# Electron in a uniform magnetic field

• Gogsey
So never leave them out again! :tongue:In summary, an electron moving in the +x direction with speed vx in a uniform magnetic field B = Bˆz will have a clockwise circular trajectory with a constant kinetic energy and decreasing velocity in the x direction as its velocity in the y direction increases. If the same electron is in a conducting material obeying Ohm's law and is subject to an electric field that maintains its x velocity at vx, it will have a clockwise quasi-parabolic trajectory away from its origin with increasing velocity in the y direction and a constant kinetic energy. The work done on the electron in both scenarios is by the electric field.
Gogsey
5. Consider an electron moving in the +x direction with speed vx, in a uniform magnetic
field B = B ˆz.
(a) What type of trajectory does the electron take? What happens qualitatively to
its velocity in the x direction as its velocity in the y direction increases? How
does its kinetic energy change?
(b) Now imagine that the electron is in a conducting material that obeys Ohm’s law,
and is subject to an electric field that ensures that its x velocity is always exactly
vx. What type of trajectory does it take? What happens to its velocity in the x
direction as its velocity in the y direction increases? How does its kinetic energy
change?
(c) When the electron’s energy changes, what force does the work?

Ok, really just want to confirm if my answers are correct.

a). Clockwise circular trajectory, as Vy increases Vx decreases to the total veloty remains constant, kinetic energy remains constant.

b). Clockwise spiral trajectory, Vx remains constant, total velocity increases and therefore kinetic energy increases. Not too sure if the trajecoty is correct though. Since Vx is constant, then it V total may return to Vx then its directly on the x-axis, and then it woulb be like a harmonic oscillator in the y direction.

c). The electric field.

a) and c) look correct to me but your trajectory for b) seems off

Perhaps you can take me through your reasoning for why you chose that trajectory...what direction does the magnetic force point when the electron just starts its motion? How about a little later? Do the x and z-components of its velocity ever change? what happens to the y-component?

Yeah, this was the part that was the trickiest to me.

I'm not really sure what they mean about the x component always being Vx. I guess what I though at first was that Vx was constant, and that Vy changed dependnet on the abngle from the y-axis aroung the xy-plane.

Now I'm inclined to think its more of a clockwise eliptical trajectory.

Gogsey said:
I'm not really sure what they mean about the x component always being Vx. I guess what I though at first was that Vx was constant, and that Vy changed dependnet on the abngle from the y-axis aroung the xy-plane.

That seems reasonable to me!...but that isn't really a spiral path is it?

No, I meant like a planets orbit style, not spiral.

Gogsey said:
No, I meant like a planets orbit style, not spiral.

It isn't an orbit either...v_x is constant, so it's always getting further away from where it started in the x-direction...

If E is in the x direcition, B is in the z-direction and v_x is a constant, what does the Lorentz force law tell you about the acceleration in the y and z directions?

Hint: $$\vec{F}=m\vec{a}=-e(\vec{E}+\vec{v}\times\vec{B})=-e\left( E(x)\hat{x}+\begin{vmatrix}\hat{x} & \hat{y} & \hat{z}\\ v_x & v_y & v_z \\ 0 & 0 & B \end{vmatrix}\right)$$

So if Vx is constant, and as you say it trvels further in the x direction everytime it passes the x-axis, and I think its Vy is changing such that it never gets any further away in the y direction, then its kinda like an ellipse, but that growing in the x direction.

Gogsey said:
So if Vx is constant, and as you say it trvels further in the x direction everytime it passes the x-axis, and I think its Vy is changing such that it never gets any further away in the y direction, then its kinda like an ellipse, but that growing in the x direction.

Huh?!

Take my advice and actually calculate the y-component of the acceleration.

ok,

-e(Ex + (VyB - VxB), but I'm not sure what that says, unless Vx and Vy are both constant. Sorry I don't know how to interpret this.

Gogsey said:
ok,

-e(Ex + (VyB - VxB)

Huh?! How did you get this?

$$\vec{F}=m\vec{a}=-e(\vec{E}+\vec{v}\times\vec{B})=-e\left( E(x)\hat{x}+\begin{vmatrix}\hat{x} & \hat{y} & \hat{z}\\ v_x & v_y & v_z \\ 0 & 0 & B \end{vmatrix}\right)=-e\left( (E(x)+v_y B)\hat{x}-v_x B\hat{y}+0\hat{z} \right)$$

$$\implies a_x=\frac{-e}{m}\left(E+v_y B \right), \quad a_y=\frac{e v_x B}{m}, \quad a_z=0$$

Now your told that $v_x$ is constant, so evidently $E$ is varied in such a way that $$a_x=\frac{-e}{m}\left(E+v_y B \right)=0$$ always.

What can you say about $a_y$...what does that mean for $v_y$?

So that means that Vx is constant, so the distance from the origin in the x direction is increasinfg at a constant rate, where as the acceleration in the y direction is constant, so Vy is increasing at a constant rate, and distance from the origin in the y direction is increasing faster than in the x direction. Everytime it pases the positive y-axis, its distance has grown exponentially?

As for what I got with that equation, I just did not write in the x hat and y hat.

Gogsey said:
So that means that Vx is constant, so the distance from the origin in the x direction is increasinfg at a constant rate,
Correct, but that's exactly what the question told you to assume.

where as the acceleration in the y direction is constant, so Vy is increasing at a constant rate, and distance from the origin in the y direction is increasing faster than in the x direction. Everytime it pases the positive y-axis, its distance has grown exponentially?

Why would it pass the y-axis more than once? And why would it's distance increase exponentially?

If it starts at the origin, doesn't it just get further and further away? It is moving at a constant speed in the x-direction and constantly accelerating in the y-direction. If you want to give a name to that sort of trajectory, I would call it a "quasi-parabolic trajectory away from its origin" --- x(t) is a straight line, while y(t) is a parbola; hence the word "quasi-parabolic" describing r(t)

As for what I got with that equation, I just did not write in the x hat and y hat.

You should always write the unit vectors when talking about a vector. a=-e(Ex + (VyB - VxB) is meaningless, and a_y=-e(Ex + (VyB - VxB) is incorrect.

So how would you know that the electron makes at least one revolution around the origin?

## 1. What is the direction of the force on an electron in a uniform magnetic field?

The direction of the force on an electron in a uniform magnetic field is perpendicular to both the direction of the magnetic field and the velocity of the electron.

## 2. How does the strength of the magnetic field affect the motion of an electron?

The strength of the magnetic field determines the magnitude of the force on the electron. A stronger magnetic field will result in a greater force and therefore a larger change in the electron's motion.

## 3. What is the equation for calculating the force on an electron in a uniform magnetic field?

The equation is F = qvBsinθ, where F is the force, q is the charge of the electron, v is the velocity of the electron, B is the strength of the magnetic field, and θ is the angle between the velocity and the magnetic field.

## 4. How does the motion of an electron in a uniform magnetic field differ from its motion in a uniform electric field?

In a uniform electric field, the force on an electron is parallel to the direction of the electric field, while in a uniform magnetic field, the force is perpendicular to both the direction of the magnetic field and the velocity of the electron.

## 5. What is the cyclotron frequency and how is it related to the strength of the magnetic field?

The cyclotron frequency is the frequency at which a charged particle, such as an electron, will complete a circular orbit in a uniform magnetic field. It is directly proportional to the strength of the magnetic field, with a larger field resulting in a higher cyclotron frequency.

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