IIT JEE , AIEEE Forum - Physics , Electricity

pinnaka

lets us solve for the case when the particle is uncharged
F_net = Eq+qvB---1
remember the addition implies a vector product
since we have q =0 (this is done under the assumption that the charge represents a point particle)
therefore the force on it is zero
And the particle doe not move at all
now when B= 0 u have q=0 therefore 1 is also zero
therfore the particle doesn't move even when E
when they are perpendicular q=0,then the xharge doesn't move. The charge doesn't move for 4 also
But if your intial condition is such that the particle is movign with veolcity v, then it a straight line since there is no accleration

No consider a charged particle
Now the initial conditions being that the particle is at rest. It starts off from rest. Assuming that the particle is positively charged one we have F_b is zero at the start of the motion. So therefore the Electric field gives it an acceleration in the direction of the electric field. Now since B and E are parallel and q travels in the direction of that of E. That is also the direction of B. So therfore F_b at some instant in time when the particle is moving with a velcity vis
F_b = qv *cross B
v cross B =o
since they are parallel
the path is a staright line

Now the case when they are perpendicular we have F_E and F_B opposite to each other. If thet are equal then the particle is at rest. This is alos under the assumption that v_0 = 0. If the initial condition is such that v_0 = v the it follows a straight line
Now the case when there is E only then we have the path of the particle along the electric field
now for the magentic field
if v =0 then it is at rest
if v = v
If B is along the direction of B then we have
a straight line for v= v and for v=o it is at rest
now the B perpendicular to v
then we have F perpendicualr to B
thereby it follows a path of a cirlce

--Hope that Helps

Ask & Discuss Questions with Community & Experts