|
|
|
|
|
| Author |
Message |
![[Post New]](/templates/default/images/icon_minipost_new.gif) 1 Aug 2007 07:18:15 IST
|
|
|
a conducting rod of mass m and length l is placed over a smooth horizontal surface.a uniform magnetic field B is acting perpendicular to rod.charge q is passed suddenly throug the rod and it acquires initial velocity v on the surface then q is equal to............
2mv/ (Bl)
|
B.Tech CSE, ISMU |
|
|
|
|
|
|
|
Let charge q is passed in time t. So current = q/t
Force on rod= iBl=(q/t)Bl
Acc = F/m = (q/t)*Bl/m
Final velocity = Acc*t = qBl/m
Or q= mv/Bl
I am getting mv/(Bl). Please check question once because i am not sure how it can be 2mv/(Bl)
|
Krishna Gopal Singh
B.Tech Chemical Engg
IIT Delhi 2002
Currently doing PhD from IIT Delhi |
this reply: 5 points
(with 1 
in 1 votes ) [?]
|
|
You have to be logged on to rate
|
|
|
|
|
|
|
|
|
|
Moderation Team
![[Up]](/templates/default/images/icon_up.gif "[Up]"){kind=icon}
2
