A proton is moving with uniform velocity in x-z plane at an angle of 30⁰  with x-axis, in the presence of an electrostatic field E = (4 kV/m)j and magnetic field B = (50 mT)k. Find the pitch of the helical trajectory followed by the proton when the electric field is switched off. 

Dear student,

proton is moving with uniform velocity so its acc is zero , hence net force on it is zero
 force by electric field = force by magnetic field

qE = q*v*B* sin($\theta$)    (angle between v and B is $\theta$= 60 degree)
v = $\frac{E}{B \sin \left(\theta \right)}$


when electric field is switch off
pitch = $$\begin{gathered} pitch = v\cos \left(\theta \right) *T \\ = v\cos \left(\theta \right)* \frac{2\mathrm{\pi m}}{qB\sin \left(\theta \right)} \\ = \frac{E}{B\sin \left(\theta \right)}\cos \left(\theta \right)* \frac{2\mathrm{\pi m}}{qB\sin \left(\theta \right)} \\ put the values \\ pitch = \frac{4*{10}^3}{50*{10}^{-3}\frac{\sqrt[]{3}}{2}}*\frac{1}{2}*\frac{2*3.14*1.67*{10}^{-27}}{1.6*{10}^{-19} *50*{10}^{-3}{\displaystyle \frac{\sqrt[]{3}}{2}}} \end{gathered}$$
= .035 m approx


Regards.