A charged particle with charge $q$ and mass ' $m$ ' is moving with velocity $160 \mathrm{~ms}^{-1}$ in the region of magnetic field $\mathrm{B}$ at an angle $60^{\circ}$ with the direction of $\overrightarrow{\mathrm{B}}$. The pitch of helix formed by particle will be:

\(\frac{160 \pi m}{q B} \)

The correct answer is Option (3) → \(\frac{160 \pi m}{q B} \)

Given,

B = magnetic field

q = charge

v, velocity = 160 m/s

θ, Angle between $\vec B$ and $\vec v$ = 60°$

Pitch (P) = $\frac{2\pi mv\cos\theta}{qB} = \frac{2\pi m\times 160\times \cos 60°}{qB}$

$=\frac{2\pi m\times 160\times \frac{1}{2}}{qB}$ $[\cos 60°=\frac{1}{2}]$

$= \frac{160\pi m}{qB}$