# Determining particle position in a magnetic field given initial conditions [closed]

I am trying to create MATLAB code that will simulate a charged particle's path in a magnetic field. I have the user enter vectors for velocity and magnetic field, and then calculate the force vector.

Given these initial conditions, how can I calculate the position vector $$[x,y,z]$$ of the particle after time t? I'm aware of $$\vec{F}=q\vec{V}×\vec{B}$$, but am not sure how to determine position once force is found.

Thanks!

• Have you considered maybe using Newton's law? Mar 9 at 2:41
• Just a note: the vector product symbol can be written: \times Mar 9 at 4:09

Newton's second law of motion is one way to derive a set of second order differential equations known as the equations of motion. We can write Newton's second law as: $$\vec F=m\vec a(t)=m{d^2\vec r(t)\over dt^2},$$ where, $$\vec a(t)$$ and $$\vec r(t)=x(t)\mathbf{\hat i}+y(t)\mathbf{\hat j}+z(t)\mathbf{\hat k},$$ are the acceleration and position respectively. The velocity is given by: $$\vec v(t)=x'(t)\mathbf{\hat i}+y'(t)\mathbf{\hat j}+z'(t)\mathbf{\hat k}.$$ In the case of a charged particle in a magnetic field: $$\vec F=q\vec v\times\vec B\implies m{d^2\vec r(t)\over dt^2}=q\vec v\times\vec B.$$ So the vector product of $$\vec v$$ and $$\vec B$$ is given by: $$\vec v\times \vec B=(y'(t)B_z-z'(t)B_y)\mathbf{\hat i}-(x'(t)B_z-z'(t)B_x)\mathbf{\hat j}+(x'(t)B_y-y'(t)B_x)\mathbf{\hat k}.$$ And the second derivative of the position $$\vec r(t)$$ as: $${d^2\vec r(t)\over dt^2}=x''(t)\mathbf{\hat i}+y''(t)\mathbf{\hat j}+z''(t)\mathbf{\hat k}.$$ Putting this all together gives a set of coupled differential equations for the respective vector components: \begin{align} x''(t)&={q\over m}(y'(t)B_z-z'(t)B_y)\\ y''(t)&={q\over m}(x'(t)B_z-z'(t)B_x)\\ z''(t)&={q\over m}(x'(t)B_y-y'(t)B_x) \end{align} Now armed with the initial conditions for the problem, viz. the initial positions $$x(0)=x_0$$, $$y(0)=y_0$$, $$z(0)=z_0$$, and velocities $$x'(0)=v_{x0}$$, $$y'(0)=v_{y0}$$, $$z'(0)=v_{z0}$$, of the charged particle you can solve for the position as a function of time.