How Do You Solve for Circular Motion in a Uniform Magnetic Field?

[matt222]

Homework Statement

particle of mass m carrying charge q, moving in a uniform magnetic of strength B. the field positive in z direction, the equation of motion are:

mx''=qBy'
my''=-qBx'
mz''=0

find general solution and apporopriate initial condition to have a circular motion on a plane

Homework Equations

The Attempt at a Solution

nothing on z direction so by eliminating m in the first equation and subtitute it in the second equation we got y'''=x''',at this point i really confused

 

[HallsofIvy]

You don't want an equation that involves both x and y!


From mx"= qBy', differentiating again gives you mx'''= qBy''= qb(-qBx'/m)= -(q^2B^2/m)x' or m^2x'''+ (q^2B^2)x'= 0.

 

[matt222]

in this case I have 3rd order differential equation and by using laplace transform I am able to solve it. But I am not sure about the initial condition let's say x''=x'=0 for the circle but should I assume x=1

 

[HallsofIvy]

You shouldn't assume anything. Also, you do not want initial conditions on x alone. Since you have three second derivative equations on x, y, and z, you want two initial conditions, say position and speed, for each of x, y, z.

Yes, as long as z'(0)= 0, z will be constant so you can take z(0)= 0 as well and get motion in the x, y plane.

Now, solving $m^2x'''+ (q^2B^2)x'= 0$ for x will give you a general solution for x(t) with three undetermined constants. You can then use that x(t) in $mx''=qBy'$ to solve for y' and then integrate to get y(t) introducing one more undetermined coefficient to make a total of 4.

Choose x(0), x'(0), y(0), and y'(0) so those coefficients will give you circular motion.