A proton enters a uniform magnetic field \( \vec{B} \) at an angle \( \theta\ (0^\circ \lt \theta \lt 90^\circ) \), its time period of motion in helical path is \( T \). If now the proton is replaced by a deuteron and the angle of entrance in the magnetic field is made three times i.e. \( 3\theta\ (0^\circ \lt 3\theta \lt 90^\circ) \), then the time period of motion of deuteron in the magnetic field will be
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → \( 2T \)
In a uniform field, the component of velocity perpendicular to \( \vec{B} \) gives circular motion and the parallel component gives uniform drift, so the path is a helix. The time period of the circular part does not depend on the speed or on the angle of entry:
\( \displaystyle T = \frac{2\pi m}{qB} \)
A deuteron has the same charge as a proton and twice its mass. The change of angle to \( 3\theta \) changes only the radius and the pitch of the helix, not the period.
\( \displaystyle T_{d} = \frac{2\pi(2m)}{qB} = 2T \)