·2008-10-23 08:17:04
Q1.
Look,magnetic force is qVXB (which is perpendicular to the direction of motion & hence causes the charged particle to revolve in circle)
But as there is friction, it will reduce the velocity to zero in some time.
So how do u think the radius gonna change ??
Note: It cant make a helix, coz there is no force directed out of the plane
1
kushal khandelwal
·2008-10-23 08:19:45
hmm.. i think radius should decrease
1
kushal khandelwal
·2008-10-23 08:22:22
then will it follow a spiral path with decreasing radius ?
if yes the why will it follow a spiral path ?
·2008-10-23 08:28:06
At any velocity V, frictional force f=b*V
if mass is m, => a = -bV/m (friction will oppose the motion)
so, dV/dt = -bV/m
=> dV/V = -b/m dt
Integrating this equation, we get
ln(V) = -bt/m + k'
V = K * e-bt/m
1
kushal khandelwal
·2008-10-23 08:34:49
so the answer to the first ques is d
·2008-10-23 08:38:52
Look if there were no frictional force, it would had kept following a circular path.
but due to presence of friction, the velocity keeps on decresing.
Try to visualise, at any time, say the velocity is V, so the radious of circle will be given by R = mV/qB
after very small time interval dt, the velocity will be something smaller than V, so the radious of the circle at that time will also be something smaller than previous R
did you get it ?
·2008-10-23 08:43:11
Try to solve Q 2 & Q3 yourself.
It just involve some calculations. If you can't then I'll post the solution in free time.
1
kushal khandelwal
·2008-10-23 08:47:14
just give me some hints .. from where do i start cause i was not able to figure out that..
hey thanks for your help bro !!
i will try the question again after having dinner :-)