#
A proton is moving with uniform velocity in x-z plane at an angle of 30^{0} ^{ }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.

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 = $\frac{E}{B\mathrm{sin}\left(\theta \right)}$

when electric field is switch off

pitch = $pitch=v\mathrm{cos}\left(\theta \right)*T\phantom{\rule{0ex}{0ex}}=v\mathrm{cos}\left(\theta \right)*\frac{2\mathrm{\pi m}}{qB\mathrm{sin}\left(\theta \right)}\phantom{\rule{0ex}{0ex}}=\frac{E}{B\mathrm{sin}\left(\theta \right)}\mathrm{cos}\left(\theta \right)*\frac{2\mathrm{\pi m}}{qB\mathrm{sin}\left(\theta \right)}\phantom{\rule{0ex}{0ex}}putthevalues\phantom{\rule{0ex}{0ex}}pitch=\frac{4*{10}^{3}}{50*{10}^{-3}\frac{\sqrt[]{3}}{2}}*\frac{1}{2}*\frac{2*3.14*1.67*{10}^{-27}}{1.6*{10}^{-19}*50*{10}^{-3}{\displaystyle \frac{\sqrt[]{3}}{2}}}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}$

= .035 m approx

Regards.

**
**