Charged particle in Electric and Magnetic field

[Vibhor]

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

 

[BvU]

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 !

 

[Vibhor]

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

Ok . So what should be my next step ?

 

[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 ?

 

[Vibhor]

Sorry , I have no idea .

 

[ehild]

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.

 

[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]

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]

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 v_x,v_y,v_z ? How do I impose the condition that particle passes the origin ? Should I need to find general coordinates of the particle ?

 

[ehild]

Ok . So what should I do once I get v_x,v_y,v_z ? 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.

 

[Vibhor]

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

What is same at a later time ?

 

[ehild]

What is same at a later time ?

x and y and z all are zero again.

 

[Vibhor]

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

 

[ehild]

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

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

 

[Vibhor]

You can write it as v_x=v₁ cos (wt).

Please explain .

 

[ehild]

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

 

[Vibhor]

Ok .

So, $x=\frac{v_1}{ω}sinωt$ ??

 

[ehild]

Ok .

So, $x=\frac{v_1}{ω}sinωt$ ??

Yes. Go ahead.

 

[Vibhor]

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

 

[ehild]

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]

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 .

 

[ehild]

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

x is correct. I meant z, sorry.

 

[Vibhor]

$z = - \frac{v_1m}{qB}(1-cosωt)$ ??

 

[ehild]

$z = - \frac{v_1m}{qB}(1-cosωt)$ ??

Yes.

 

[Vibhor]

Ok .

x will be 0 when $\frac{qBt}{\pi m }$ is an integer , y will be 0 at $t = \frac{2mv_2}{qE}$ , z will be 0 when $\frac{qBt}{2\pi m }$ is an integer

 

[ehild]

Ok .

x will be 0 when $\frac{qBt}{\pi m }$ is an integer , y will be 0 at $t = \frac{2mv_2}{qE}$ , z will be 0 when $\frac{qBt}{2\pi m }$ is an integer

What integer is it? When is z = 0? And all of them should be zero at the same time.