# 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 .

Please help me .

Thanks

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## Answers and Replies

BvU
Science Advisor
Homework Helper
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
Science Advisor
Homework Helper
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 ,

Thanks for replying.

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).

Please explain .

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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## ??
Yes. Go ahead.

• 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
Are you sure in the minus sign in x ?

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 .