# Charged particle in Electric and Magnetic field

## Homework Statement

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

## Homework Equations

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

## 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

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BvU
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2019 Award
There will be a force on the particle in negative z- direction
If so, it acquires a velocity component in that direction and your Lorentz force changes !

If so, it acquires a velocity component in that direction and your Lorentz force changes !
Ok . So what should be my next step ?

BvU
Homework Helper
2019 Award
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 ?

Sorry , I have no idea .

ehild
Homework Helper

## Homework Statement

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

## Homework Equations

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

## The Attempt at a Solution

The force on the charged particle will be $\vec{F} = qBv_y\hat{i} - qE\hat{j}- qBv_x\hat{k}$ .
Check the vector product. Y component of velocity does not give component of force with y component of magnetic field.
There will be a force on the particle in negative z- direction which means particle will never return to origin .
Thanks
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.

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Hello ehild ,

Check the vector product. Y component of velocity does not give component of force with y component of magnetic field.
Sorry . That was a typo . $\vec{F} = qBv_z\hat{i} - qE\hat{j}- qBv_x\hat{k}$

Without the electric field, the particle would move in the ( x,z) plane, perpendicular to the magnetic field, along a circle.
Why ? There is an initial component of velocity along Y-axis .

Recall that $\vec F= m \vec {\dot v}$. Write it in components.
$m a_x = qBv_z$
$m a_y = -qE$
$m a_z = -qBv_x$

ehild
Homework Helper
Hello ehild ,

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

$m a_x = qBv_z$
$m a_y = -qE$
$m a_z = -qBv_x$
Write the acceleration components as time derivatives of the velocity components, and solve the system of differential equation.

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

ehild
Homework Helper
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 ?
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.

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 ?

ehild
Homework Helper
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 ?
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.

Find the condition that they are the same at a later time.
What is same at a later time ?

ehild
Homework Helper
What is same at a later time ?
x and y and z all are zero again.

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 ?

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

You can write it as vx=v1 cos (wt).

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

Ok .

So, $x=\frac{v_1}{ω}sinωt$ ??

ehild
Homework Helper
Ok .

So, $x=\frac{v_1}{ω}sinωt$ ??

The general coordinates are $x=\frac{v_1m}{qB}sinωt$ , $y = v_2t - \frac{qE}{2m}t^2$ , $z = \frac{v_1m}{qB}(1-cosωt)$

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ehild
Homework Helper
The general coordinates are $x=\frac{v_1m}{qB}sinωt$ , $y = v_2t - \frac{qE}{2m}t^2$ , $z = \frac{v_1m}{qB}(1-cosωt)$
Are you sure in the minus sign in z? Check the acceleration.

Then find the time when x, y, z are zero and the condition that this can happen.

You mean there should be a minus sign ??

ehild
Homework Helper
You mean there should be a minus sign ??
No, in z. I think the minus sign is at the wrong place. Check.

Sorry . I think It should be $x=\frac{v_1m}{qB}sinωt$ only . I don't see why there should be a minus sign .