Understanding the Motion of Electrons in a Magnetic Field

[Saitama]

Homework Statement

Homework Equations

The Attempt at a Solution

Let magnetic field be inside the plane of sheet (towards -ve z-axis)
At any instant of time, let the components of velocity of electron be $v_x$ and $v_y$.
Then $F_x=ev_yB_o$ (i) and $F_y=-ev_xB_o$ (ii)
Substituting $v_x=dx/dt, v_y=dy/dt, F_x=md^2x/dt^2$ and $F_y=md^2y/dt^2$ and dividing (i) and (ii), I end up with $(d^2x)dx=-(d^2y)dy$ and solving this with wolfram alpha gives $y^2=k-x^2$. k=0 as this curve passes through origin. This gives $y^2=-x^2$ which is not possible.

 

[ehild]

What are the trajectories of the electrons in magnetic field? Can you find them without Wolframalpha? And what is k?

ehild

 

[Saitama]

Why is k=0?

ehild

Because the curve passes through origin. :)

 

[ehild]

You did nonsense with the second derivatives.
d²y/dx² is not (dx)² divided by (dy)², that is, square of the tangent. Think of a parabola y=x². y'=dy/dx=2x, y"=d²y/dx²=2, (y')²=4x²...

Remember your high-school studies. What are the trajectories in magnetic field?

 

[Saitama]

Remember your high-school studies. What are the trajectories in magnetic field? You have completely ignored the velocity of the electrons.

The trajectory depends on the velocity of electrons and direction of magnetic field. In this case, if I consider an electron which is projected along the y-axis, it will hit the x-axis at a distance of $mv_F/{eB}$ from the origin (the trajectory would be of a semicircle). If an electron is projected long the axis, it will disappear as it goes in $y<0$ region. But I am confused as to what happens when it is thrown at an angle (except 90) with x-axis. I am still clueless. :(

 

[ehild]

The velocity, that is tangent to the trajectory makes and angle with the x axis. The radius is perpendicular to it. You get the radius of the circle from the speed and B. Can you find the position of centre of the circle?

I edited my previous post, read it.

 

[Saitama]

d²y/dx² is not (dx)² .

And I never said that it is. I had the following:
m\frac{d^2x}{dt^2}=e\frac{dy}{dt}B_o
m\frac{d^2y}{dt^2}=-e\frac{dx}{dt}B_o
I divided these two equations and ended up with
(d^2x)dx=-(d^2y)dy
Plugging this in wolfram alpha gave me $y^2=k-x^2$.

The velocity, that is tangent to the trajectory makes and angle with the x axis. The radius is perpendicular to it. You get the radius of the circle from the speed and B. Can you find the position of centre of the circle?

I edited my previous post, read it.

Let the electron be projected at an angle $\theta$, then is the radius $mv_F\sin\theta/(eB_o)$?

 

[ehild]

Why? What angle does the velocity enclose with the magnetic field?

ehild

 

[Saitama]

Why? What angle does the velocity enclose with the magnetic field?

ehild

Sorry, the angle with magnetic field is $\pi/2$ but I still have no idea. :(

 

[ehild]

So what is the radius of the trajectory with the given speed of the electrons?

I have to leave. Hope, you can proceed. Make a drawing.

ehild

 

[Saitama]

So what is the radius of the trajectory with the given speed of the electrons?

I still don't have any idea. Please give me a few hints. :(

 

[ehild]

And I never said that it is. I had the following:
m\frac{d^2x}{dt^2}=e\frac{dy}{dt}B_o
m\frac{d^2y}{dt^2}=-e\frac{dx}{dt}B_o
I divided these two equations and ended up with
(d^2x)dx=-(d^2y)dy
Plugging this in wolfram alpha gave me $y^2=k-x^2$.

What do you mean with the d-s in (d^2x)dx=-(d^2y)dy?

Wolframalpha interpreted d as constant. You can cancel them, and get xdx=-ydy, which is far away from the original problem.

As for electron in magnetic field: what angle does the magnetic force enclose with the velocity? What is the trajectory? Maybe a circle? If it is circle, what is the centripetal force? Given the velocity, what is the radius that corresponds to that centripetal force?

 

[Saitama]

What do you mean with the d-s in (d^2x)dx=-(d^2y)dy?

Wolframalpha interpreted d as constant. You can cancel them, and get xdx=-ydy.

Looks here: Wolfram Alpha

EDIT: Okay, I get, but what should I do now?

It follows the path of circle only when it is projected along the y-axis. I don't see how can I find the centripetal force for the general case. :(

 

[ehild]

You have a velocity vector in the x,y plane. Magnitude VF. You have a magnetic field B in the z direction.
What is the direction of the Lorentz force with respect to the velocity and magnetic field?

 

[Saitama]

You have a velocity vector in the x,y plane. Magnitude VF. You have a magnetic field B in the z direction.
What is the direction of the Lorentz force with respect to the velocity and magnetic field?

The direction of Lorentz force is perpendicular to both velocity and magnetic field. Do you ask me this:
ev_FB_o=\frac{mv_F^2}{R}
R=\frac{mv_F}{eB_o}

 

[ehild]

exaxtly. So is the trajectory an arc of circle at arbitrary angle?

 

[Saitama]

So is the trajectory an arc of circle at arbitrary angle?

Yes?

 

[ehild]

Yes, of course. If a force of constant magnitude acts perpendicularly to the velocity the trajectory is a circle. Now it is only part of a circle as the electron disappears after reaching the x axis. And where is its centre? Make a drawing.

 

[Saitama]

Yes, of course. If a force of constant magnitude acts perpendicularly to the velocity the trajectory is a circle. Now it is only part of a circle as the electron disappears after reaching the x axis. And where is its centre? Make a drawing.

If an electron is projected along y-axis, the centre of the circle is on the x-axis at a distance $R$ from the origin and if it is projected at an angle, the centre lies in $y<0$ region. Right?

If an electron is projected at angle $\theta$ with the x-axis, it reaches the x-axis again at a distance of $2R\sin \theta$.

EDIT: Got it! Thanks! The question asked for the greatest value of x and it is greatest when projected along y-axis. Thank you ehild!

 

[ehild]

The problem asked the position of the detector where it gets the greatest signal. How can be the magnitude of the signal depend on the angle the electron is projected at?

 

[Saitama]

The problem asked the position of the detector where it gets the greatest signal. How can be the magnitude of the signal depend on the angle the electron is projected at?

Sorry, didn't read the problem statement properly.

I don't know of any equation which relates the magnitude of signal with the angle. I can only think that the question asks where the maximum number of electrons hit, right?

 

[ehild]

The problem also says that the direction of projection is totally random. It means that the number of electrons projected between angles θ1 and θ2 is proportional to the difference of the angles, Δθ.
The width of the detector is W and you can assume that W<<R. Find the angle which results in the greatest Δθ, that is the highest number of electrons reaching the detector.

At projection angle θ, x=2Rsinθ. If the projection angle is θ-Δθ, x2=x-ΔW=2Rsin(θ-Δθ). --->

2r(sinθ-sin(θ-Δθ))=W *

When Δθ is maximum, its derivative with respect θ must be zero: θ'=0

Apply implicit differentiation on eq(*) and replace θ'=0.

ehild

 

[Saitama]

Sorry for the late reply.

The problem also says that the direction of projection is totally random. It means that the number of electrons projected between angles θ1 and θ2 is proportional to the difference of the angles, Δθ.
The width of the detector is W and you can assume that W<<R. Find the angle which results in the greatest Δθ, that is the highest number of electrons reaching the detector.

At projection angle θ, x=2Rsinθ. If the projection angle is θ-Δθ, x2=x-ΔW=2Rsin(θ-Δθ). --->

2r(sinθ-sin(θ-Δθ))=W *

When Δθ is maximum, its derivative with respect θ must be zero: θ'=0

Apply implicit differentiation on eq(*) and replace θ'=0.

ehild

Differentiating the equation you gave me,
\frac{dW}{d\theta}=2r(\cos \theta-\cos(\theta-\Delta \theta)(1-\frac{d\Delta \theta}{d\theta}))
The term $\frac{d\Delta \theta}{d\theta}=0$,
\Rightarrow \frac{dW}{d\theta}=2r(\cos \theta-\cos(\theta-\Delta \theta))
What am I supposed to do with this?

 

[ehild]

Sorry, Pranav. I did not think it over carefully enough.

Assume angles between 0 and 90 degrees, and a given angle interval, Δθ. You need the narrowest detector at projection angles near 90 degrees. In the figure, Δθ=10°.

It is a bit more complicated if you allow angles between 0 and 180°. I will think about it.


ehild

 

[ehild]

Electrons projected at angles greater than 90° also fall to the detector area W if they are in the range between θ and θ+Δθ and 2r(sinθ-sin(θ+Δθ))=W

So you have the piecewise continuous function, see attachment.

W(θ) = 2r(sinθ-sin(θ-Δθ)) if θ≤90°
and
W(θ) = 2r(sinθ-sin(θ+Δθ)) if θ≥90°.

Electrons projected at a range Δθ fall on the smallest width W if the angle is 90°.
Or you can conclude from x=2rsinθ that the electron density along the x-axis (proportional to dθ/dx) is highest when cosθ=0, that is at x=2r: you should place the detector there.

ehild

 

[Saitama]

W(θ) = 2r(sinθ-sin(θ+Δθ)) if θ≥90°.

Won't this give a negative value for W?

 

[ehild]

will it?

 

[Saitama]

will it?

Oops, that was really silly of me.

Thank you ehild for the explanation!