Question

A charged particle is fired at an angle $\theta$ to a uniform magnetic field directed along the $x–$axis. During its motion along a helical path, if the pitch of the helical path is equal to the maximum distance of the particle from the $x–$ axis, then $\tan \theta$ is equal to

Answer 1

The velocity perpendicular to magnetic field provides the circular motion, therefore
$\frac{m(v\sin \theta {)}^2}{R}=qvB\sin \theta$
$\Rightarrow R=\frac{mv\sin \theta }{qB}$
Pitch: $P=2R=\frac{2mv\sin \theta }{qB}........\left(1\right)$ (This is the maximum distance of the particle from the x-axis)
Time period: $T=\frac{2\pi R}{v\sin \theta }=\frac{2\pi }{v\sin \theta }\left(\frac{mv\sin \theta }{qB}\right)$
$\Rightarrow T=\frac{2\pi m}{qB}$
Again, pitch can be written as
$P=\frac{2\pi m}{qB}\times v\cos \theta ......\left(2\right)$
From $\left(1\right)$ and $\left(2\right)$,
$\frac{2\pi m}{qB}\times v\cos \theta =\frac{2mv\sin \theta }{qB}$
$\Rightarrow \frac{\sin \theta }{\cos \theta }=\pi$
$\Rightarrow \tan \theta =\pi$