A long wire of radius ‘a’ is carrying a direct current I.
From its surface at point A, an electron of charge −e (e > 0) escapes with velocity v0 perpendicular to this surface.
(see Figure)
Analyze the motion to check whether the electron
- escapes to x -> ꝏ, or
- approaches some x -> x0 (find it if yes), or
- comes back, after reaching an xmax (find it if yes) .
Ignore gravity.
- escapes to x -> ꝏ, or
- approaches some x -> x0 (find it if yes), or
- comes back, after reaching an xmax (find it if yes) .
Ignore gravity.
< Courtesy: INPhO 2011 >
I remember this one. (I was able to solve it using a fairly conventional approach).
ReplyDeleteI now see a tendency among readers to not try the hard problems. Remember that these problems will eventually shape your physics.
Good luck and I'm waiting for the trick!
I am able to find velocity along x,y as a function of x and y but not able to integrate for the trajectory.Is the trick that the velocity remains along x,y only and vx^2+vy^2=vO^2?i have a doubt whether we can analyse the motion without an equation for trajectory?
ReplyDeletethis case.. we cannot analyze w/o eqns of motion..
ReplyDeletedont bother about the complexity.. its quite complex..(but u can solve it if u want to - after spending hours)
ReplyDeletebtw.. i changed the question a bit..
I think that the particle comes back, after reaching an x(max)= ae^(2πv0m/eµI)
ReplyDeletewhere e(in the exponent of the exponential)=charge of 1 electron
m = mass of electron
i think it comes after a Xmax,
ReplyDelete(Xmax)= ae^[{(mv/ke)^2}/2];
where k=µI/2π
correct.. (:
ReplyDeletepls give a solution/hint?
ReplyDeleteWhat was the Trick?
ReplyDeleteto solve use: v_o^2 = v_x^2 + v_y^2
ReplyDeleteand solve a d.e. in v_x and x
important: a_x = v_x dv_x/dx
I wrote F_x as a function of x.Put a_x=v_xdv_x/dx and integrate x (a to xmax) ,v_x (v to 0)
ReplyDeletecorrect?