A charged particle with charge \( q \) and mass 'm' is moving with velocity \( 160\,\mathrm{ms^{-1}} \) in the region of magnetic field \( B \) at an angle \( 60^\circ \) with the direction of \( \vec{B} \). The pitch of helix formed by particle will be:
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → \( \displaystyle\frac{160\pi m}{qB} \)
The velocity component perpendicular to \( \vec{B} \) makes the particle go round a circle, while the component parallel to \( \vec{B} \) carries it along the field. The pitch is the distance moved along the field in one revolution.
\( \displaystyle T = \frac{2\pi m}{qB} \)
\( \displaystyle v_{\parallel} = v\cos 60^\circ = 160 \times \frac{1}{2} = 80\,\mathrm{m\,s^{-1}} \)
\( \displaystyle p = v_{\parallel}\,T = 80 \times \frac{2\pi m}{qB} = \frac{160\pi m}{qB} \)