A charged particle moving in a magnetic field | vedclass

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(C) Let $v_1$ be the component of velocity parallel to the magnetic field $B$ and $v_2$ be the component perpendicular to $B$.$1$. Due to the parallel

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helical path with the axis along magnetic field $B$

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(C) Let $v_1$ be the component of velocity parallel to the magnetic field $B$ and $v_2$ be the component perpendicular to $B$.$1$. Due to the parallel component $v_1$,the magnetic force $F = q(v_1 	imes B) = q v_1 B \sin(0^{\circ}) = 0$. Thus,the particle continues to move with constant velocity $v_1$ along the direction of $B$.$2$. Due to the perpendicular component $v_2$,the magnetic force $F = q(v_2 	imes B)$ acts perpendicular to both $v_2$ and $B$,providing the necessary centripetal force for circular motion in a plane perpendicular to $B$.$3$. The combination of uniform linear motion along $B$ and uniform circular motion in the plane perpendicular to $B$ results in a helical path,where the axis of the helix is parallel to the magnetic field $B$.

$A$ charged particle moving in a magnetic field $B$ has velocity components both along $B$ and perpendicular to $B$. The path of the charged particle will be:

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