# 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
Homework Helper
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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• Vibhor
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.

• Vibhor
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.

• 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 ?

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.

• BvU and Vibhor
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.

• 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 ?

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?

• Vibhor
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.

• Vibhor
Ok .

So, $x=\frac{v_1}{ω}sinωt$ ??

ehild
Homework Helper
Ok .

So, $x=\frac{v_1}{ω}sinωt$ ??

• Vibhor
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.

• Vibhor
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.

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