Motion in Magnetic Field : Co-Ordinates

[HIGHLYTOXIC]

Is there any way we can find the co-ordinates of a charged particle undergoing uniform circular motion in uniform magnetic field, in space ?

I tried it using the angular velocity of the particle but it becomes quite complex..

Can anyone help? How about co-ordinates in Helical Motion? Any Chance?

 

[Tide]

Yes - it's pretty straightforward. For example, take the magnetic field to be in the \hat z direction and your equations of motion become
\frac {d v_x}{dt} = \Omega v_y
\frac {d v_y}{dt} = - \Omega v_x
\frac {d v_z}{dt} = 0

where \Omega is all the magnetic field, charge and mass folded into a single parameter. You shouldn't have any difficulty solving them if you have any experience with differential equations.

 

[Tide]

Oh, and once you have the velocity you can find the position by integrating
\frac {d \vec x}{dt} = \vec v

 

[HIGHLYTOXIC]

Yeah, that does make it simple..Thanx for the help!

 

[SoberSteve2121]

How would you prove using Newton's Second law that the trajectory of a charged particle with Velocity in the i direction and magnetic field in the -k direction must be circular?

 

[Tide]

How would you prove using Newton's Second law that the trajectory of a charged particle with Velocity in the i direction and magnetic field in the -k direction must be circular?

The preceding posts showed exactly how to do that.

 

[SoberSteve2121]

Would you be able to lay it out for me because I don't understand that?

 

[Tide]

Steve,

The equations I wrote in Post #2 in this thread ARE Newton's Law! Acceleration is force divided by mass and you see the left side of the equations represent the acceleration vector. The right side of the equations are the force vector q \vec \times \vec B divided by the mass (I rolled all the constants into the constant \Omega.

If you can solve those equations then you have your answer. You should at least be able to convince yourself that \cos \Omega t and \sin \Omega t are solutions of the first two equations so all you would have to do is apply initial conditions to the general solution

\vec x = \hat i x + \hat j y = \vec a \cos \Omega t + \vec b \sin \Omega t

to find the constants a and b. The result is a parametric representation of a circle!