Vishal Vaibhav
Last Activity: 15 Years ago
(Note: initial speed is also needed to solve the problem)
Since E is prependicular to B
one can write the velocity of the proton as
V = alpha(t)*E+beta(t)*(ExB)+gamma(t)*B
where E,(ExB),B constitue the mutually orthogonal axes.
Equation of motion of the proton in
m*dV/dt=q*(E+VxB)=q*[(1-beta*|B|^2)*E+alpha*(ExB)].------(1)
which leads to following equations for the alpha(t),beta(t),gamma(t)
a) gamma(t)=const.=gamma.
b) m*d(alpha)/dt=q*(1-beta*|B|^2).
c) m*d(beta)/dt=q*alpha.
Given that the initial motion of the Proton in rectilinear
we must have alpha=const., beta=const.. On inspection of b) and c)
we obtain beta = 1/|B|^2 and alpha=0.
Hence before E is switched off: V_0=(ExB)/|B|^2+gamma*B, where gamma is a constant(can be determined from the condition given in the question before switching off E).
When E is switched off: keeping E,(ExB),B as the axes
so that V = alpha(t)*E+beta(t)*(ExB)+gamma*B
m*dV/dt=q*(VxB)=q*[-beta(t)*|B|^2*E+alpha(t)*(ExB)],
b) m*d(alpha)/dt=-q*beta*|B|^2.
c) m*d(beta)/dt=q*alpha.
elimanating alpha we get
d^2(beta)/dt=-(q*|B|/m)^2*beta -->beta(t)=A1*cos(w*t)+A2*sin(w*t),w=(q*|B|/m),
and alpha(t)=-A1*|B|*sin(w*t)+A2*|B|*cos(w*t), where A1 and A2 are constants.
Using the initial condition V_0 we have
alpha(0)=0 ==> A2=0.
beta(0)=1/|B|^2 ==>A1=1/|B|^2.
Hence
V = {cos(w*t)/|B|^2}E+{-sin(wt)/|B|}*(ExB)+gamma*B
position vector works out to be
R = {cos(w*t)/(w*|B|^2)}E+{-sin(wt)/(w*|B|)}*(ExB)+gamma*t*B
let us call
X(t)=|E|*cos(w*t)/(w*|B|^2), Y(t)=-|E|*sin(w*t)/w, Z(t)=gamma*t*|B|
the base of the helix is: (X/|B|)^2+Y^2=(|E|/w)^2;
the pitch H is given by taking t=2*pi/w
H=Z(2*pi/w)=2*pi*gamma*|B|/w=2*pi*m*gamma/q
if the proton makes an angle "theta" with the direction of B
|B|*gamma =|V_0|cos(theta)
and H = 2*pi*|V_0|*cos(theta)/w.
