Finding a Parametric Solution for Particle Trajectory in Magnetic Field

AI Thread Summary
A parametric solution for the trajectory of a charged particle in a uniform magnetic field, considering air resistance, is derived using Newton's second law. The model predicts a constant angle between acceleration and velocity, expressed mathematically. The equations of motion are transformed into coupled differential equations, which are solved using complex variables. The resulting velocity components are derived, leading to expressions for the particle's position over time. This approach provides a comprehensive framework for analyzing particle motion in magnetic fields, highlighting the interplay of forces involved.
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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.

Trajectory.png
 
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It's going to take some dissecting to see how far off I was. Thanks for sharing!
 
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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!
 
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.
 
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