Question
Explain the condition under which a charged particle will travel through a uniform magnetic field in a helical path.###Describe the general motion of charged particle in a uniform magnetic field.
✓
Answer
Suppose a particle of mass $m$ and charge $q$ starts in a region of uniform magnetic field of induction $\vec{B}$ with a velocity $\vec{v}$ which has a non-zero component $v _{\|}$in the direction of $\vec{B}$, From below figure. The magnetic force $\vec{F}_{\text {II }}$ on the particle is always perpendicular to $\vec{v}$ and provides a centripetal acceleration such that
The parallel component of the motion $\vec{v}_{\|}$is unaffected by the magnetic field, so that the motion of the particle is a composite motion: an UCM with speed $v_{\perp}$-the speed perpendicular to $\vec{B}$ and a translation with a constant speed $v_{\|}$. Therefore, the particle moves in a helix. Thus, the perpendicular component $v_{\perp}$ determines the radius of the helix while the parallel component $v_{\|}$determines the pitch $x$ of the helix, i.e, the distance between adjacent turns. $x=v_{\mid l} / T$.
[Notes: (1) At non-relativistic speeds ( $v$ much less than the speed of light), the period $T$ is independent of the speed of the particle. For all particles with the same charge-to-mass ratio $( q / m )$, faster particles move in larger circles than the slower ones, but all take the same time T to complete one revolution. (2) Looking in the direction of $\vec{B}$ a positive charge always revolves anticlockwise, and a negative charge always clockwise.]
The parallel component of the motion $\vec{v}_{\|}$is unaffected by the magnetic field, so that the motion of the particle is a composite motion: an UCM with speed $v_{\perp}$-the speed perpendicular to $\vec{B}$ and a translation with a constant speed $v_{\|}$. Therefore, the particle moves in a helix. Thus, the perpendicular component $v_{\perp}$ determines the radius of the helix while the parallel component $v_{\|}$determines the pitch $x$ of the helix, i.e, the distance between adjacent turns. $x=v_{\mid l} / T$.
[Notes: (1) At non-relativistic speeds ( $v$ much less than the speed of light), the period $T$ is independent of the speed of the particle. For all particles with the same charge-to-mass ratio $( q / m )$, faster particles move in larger circles than the slower ones, but all take the same time T to complete one revolution. (2) Looking in the direction of $\vec{B}$ a positive charge always revolves anticlockwise, and a negative charge always clockwise.]
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