yes thats right...
as mV2/R = qVB
which gives R = mV/qB
A charged particle have a charge q enters a region in which there is a friction force which is proportional to the speed of particle. THe constant of proportionality of friction force is b. The particle stops traveling a distance of 10 m in a straight line.
Q-1
A magnetic field of unknown strength, in a direction perpendicular to entry velocity of particle is turned on, the particle will follow :
(a)circular path
(b)helical path
(c)spiral path of increasing radius of curvature
(d)spiral path of decreasing radius of curvature
Q-2
In the presence of above magnetic field if the particle stops at a point 6m away from the point of entry, the magnetic field is given by :
(a) 4b/3q
(b) 3b/4q
(c) 4q/3b
(d) 3q/4b
Q-3
The magnetic field is now doubled in magnitude. how far from the point of entry the particle will come to rest :
(a)√73/2m
(b)8/√73m
(c)20/√73m
(d)30/√73m
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10 Answers
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
1then will it follow a spiral path with decreasing radius ?
if yes the why will it follow a spiral path ?
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
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 ?
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.
1just 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 :-)