# Magnetic Fields of a charged particle

1. Jul 22, 2010

### ninetyfour

The problem statement, all variables and given/known data

I have a couple questions from my textbook that I was looking at when studying, and since there are no answers, I'm not sure whether what I'm thinking is right.

1. A charged particle is moving in a circle in a uniform magnetic field. A uniform electric field is suddenly created, running in the same direction as the magnetic field. Describe the motion of the particle.

2. An electron is at rest. Can this electron be set into motion by applying a) a magnetic field, and/or b) an electric field?

3. A negatively charged particle enters a region with a uniformly magnetic field perpendicular to the velocity of the particle. Explain what will happen to the kinetic energy of the particle.

The attempt at a solution

1. The charged particle would accelerate towards the opposite charged plate?

2. a) Yes, the magnetic field would set the electron into motion because a magnetic field will exert a force on the electron?

b) Yes, because an electric field will also exert a force on the electron and the electron will move towards any positive charge?

3. The kinetic energy of the particle will be converted to magnetic energy or work done by the magnetic field?

I want to get down the basics, because I'm bad at that. It's kind of easy to just plug in numbers into a formula, and I want to make sure I understand the concepts. Help? :)

2. Jul 22, 2010

### collinsmark

Yes, that's technically correct more-or-less. However, can you think of a more detailed description of its motion?
Think about this one. What is the magnetic force on a charged particle (in a given magnetic field)? (I know you don't want to just plug numbers into an equation, but for this one it might be a good time to bring out that equation and think about what it means.)
This is another one that deserves some thought. I'm sure you know that kinetic energy (I'll call it T here, but some express it as K.E. or K, or whatever) is

$$T = \frac{1}{2}mv^2$$

In other words, kinetic energy is a function of the speed of the particle (as opposed to velocity). Remember, speed is a scalar, and the kinetic energy is a function of this speed.

When the charged particle enters the perpendicular magnetic field, does its speed change? (Hint: you are able to calculate the magnetic force applied to the particle. And since there is a force on the particle it has an acceleration. But what is the angle between this acceleration and the particle's velocity direction? Will the particle ever speed up or slow down?)

Last edited: Jul 22, 2010
3. Jul 22, 2010

### ninetyfour

I'm thinking that due to the force of both the electric field and the magnetic field on the charged particle, since Fe = qE, and Fm = qv x B, the particle will move faster, and in the direction of it's opposite charge?

Oh, is it that because the velocity is zero the force will also be zero, since Fm = qv x B? I just realized that... So if v is zero Fm has to be zero as well, therefore an electron cannot be set into motion by a magnetic field?

:)

This one confuses me a little When the charged particle enters the magnetic field, it's velocity vector will be perpendicular to the magnetic field...and it's kinetic energy should be conserved...but...I.. ..uh..yeah, I don't get it :uhh:

4. Jul 22, 2010

### collinsmark

Again, yes, technically you are correct. But how does the direction of the individual forces affect the motion? How do these different forces affect the particle's movement. Will the particle move toward the oppositely charged plate in a straight line, or will it squiggle about on its way? And if it squiggles about in some way, can you describe this squiggly like motion?
Sounds correct.
Yes, the velocity vector is perpendicular to the magnetic field vector.

And because you have Fm = qv x B, the force vector is perpendicular to both of them!

Take a step back and think about Newtonian gravity. The moon can be approximated to move around the Earth in a perfect circle. For now, let's use this approximation. The moon stays in this circular orbit due to the force of gravity. From the moon's perspective, what is this gravitational force's direction compared to the moon's velocity vector? Does this gravitational force cause a change the moon's speed?

(Hint: it's obvious the moon's direction is changing, but what about its speed? Thus what about its kinetic energy?)

5. Jul 22, 2010

### ninetyfour

Okay, so the force exerted by the electric field will be tangent to the field line, whereas the force exerted by the magnetic field will be perpendicular to the field line, but the particle is already moving in a circle....so would it continue towards the opposite charge in this circular motion?

Am I close, or just babbling?:tongue2:

:D

Okay, so the moon's speed is not changing, but its velocity is. So the kinetic energy is the same because the speed remains the same? In relation to the particle, the particle's kinetic energy will remain the same because the speed does not change? From what you said before, if there is a force there is an acceleration...but then is this centripetal acceleration caused by centripetal force? Since it's moving in a circle, it's velocity would be changing, but the magnitude of the velocity, the speed, remains the same, therefore the kinetic energy remains the same??
Again, am I close? :P

6. Jul 22, 2010

### collinsmark

Yeah, that's about it in a nutshell.
You got it!

7. Jul 22, 2010

### ninetyfour

Okay, thank you so much ! :D

8. Jul 23, 2010

### collinsmark

One last thing to think about, before I forget.

In the example with the moon, the moon happened to be moving in a circle, perpendicular to the gravitational force. But that was just in the example. Not all objects need to move perpendicular to a gravitational force. And if an object is moving at an angle that is not perpendicular to the force, the kinetic energy will change.

But with magneto-statics, we have this $\mathbf{F_m} = q \mathbf{v} \times \mathbf{B}$ equation. That cross product means that the force will always be perpendicular to the velocity (unlike gravity). That has profound implications on how magnetic fields can effect the kinetic energy of charged particles.