# Cycloidal Path of Charged Particle

• theQmechanic
In summary: I was just guessing that it might be easier to first solve for y(t) and then get h from the y max and min values. (And, to aviod a potential error in computing the derivative with respect to t of cosine, I would use the complex exponential form of cosine).@BvU...yep that's okay, but for the sake of simplicity let's solve in the original coordinate system.Let's not get confused with the first l distance.The free body diagram of the mass while it is sliding down the incline shows that the magnetic force is directed towards the center of the circle and the normal reaction of the incline acts normal to the surface of the incline. Thus, the magnetic force acts as
theQmechanic

## Homework Statement

A charged particle (q), mass (m) is released from a frictionless inclined plane of angle θ under influence of Earth's acceleration (g) and magnetic field (B) perpendicular to (g) and plane of motion of particle. The particle slides down distance "l" along incline and then follows a cycloidal path with vertical displacement between highest and lowest point (h). Prove that $l = h\frac{cot^{2}θ}{4}$

## Homework Equations

F= qv$\times$B

## The Attempt at a Solution

In vector notation I took g as -g j. B as -B k.
Took instantaneous velocity as v= vxi + vy j.
Then formed a differential equation to vx and vy in terms of time. The point where it leaves the incline is the point of inflection of the curve. I bashed whatever equations I had to finally get the answer, but after doing this I think there's got to be an easy way to look at this. Can someone help me out?

Wouldn't life be a lot easier for you if the particle does NOT leave the incline ?
 Hmmm... weak moment. B has to be ##\bot## g AND the plane of motion. You don't happen to have a picture at hand ?

Anyway, if the particle stays on the plane and also if it leaves, there has to be a g in one of the relevant equations.
And also in 3: what is your differential eqn? Can you show some of the bashing?

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Looks like B is to be perpendicular to g and to the NORMAL of the inclined plane. This appears to be what the OP has assumed, and it would indeed seem to produce a spiral path as it works its way down the incline.

For the first l distance there are 3 forces on the mass: gravity, magnetic and the plane pushing up on the mass. After that, when the mass becomes airborne, the last of these forces is absent. So it seems like you first have to determine l which is when the inclined components of magnetic and gravitational forces add up to zero. That should not be too bad.

Then, determine x and y for the airborne path to get h. I would write F = ma equations in x and y and solve for the locus. That sounds more imposing ...

EDIT: you can solve for just the y motion & then the max - min of this motion should be h I guess. Then compute l/h.

Getting late, may look some more tomorrow but probably others will have joined in by then.

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@BvU...h and l both come in terms of m,q,g,B,θ etc. then u take ratio and only θ remains. I'll Show the bashing.

@rude man...it doesn't come to be spiral, i'll show that mathematically.

The F=qvb=ma equation yield the following
$m\frac{d}{dt}(v_{x}i + v_{y}j)=-qBv_{y}i + (qBv_{x}-mg)j$.

Thus,
$\frac{dv_{x}}{dt}= -\frac{qBv_{y}}{m}$ and $\frac{dv_{y}}{dt}=\frac{qBv_{x}}{m} -g$

$\Rightarrow \frac{d^{2}v_{x}}{dt^{2}}=-(\frac{qB}{m})^{2}(v_{x}-\frac{mg}{qB})$.
Let t=0 when particle is at bottom most point and velocity be $v_{0}i \Rightarrow v_{x}= (v_{0} -\frac{mg}{qB})cos(\frac{qBt}{m}) + \frac{mg}{qB}$.

Further, $y= (v_{0} - \frac{mg}{qB})\frac{m}{qB}(1-cos(\frac{qBt}{m})))$.
Clearly, $h=\frac{2m}{qB}(v_{0} - \frac{mg}{qB})$

As can be seen $x= aθ + bsin(kθ) and y=ccos(kθ)$. Thus it happens to be a cycloid.

#### Attachments

• cycloid.jpg
6.8 KB · Views: 485
OK, I'm inclined to do away with the particle staying on the incline
Argument: B doesn't do any work, so potential energy at highest point = potential energy at releasing meaning h = l sinΘ, so not the cot2/4. I took ##\vec B \bot## the plane -- as I was told.
So we continue with
• the assumption ##\vec B## is NOT ##\bot## the plane but in the horizontal negative z direction
• we look at free fall in a magnetic field with inital condition ##v_0^2 = 2 l g \,sin\theta/ m##. (So ##\vec v_0 \cdot \hat\imath = v_0 \cos\theta## and ##\vec v_0 \cdot \hat\jmath = -v_0 \sin\theta##)
Do we agree so far ?

I agree with BvU's v2 = 2l sinθ/m. This is the 'takeoff point'. I now see a system of two diff. eq. in x and y with new initial conditions on vx and vy. I would impose a new coordinate system with x=y=0 at this point to clean up the equations in x and y.

I also got the OP's dvx/dt and dvy/dt equations, but then I don't follow the rest to where he/she gets h. If that h is correct then we have a problem getting the answer, seems like.

I would pursue solving the system of two diff. eq. in x and y, solving for y(t) and getting h that way. (I would use Laplace, but whatever).

@theQmechanic: I used the word 'spiral' loosely. The exact shape is of course determined by the equations. It's undoubtedly a cycloid as you and the question state.

## What is the cycloidal path of a charged particle?

The cycloidal path of a charged particle refers to the trajectory or path that a charged particle follows when it is subjected to a constant magnetic field. This path is shaped like a cycloid, which is a curve created by a point on a circle rolling along a straight line.

## What factors determine the shape of the cycloidal path?

The shape of the cycloidal path is determined by the strength of the magnetic field, the charge and mass of the particle, and the initial velocity of the particle. These factors interact to create the cycloid shape and determine the size and frequency of the oscillations.

## Can the cycloidal path be altered or controlled?

Yes, the cycloidal path can be altered or controlled by changing the strength of the magnetic field or the initial velocity of the particle. Additionally, the path can be manipulated by using electric or magnetic fields to deflect the particle.

## What applications does the cycloidal path have in science and technology?

The cycloidal path of charged particles has many applications in various fields of science and technology. For example, it is used in particle accelerators to manipulate and control the movement of particles. It is also used in medical imaging techniques such as MRI machines and in the production of electronic devices such as cathode ray tubes.

## Can the cycloidal path be observed in nature?

Yes, the cycloidal path can be observed in nature. In fact, cosmic rays and charged particles from the sun follow cycloidal paths as they travel through the Earth's magnetic field. This phenomenon is also observed in the Northern and Southern Lights, where charged particles from the sun collide with gases in the Earth's atmosphere, creating beautiful cycloid-shaped light displays.

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