Question

A charged particle of specific charge (charge/mass) $\alpha$ is released from origin at time $t=0$ with velocity $\overrightarrow{\upsilon }={{\upsilon }_{0}}\left( \widehat{i}+\widehat{j} \right)$ in a uniform magnetic field $\overrightarrow{B}={{B}_{0}}\widehat{i}$. The coordinates of the particle at time $t=\dfrac{\pi }{{{B}_{0}}\alpha }$ are
$\begin{aligned} & A.\left( \dfrac{{{\upsilon }_{0}}}{2{{B}_{0}}\alpha },\dfrac{\sqrt{2}{{\upsilon }_{0}}}{{{B}_{0}}\alpha },\dfrac{-{{\upsilon }_{0}}}{{{B}_{0}}\alpha } \right) \\\ & B.\left( \dfrac{-{{\upsilon }_{0}}}{2{{B}_{0}}\alpha },0,0 \right) \\\ & C.\left( 0,\dfrac{2{{\upsilon }_{0}}}{{{B}_{0}}\alpha },\dfrac{{{\upsilon }_{0}}\pi }{2{{B}_{0}}\alpha } \right) \\\ & D.\left( \dfrac{{{\upsilon }_{0}}\pi }{2{{B}_{0}}\alpha },0,\dfrac{-2{{\upsilon }_{0}}}{{{B}_{0}}\alpha } \right) \\\ \end{aligned}$

Answer 1

We must know that any charged particle, in the presence of a magnetic field, experiences a force in the direction perpendicular to both the velocity and the magnetic field. In this case, there are two kinds of motion, translational and rotational, which results in a combined helical motion of the particle. Here, the particle will be having translational motion through the x-axis and circular motion through the z-axis.

Formula used:
$\begin{aligned} & T=\dfrac{2\pi m}{{{B}_{0}}q} \\\ & R=\dfrac{m{{v}_{0}}}{{{B}_{0}}q} \\\ \end{aligned}$

Complete step by step answer:
In this question, a particle of specific charge $\alpha$ is released at the origin. At time t=0, the particle is moving in the x-y plane But the magnetic field intensity is acting in the direction of the positive x-axis. So, if we take the components of the velocity, according to right hand rule, the magnetic field will keep the charged particle in translational motion through the x-axis and rotational motion through the z-axis. It will result in the helical motion of the particle.

Now, it is given that specific charge, $\alpha =\dfrac{q}{m}$.
We know, the particle will follow a helical path with a time period,
$T=\dfrac{2\pi m}{{{B}_{0}}q}=\dfrac{2\pi }{{{B}_{0}}\alpha }$
We are asked to find the position of the particle at time $t=\dfrac{\pi }{{{B}_{0}}\alpha }$.
$\Rightarrow t=\dfrac{T}{2}$
Now, in order to find the coordinates of the particle at $t=\dfrac{T}{2}$, we will draw the trajectory of the particle. It will be a helical path.

So, as we are considering the position of the particle at $t=\dfrac{T}{2}$, the path will only be forming a semi-circle as it takes time T to complete a circle.
Then, the x coordinate will be equal to the linear distance covered by the particle which will be given by,
$s=v\times t$
Where s is the displacement along x axis, v is the velocity of the particle and t is the time.
So, the x coordinate will be given by,
$x={{v}_{x}}\times \dfrac{T}{2}$
The helix will be forming in the z-x plane and the y coordinate will be zero. Or we can interpret the y coordinate becoming zero as, the angle between velocity of the particle and the direction of magnetic field will be 90 degrees. So the force becomes zero and no force will be acting in y-direction. So the particle will travel in a helical path through the z-x plane. That is,

Now, at $t=\dfrac{T}{2}$, the particle will be at the position as shown in the figure.

So, at $t=\dfrac{T}{2}$, the helical path will only form a semi-circle. Therefore, the z coordinate will be the diameter of the circle. That is,
$z=-d=-2r$
Where r is the radius of the semi-circle.
Therefore, the coordinates of the particle at time $t=\dfrac{T}{2}$ would be $\left( {{v}_{x}}\times \dfrac{T}{2},0,-2r \right)$.
Now, to find the radius of the circle, we can use the concept,
The force due to the magnetic field will supply the necessary centripetal force for the circular motion. That is,
$\begin{aligned} & {{F}_{B}}={{F}_{C}} \\\ & \Rightarrow q{{v}_{0}}{{B}_{0}}=\dfrac{m{{v}_{0}}^{2}}{r} \\\ & \Rightarrow r=\dfrac{m{{v}_{0}}}{q{{B}_{0}}} \\\ & \therefore r=\dfrac{{{v}_{0}}}{\alpha {{B}_{0}}} \\\ \end{aligned}$
So, we can get the coordinates as,
$\begin{aligned} & x={{v}_{x}}\times \dfrac{T}{2}=\dfrac{{{v}_{0}}\pi }{{{B}_{0}}\alpha } \\\ & y=0 \\\ & z=-2r=\dfrac{-2{{v}_{0}}}{{{B}_{0}}\alpha } \\\ \end{aligned}$
That is, $\left( x,y,z \right)=\left( \dfrac{{{v}_{0}}\pi }{{{B}_{0}}\alpha },0,\dfrac{-2{{v}_{0}}}{{{B}_{0}}\alpha } \right)$.
So, option D is the correct answer.

Note: We must know that the helical path of the changed particle can be traced from Fleming’s right-hand rule by applying it at either end of the diameter of the circle. Here, we can observe that the velocity of the charged particle is not subject to change. Understanding the direction of propagation of the waves is most important.