Let us take axes as shown in
According to the right handed system, the z-exis is
upward in the figure and hence the magnetic field
is shown downwards. At any time, the velocity of the
electron may be written as
μ→=uxi→+uyj→.
The electric and magnetic fields may be written as
E→=−Ej→
and B→=−Bk→,
respectively . The force on the electron is
F→=−e(E→+u→XB→)
=eEJ→+eB(uyi→−uxJ→).
Thus, Fx=euyB,
and Fy=e(E−uxB).
The components of the acceleration are
ax=duxdt=eBmuy
and ay=duydt=em(E−uxB).
we have,
d2uydt2=−eBm(dux)(dt)
=eBm.eBmuy
=-omega^2u_ywhereomega= (eB)/(m) (iii)Thisequationissimilat→tfˆorasimp≤harmonicmotion.thus,u_y = A sin)omegat+delta) (iv)andhence,(du_y)(dt) = A omega cos(omegat+s). (v)At t = 0, u_y = 0and(du_y)/(dt) = (F_y)/ (m)= (eE)/(m).Pu∈g∈(iv)and(v),delta = 0 and A = (eE)/(momega) = (E)/(B).Thusu_y = (E)/(B) sin omegat.Thepathofthee≤ctronwillbeperpendica−−r→they−aξswhenuy=0.Thiswillbethecaseforthefirsttimeattwhere.sin omegat = 0or,omegat= pior,t=(pi)/(omega)=(pim)/(eB).Also,u_y = (dy)/(dt)=(E)/(B) sin omegat.or,int_0^y dy= (E)/(B) int_0^t sin omegat dt.or,y=(E)/(B omega) (1-cos omegat).Att =(pi)/(omega),y=-(E)/(B omega)(1 - cos pi) = (2E)(B omega).Thus,thedisplacementalongthey−aξsis(2E)/B omega) = (2Em)/(BeB) =(2Em)/(eB^2).`