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Magnetic Fieldsby ninetyfour
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#1
Jul2210, 08:19 PM

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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
Jul2210, 08:57 PM

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[tex] T = \frac{1}{2}mv^2 [/tex] 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?) 


#3
Jul2210, 09:37 PM

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#4
Jul2210, 10:17 PM

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Magnetic Fields
And because you have F_{m} = 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
Jul2210, 10:32 PM

P: 28

Am I close, or just babbling? Again, am I close? :P 


#7
Jul2210, 10:42 PM

P: 28

Okay, thank you so much ! :D



#8
Jul2310, 12:59 AM

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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 magnetostatics, we have this [itex] \mathbf{F_m} = q \mathbf{v} \times \mathbf{B} [/itex] 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. 


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