Finding a Parametric Solution for Particle Trajectory in Magnetic Field

• kuruman
In summary, this is a solution to a problem involving a particle moving in a uniform magnetic field with air resistance. The model predicts a constant angle between acceleration and velocity, with an expression for the angle between the two. By using the cyclotron frequency and an auxiliary variable, the equations for the components of acceleration can be simplified and solved to find the trajectory. The final solution involves parametric equations for x and y, with a graph showing the trajectory.
kuruman
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Homework Statement
Find the trajectory of a particle the acceleration of which is fixed at angle ##\varphi > \pi/2## relative to the velocity.
Relevant Equations
Newton's second law.
This is a solution to a problem inspired by another thread. It is posted here to separate it from the multiple choice question which was the subject of that thread. A parametric solution for the trajectory can be found quite easily if the motion is modeled as a particle with charge ##q## moving in a uniform magnetic field ##\mathbf{B}=B~\mathbf{\hat z}##. In addition to the Lorentz force, air resistance provides retarding force ##\mathbf{F}_{\text{ret.}}=-b\mathbf{v}.##

First we prove that this model predicts a constant angle between acceleration and velocity and find an expression for the angle ##\varphi## between acceleration and velocity. We assume that the particle moves in the ##xy##-plane. From Newton's second law we have \begin{align}\mathbf{a}=\frac{1}{m}(q\mathbf{v}\times\mathbf{B}-b\mathbf{v}).\end{align}Then, noting that ##\mathbf{v}\cdot(\mathbf{v}\times\mathbf{B})=0##, $$\mathbf{v}\cdot\mathbf{a}=\frac{1}{m}\mathbf{v}\cdot[q(\mathbf{v}\times\mathbf{B})-b\mathbf{v}]=-\frac{bv^2}{m}$$Also, \begin{align} \mathbf{a}\cdot\mathbf{a} =a^2 & =\frac{1}{m}(q\mathbf{v}\times\mathbf{B}-b\mathbf{v})\cdot \frac{1}{m}(q\mathbf{v}\times\mathbf{B}-b\mathbf{v})\nonumber \\ & =\frac{1}{m^2}({q^2B^2v^2+b^2v^2})\implies a=\frac{1}{m}(q^2 B^2 v^2+b^2 v^2)^{1/2}.\nonumber \end{align}The cosine of the angle between acceleration and velocity is constant and given by \begin{align} \cos\!\varphi=\frac{\mathbf{v}\cdot\mathbf{a}}{va}=-\frac{b}{(q^2B^2+b^2)^{1/2}}\end{align}. We now write Newton's second law in two dimensions to find the trajectory. To simplify the form of the equations we use the cyclotron frequency ##\omega_c=qB/m## and auxiliary variable ##\beta=b/m##. Then from equation (1) the components of the acceleration are
\begin{align} & \dot v_x =\omega_c v_y-\beta v_x \\ & \dot v_y=-\omega_c v_x-\beta v_y. \end{align}The two coupled equations can be solved quite easily by changing variables, $$\xi=v_x+iv_y~;~~\eta = v_x-iv_y$$ in which case equations (3) and (4) become \begin{align} & \dot{\xi}+\dot{\eta}=-i\omega_c(\xi-\eta)-\beta(\xi+\eta) \nonumber \\&\dot{\xi}-\dot{\eta}=-i\omega_c(\xi+\eta)-\beta(\xi-\eta). \nonumber \end{align} Adding the equations yields \begin{align} &\dot {\xi} =-(i\omega_c+\beta)\xi\implies \xi=Ae^{-\beta t}e^{-i\omega_c t} \nonumber \\ & \dot{\eta}=\dot {\xi}^*= A^*e^{-\beta t}e^{i\omega_c t} .\nonumber \end{align}We can now employ the definitions for ##\xi## and ##\eta## and use the initial conditions ##v_x(0)=v_0## and ##v_y(0)=0## to find \begin{align} & v_x(t)=v_0e^{-\beta t}\cos(\omega_c t) \nonumber \\ & v_y(t) = v_0e^{-\beta t}\sin(\omega_c t). \nonumber \end{align}Finally, we integrate to find ##x(t)## and ##y(t)## such that ##v_x(0)=v_0## and ##v_y(0)=0##: \begin{align} & x(t)=\frac{v_0e^{-\beta t}}{\beta^2+\omega_c^2}[\omega_c\sin(\omega_c t)-\beta \cos(\omega_c t)]\nonumber \\ & y(t)=\frac{v_0e^{-\beta t}}{\beta^2+\omega_c^2}[\omega_c\cos(\omega_c t)+\beta \sin(\omega_c t)]\nonumber \end{align} The graph below is a parametric plot of the trajectory with parameters as shown.

Last edited:
AmanWithoutAscarf, berkeman, DrClaude and 6 others
It's going to take some dissecting to see how far off I was. Thanks for sharing!

member 731016 and kuruman
kuruman said:
Homework Statement:: Find the trajectory of a particle the acceleration of which is fixed at angle ##\varphi > \pi/2## relative to the velocity.
Relevant Equations:: Newton's second law.

This is a solution to a problem inspired by another thread. It is posted here to separate it from the multiple choice question which was the subject of that thread. A parametric solution for the trajectory can be found quite easily if the motion is modeled as a particle with charge ##q## moving in a uniform magnetic field ##\mathbf{B}=B~\mathbf{\hat z}##. In addition to the Lorentz force, air resistance provides retarding force ##\mathbf{F}_{\text{ret.}}=-b\mathbf{v}.##

First we prove that this model predicts a constant angle between acceleration and velocity and find an expression for the angle ##\varphi## between acceleration and velocity. We assume that the particle moves in the ##xy##-plane. From Newton's second law we have \begin{align}\mathbf{a}=\frac{1}{m}(q\mathbf{v}\times\mathbf{B}-b\mathbf{v}).\end{align}Then, noting that ##\mathbf{v}\cdot(\mathbf{v}\times\mathbf{B})=0##, $$\mathbf{v}\cdot\mathbf{a}=\frac{1}{m}\mathbf{v}\cdot[q(\mathbf{v}\times\mathbf{B})-b\mathbf{v}]=-\frac{bv^2}{m}$$Also, \begin{align} \mathbf{a}\cdot\mathbf{a} =a^2 & =\frac{1}{m}(q\mathbf{v}\times\mathbf{B}-b\mathbf{v})\cdot \frac{1}{m}(q\mathbf{v}\times\mathbf{B}-b\mathbf{v})\nonumber \\ & =\frac{1}{m^2}({q^2B^2v^2+b^2v^2})\implies a=\frac{1}{m}(q^2 B^2 v^2+b^2 v^2)^{1/2}.\nonumber \end{align}The cosine of the angle between acceleration and velocity is constant and given by \begin{align} \cos\!\varphi=\frac{\mathbf{v}\cdot\mathbf{a}}{va}=-\frac{b}{(q^2B^2+b^2)^{1/2}}\end{align}. We now write Newton's second law in two dimensions to find the trajectory. To simplify the form of the equations we use the cyclotron frequency ##\omega_c=qB/m## and auxiliary variable ##\beta=b/m##. Then from equation (1) the components of the acceleration are
\begin{align} & \dot v_x =\omega_c v_y-\beta v_x \\ & \dot v_y=-\omega_c v_x-\beta v_y. \end{align}The two coupled equations can be solved quite easily by changing variables, $$\xi=v_x+iv_y~;~~\eta = v_x-iv_y$$ in which case equations (2) and (3) become \begin{align} & \dot{\xi}+\dot{\eta}=-i\omega_c(\xi-\eta)-\beta(\xi+\eta) \nonumber \\&\dot{\xi}-\dot{\eta}=-i\omega_c(\xi+\eta)-\beta(\xi-\eta). \nonumber \end{align} Adding the equations yields \begin{align} &\dot {\xi} =-(i\omega_c+\beta)\xi\implies \xi=Ae^{-\beta t}e^{-i\omega_c t} \nonumber \\ & \dot{\eta}=\dot {\xi}^*= A^*e^{-\beta t}e^{i\omega_c t} .\nonumber \end{align}We can now employ the definitions for ##\xi## and ##\eta## and use the initial conditions ##v_x(0)=v_0## and ##v_y(0)=0## to find \begin{align} & v_x(t)=v_0e^{-\beta t}\cos(\omega_c t) \nonumber \\ & v_y(t) = v_0e^{-\beta t}\sin(\omega_c t). \nonumber \end{align}Finally, we integrate to find ##x(t)## and ##y(t)## such that ##v_x(0)=v_0## and ##v_y(0)=0##: \begin{align} & x(t)=\frac{v_0e^{-\beta t}}{\beta^2+\omega_c^2}[\omega_c\sin(\omega_c t)-\beta \cos(\omega_c t)]\nonumber \\ & y(t)=\frac{v_0e^{-\beta t}}{\beta^2+\omega_c^2}[\omega_c\cos(\omega_c t)+\beta \sin(\omega_c t)]\nonumber \end{align} The graph below is a parametric plot of the trajectory with parameters as shown.

View attachment 321236
Very nice!

kuruman
This is a fun problem. I usually have this setup in my "simulation of motion" module (it is one problem which the students can choose to study) and I present the "algebraic" solution as a bonus when the module is finished.

PeroK

What is a parametric solution in the context of a particle trajectory in a magnetic field?

A parametric solution describes the position of a particle as a function of a parameter, typically time. In the context of a particle trajectory in a magnetic field, it involves expressing the particle's coordinates (x, y, z) as functions of time (t), taking into account the forces exerted by the magnetic field on the particle.

How does the Lorentz force affect the trajectory of a charged particle in a magnetic field?

The Lorentz force is the force exerted on a charged particle moving through a magnetic field. It is given by the equation F = q(v x B), where q is the charge of the particle, v is its velocity, and B is the magnetic field. This force causes the particle to move in a helical or circular path, depending on the initial conditions of the particle's velocity and the orientation of the magnetic field.

What are the typical equations used to describe the motion of a charged particle in a uniform magnetic field?

The motion of a charged particle in a uniform magnetic field is typically described by the following set of differential equations: dx/dt = vx, dy/dt = vy, dz/dt = vz, dvx/dt = (q/m) * (vy * Bz - vz * By), dvy/dt = (q/m) * (vz * Bx - vx * Bz), dvz/dt = (q/m) * (vx * By - vy * Bx), where (vx, vy, vz) are the components of the particle's velocity, (Bx, By, Bz) are the components of the magnetic field, q is the charge, and m is the mass of the particle.

What initial conditions are needed to find a parametric solution for a particle's trajectory in a magnetic field?

To find a parametric solution, you need the initial position (x0, y0, z0) and initial velocity (vx0, vy0, vz0) of the particle at t = 0. These initial conditions are crucial because they determine the specific path the particle will follow under the influence of the magnetic field.

Can the parametric solution for a particle trajectory in a magnetic field be solved analytically?

In some simple cases, such as when the magnetic field is uniform and the initial velocity is perpendicular to the field, the equations can be solved analytically to give a clear parametric form. However, in more complex scenarios involving non-uniform fields or additional forces, numerical methods are often required to obtain the solution.

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