Question

Pitch of the helical path described by the particle is:
A. $\dfrac{{2\pi m{v_0}}}{{{B_0}q}}$
B. $\dfrac{{\sqrt 3 \pi m{v_0}}}{{2{B_0}q}}$
C. $\dfrac{{\pi m{v_0}}}{{{B_0}q}}$
D. $\dfrac{{2\sqrt 3 \pi m{v_0}}}{{{B_0}q}}$

Answer 1

A charged particle in a magnetic field experiences two forces acting on it. These forces are the forces due to the magnetic field and the centripetal force. As the velocity of the particle is constant in the direction parallel to the magnetic field and its motion accelerates in the perpendicular direction to the magnetic field, the particle traces a helical path. We need to find the equation for this path.

Complete step by step answer:
Let us define the variables used in the options and then see the forces acting on the particle and then balance them.
Let $m$ be the mass of the particle, ${v_0}$ be its velocity parallel to the magnetic field, ${B_0}$ be the magnitude of the magnetic field and $q$ be the charge on the particle.
The acceleration of the particle as it performs circular motion is given as:
$\dfrac{{{v^2}}}{r}$
Here, $v$ is the velocity of the particle moving perpendicular to the magnetic field.
The centripetal force ${F_C}$ will be given as
${F_C} = \dfrac{{m{v^2}}}{r}$
This acceleration is always directed towards the centre. The acceleration is always perpendicular to the particle’s instantaneous direction of motion. Also, the force ${F_B}$ exerted on the moving charge under the magnetic field is given as:
${F_B} = q(v \times B)$
This magnetic force is always perpendicular to the instantaneous direction of motion.
The particle is charged and it is performing circular motion perpendicular to the direction of the field. The combination of circular motion in the plane perpendicular to the magnetic field and uniform motion along the direction of magnetic field leads to a spiral trajectory which is known as helical path.
The charged particle is having a constant speed in the region parallel to the magnetic field. As a result, no external unbalanced force is applied. Therefore, the magnetic force and centripetal force must be equal.
$\dfrac{{m{v^2}}}{r} = q(v \times B)$
$\Rightarrow r = \dfrac{{mv}}{{qB}}$
Now, the time period $T$ of this oscillation will be
$T = \dfrac{{2\pi r}}{v}$
Pitch $P$ is the distance travelled by the particle along the direction of the magnetic field in one rotation and it is given as
$P = {v_0}T$
Substituting the value of time period and radius, we get
$\therefore P = \dfrac{{2\pi m}}{{qB}}{v_0}$

Thus, option A is the correct option.

Note: The particle is having two velocities, one in the direction parallel to the magnetic field and one in the direction perpendicular to the magnetic field. Due to the velocity parallel to the direction of the magnetic field, the particle is tracing the helical path. The centripetal force and magnetic force must be balanced.