Question

Non-relativistic protons move reactilinearly in the region of space where there are uniform mutually perpendicular electric and magnetic fields with E and B. The trajectory of the protons lie in the plane X-Y as shown in fig. and forms an angle ϕϕ with X-axis. If the pitch of the helical trajectory along which the protons will move after the electric field is switched off is $Pitch=\frac{x\pi mE}{q{B}^2}tan\varphi$. Find x.

Answer 1

When electric field is switched off, the path followed by the particle will be helical.
$Pitch=v_{\parallel }T$ .... (i)
where, $v_{\parallel }=$ velocity of the particle parallel to B, and $T=$ time of revolution $=\left(\frac{2\pi m}{qB}\right)$
Here $v_{\parallel }=vcos(90-\varphi )=vsin\varphi$
Hence, from Eq. (i)
$Pitch=\left(vsin\varphi \right)\left(\frac{2\pi m}{qB}\right)$ .... (ii)

When both fields are present:
$qE=qvBsin({90}^o-\varphi )=qvBcos\varphi$
$v=\frac{E}{Bcos\varphi }$ ..... (iii)

Substituting the value of v from Eq. (iii) in Eq. (ii), we get
$Pitch=(\frac{E}{Bcos\varphi }\times sin\varphi )\frac{2\pi m}{qB}=\frac{2\pi mE}{q{B}^2}tan\varphi$