|
1 |
2 |
qE |
m |
U |
2d |
|
m
| ||
R |
2mv0U | ||
d(qU+4m
|
2mv0U | ||
d(qU+4m
|