A charged particle is moving through a uniform magnetic field. At a...

Let the magnetic field be $\vec{B} and the velocity of the charged particle be$\vec{v}$.<br><br>The velocity$\vec{v}$can be resolved into two components:$v_{\parallel} = v \cos \theta$parallel to the magnetic field and$v_{\perp} = v \sin \theta$perpendicular to the magnetic field, where$\theta$is the acute angle between$\vec{v}$and$\vec{B}$.<br><br>The perpendicular component$v_{\perp}$provides the necessary centripetal force, causing the particle to move in a circular path of radius$r = \dfrac{mv_{\perp}}{qB}$.<br><br>The parallel component$v_{\parallel}$remains constant since there is no force acting along the direction of the magnetic field.<br><br>The combination of uniform circular motion and uniform linear motion results in a helical path.<br><br>The pitch of the helix is the distance traveled along the magnetic field in one time period$T$, given by$p = v_{\parallel} T = v \cos \theta \left(\dfrac{2\pi m}{qB}\right).

Since $v$, $\theta$, $m$, $q$, and $B$ are all constant, the pitch of the helix is uniform.