91Ó°ÊÓ

The given data is listed below as:

• The charge of the particle is,$q$
• The mass of the particle is,$m$
• The particle carries a charge$I$ .

Themotion of a particle is described as the direction at which the particle’s velocity vector is mainly tangent to the path of the particle. The velocity vector’s magnitude equals the particle’s speed.

From the given data, it can be identified that the magnetic force does not work around the particle. Hence, the kinetic energy is constant.

Thus, the kinetic energy is conserved.

The equation of the force on the particle is expressed as:

$F=qv×B$ …(¾±)

Here,$q$ is the particle’s charge,$v$ is the velocity and$B$ is the magnetic field of the particle.

The equation of the magnetic field can be expressed as:

$B=\frac{{\mathrm{μ}}_{0}I}{2\mathrm{Ï€²õ}}\stackrel{Áåœ}{\mathrm{Ï•}}$ …(¾±¾±)

Here,${\mathrm{μ}}_0$ is the permittivity,$I$ is charge carried by the particle,$s$ is the distance moved and$\stackrel{Áåœ}{\mathrm{Ï•}}$ is the position vector of the particle.

The equation of the velocity of the particle is expressed as:

$v=s\stackrel{Áåœ}{s}+s\stackrel{.}{\mathrm{Ï•}}\stackrel{Áåœ}{\mathrm{Ï•}}+\stackrel{.}{z}\stackrel{Áåœ}{z}$ …(¾±¾±¾±)

Here,$\stackrel{Áåœ}{z}$ is the position vector in the z axis.

Substitute the value of the equation (ii) and (iii) in equation (i).

$$\begin{gathered} F=q\left(s\stackrel{Áåœ}{s}+s\stackrel{.}{\mathrm{Ï•}}\stackrel{Áåœ}{\mathrm{Ï•}}+\stackrel{.}{z}\stackrel{Áåœ}{z}\right)×\frac{{\mathrm{μ}}_{0}I}{2\mathrm{Ï€²õ}}\stackrel{Áåœ}{\mathrm{Ï•}} \\ =-q\frac{{\mathrm{μ}}_{0}I}{2\mathrm{Ï€²õ}}\stackrel{.}{z}\stackrel{Áåœ}{s}+q\stackrel{.}{s}\frac{{\mathrm{μ}}_{0}I}{2\mathrm{Ï€²õ}}\stackrel{Áåœ}{z} \end{gathered}$$

Thus,the force on the particle is$-q\frac{{\mathrm{μ}}_{0}I}{2\mathrm{Ï€²õ}}\stackrel{.}{z}\stackrel{Áåœ}{s}+q\stackrel{.}{s}\frac{{\mathrm{μ}}_{0}I}{2\mathrm{Ï€²õ}}\stackrel{Áåœ}{z}$.

The equation of the force of the particle is expressed as:

$F=ma$ …(¾±)

Here, $m$is the mass and $a$is the acceleration of the particle.

The equation of the force can also be expressed as:

$F=\left(\stackrel{.}{s}-s{\stackrel{.}{\mathrm{Ï•}}}^2\right)\stackrel{Áåœ}{s}+\left(\stackrel{.}{s}\stackrel{..}{\mathrm{Ï•}}+2\stackrel{.}{s}\stackrel{.}{\mathrm{Ï•}}\right)\stackrel{Áåœ}{\mathrm{Ï•}}+\stackrel{..}{z}\stackrel{Áåœ}{z}$ …(¾±¾±)

Here, $s$is the distance moved by the particle, $\mathrm{Ï•}$is the angle subtended at the $z$axis and $z$is the coordinate of the particle in the $z$axis.

The product of the mass and the acceleration of the particle can be expressed as:

$ma=\frac{q{\mathrm{μ}}_0}{2\mathrm{Ï€³¾s}}\left(-\stackrel{.}{z}\stackrel{Áåœ}{s}+\stackrel{.}{s}\stackrel{Áåœ}{z}\right)$ …(¾±¾±¾±)

Here,$q$is the charge of the particle and ${\mathrm{μ}}_0$is the permeability.

Substitute the values of the equation (ii) and (iii) in equation (i).

$\left(\stackrel{.}{s}-s{\stackrel{.}{\mathrm{Ï•}}}^2\right)\stackrel{Áåœ}{s}+\left(\stackrel{.}{s}\stackrel{..}{\mathrm{Ï•}}+2\stackrel{.}{s}\stackrel{.}{\mathrm{Ï•}}\right)\stackrel{Áåœ}{\mathrm{Ï•}}+\stackrel{..}{z}\stackrel{Áåœ}{z}=\frac{q{\mathrm{μ}}_0}{2\mathrm{Ï€³¾s}}\left(-\stackrel{.}{z}\stackrel{Áåœ}{s}+\stackrel{.}{s}+\stackrel{.}{s}\stackrel{Áåœ}{z}\right)$

From the above equation, three equations can be obtained such as;

$$\begin{gathered} \stackrel{..}{s}-s{\stackrel{.}{\mathrm{Ï•}}}^2=-\frac{q{\mathrm{μ}}_0}{2\mathrm{Ï€³¾}}\frac{{\displaystyle \stackrel{.}{z}}}{s} \\ s\stackrel{.}{\mathrm{Ï•}}+2\stackrel{.}{s}\stackrel{.}{\mathrm{Ï•}}=0 \\ \stackrel{..}{z}=\frac{q{\mathrm{μ}}_0}{2\mathrm{Ï€³¾}}\frac{{\displaystyle \stackrel{.}{s}}}{s} \end{gathered}$$ …(¾±±¹)

Thus, the equations of motion are $\stackrel{..}{s}-2{\stackrel{.}{\mathrm{Ï•}}}^2=-\frac{q{\mathrm{μ}}_0}{2\mathrm{Ï€³¾}}\frac{{\displaystyle \stackrel{.}{z}}}{s},s\stackrel{..}{\mathrm{Ï•}}+2\stackrel{.}{s}\stackrel{..}{\mathrm{Ï•}}=0\mathrm{and}\stackrel{..}{z}=\frac{q{\mathrm{μ}}_0}{2\mathrm{Ï€³¾}}\frac{{\displaystyle \stackrel{.}{s}}}{s}$respectively.

From the equations of motion gathered in the above step, if the value of $\stackrel{.}{z}$is constant, then $\stackrel{..}{z}$will be zero. Like this, if the value of s is constant, then $\stackrel{.}{s}$and $\stackrel{..}{s}$will be zero.

Substitute 0 for$\stackrel{..}{s}$ in the equation (iv).

$$\begin{gathered} -s\stackrel{.}{\mathrm{Ï•}}=-\frac{q{\mathrm{μ}}_0}{2\mathrm{Ï€³¾}}\frac{{\displaystyle \stackrel{.}{z}}}{s} \\ {\stackrel{.}{\mathrm{Ï•}}}^2=\frac{q{\mathrm{μ}}_0}{2\mathrm{Ï€³¾}}\frac{{\displaystyle \stackrel{.}{z}}}{s^2} \\ \stackrel{.}{\mathrm{Ï•}}=Â\pm \frac{1}{s}\sqrt{\frac{q{\mathrm{μ}}_0}{2\mathrm{Ï€³¾}}}\stackrel{.}{z} \end{gathered}$$

From the above equation, it can be identified that the charge mainly moves like a helix.

Thus, the motion of$\stackrel{.}{z}$ is helix.