The force on a proton, which has charge 1, traveling with velocity vector v through a magnetic field with vector u is the cross product of u and v. That is mdv/dt= v x u.
If the magnetic field vector is $\displaystyle \left<M_x, M_y, M_z\right>$ and the proton's velocity vector is $\displaystyle \left<v_x, v_y, v_z\right>$ then
$\displaystyle \left<m\frac{dv_x}{dt}, m\frac{dv_y}{dt}, m\frac{dv_z}{dt}\right>= \left<M_yv_z- M_zv_y, M_zv_x- M_xv_z, M_xv_y- M_yv_x\right>$.
So you need to solve the three equations
$\displaystyle m\frac{dv_x}{dt}= M_Yv_z- M_zv_y$
$\displaystyle m\frac{dv_y}{dt}= M_zv_x- M_xv_z$
$\displaystyle m\frac{dv_z}{dt}= M_xv_y- M_yv_x$
If you set up a coordinate system so that the magnetic field vectors are in the y direction $\displaystyle \left<M_x, M_y, M_z\right>= \left<0, M_y, 0\right>$ so those equations become
$\displaystyle m\frac{dv_x}{dt}= M_yv_z$
$\displaystyle m\frac{dv_y}{dt}= 0$
$\displaystyle m\frac{dv_z}{dt}= -M_yv_x$
quite a lot simpler!
For example, if you differentiate the first equation again,
$\displaystyle m\frac{d^2v_x}{dt}= M_y\frac{dv_z}{dt}$
and you can replace that derivative of $\displaystyle \frac{dv_z}{dt}$ from the third equation to get $\displaystyle m^2\frac{d^2v_x}{dt^2}= -M_y^2v_x$
a relatively easy second order linear differential equation with constant coefficients- though certainly **not** secondary school level!
Probably you are given a formula relating the velocity and magnetic field strength of a "radius of curvature". Do you have such a formula?
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Last edited by HallsofIvy; Apr 19th 2018 at 05:39 AM.
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