Charged particle in Electric and Magnetic field

1. May 8, 2016

Vibhor

1. The problem statement, all variables and given/known data

A particle of specific charge q/m is projected from the origin of coordinates with initial velocity $v_1\hat{i}+v_2\hat{j}$ in space having uniform electric field and magnetic field as $-E\hat{j}$ and $-B\hat{j}$ respectively. The particle will definitely return to the origin once if

A) $\frac{v_1B}{\pi E}$ is an integer
B) $\frac{v_2B}{\pi E}$ is an integer
C) $\frac{\sqrt{{v_1}^2+{v_2}^2}B}{\pi E}$ is an integer
D) $\frac{q}{m}\frac{\sqrt{{v_1}^2+{v_2}^2}B}{\pi E}$ is an integer

2. Relevant equations

Lorentz force = $q\vec{v}$ x $\vec{B} + q\vec{E}$

3. The attempt at a solution

The force on the charged particle will be $\vec{F} = qBv_y\hat{i} - qE\hat{j}- qBv_x\hat{k}$ .

There will be a force on the particle in negative z- direction which means particle will never return to origin . I am not sure how to move forward .

Thanks

Last edited: May 9, 2016
2. May 9, 2016

BvU

If so, it acquires a velocity component in that direction and your Lorentz force changes !

3. May 9, 2016

Vibhor

Ok . So what should be my next step ?

4. May 9, 2016

BvU

What kind of trajectories can you think of in a field like that ? Which would let the partcle go through the origin more than once ?

5. May 9, 2016

Vibhor

Sorry , I have no idea .

6. May 10, 2016

ehild

Check the vector product. Y component of velocity does not give component of force with y component of magnetic field.
Without the electric field, the particle would move in the ( x,z) plane, perpendicular to the magnetic field, along a circle.
Recall that $\vec F= m \frac {d \vec v }{dt}$. Write it in components.

Last edited: May 10, 2016
7. May 10, 2016

Vibhor

Hello ehild ,

Sorry . That was a typo . $\vec{F} = qBv_z\hat{i} - qE\hat{j}- qBv_x\hat{k}$

Why ? There is an initial component of velocity along Y-axis .

$m a_x = qBv_z$
$m a_y = -qE$
$m a_z = -qBv_x$

8. May 10, 2016

ehild

Ah, yes.

Write the acceleration components as time derivatives of the velocity components, and solve the system of differential equation.

9. May 10, 2016

Vibhor

You mean I should solve the following three equations .

$$m \frac{dv_x}{dt} = qBv_z$$
$$m \frac{dv_y}{dt} = -qE$$
$$m \frac{dv_z}{dt} = -qBv_x$$

They look bit difficult to solve ?

10. May 10, 2016

ehild

No. You can solve the y component at once. The x, z components correspond to circular motion. (Why?) Take the derivative of the first equation and substitute $\frac{dv_z}{dt}$ into the third one.

11. May 10, 2016

Vibhor

Ok . So what should I do once I get vx,vy,vz ? How do I impose the condition that particle passes the origin ? Should I need to find general coordinates of the particle ?

12. May 10, 2016

ehild

Integrate the velocity components to get the coordinates with the initial condition x=y=z = 0. Find the condition that they are the same at a later time.
No need to use other coordinates.

13. May 10, 2016

Vibhor

What is same at a later time ?

14. May 10, 2016

ehild

x and y and z all are zero again.

15. May 10, 2016

Vibhor

Ok . I tried finding vx . $v_x = ±v_1sin(ωt+φ)$ where ω = qB/m and φ = nπ/2 ( n is odd) . Is this alright ?

16. May 10, 2016

ehild

You can write it as vx=v1 cos (wt). And what is vz?

17. May 10, 2016

Vibhor

Last edited: May 10, 2016
18. May 10, 2016

ehild

You left your solution with an arbitrary odd n. Choose n =1, then vx=v1cos(wt) will fulfill the initial condition.

19. May 10, 2016

Vibhor

Ok .

So, $x=\frac{v_1}{ω}sinωt$ ??

20. May 10, 2016