Calculus & magnetic force on charged particle

[marciokoko]

I'm reading how a charged particle moving through a magnetic field experiences a force (called magnetic force).
I understand that it moves the particle in a certain direction and because the greatest force is experienced when the magnetic field is perpendicular to the particle's velocity vector, then it keeps moving the particle in a circular motion, acting as a centripetal force.

My interest is specifically to apply my calculus and trigonometry to better understand their application in a related example. So I know that F = qvBsinØ and I know that I can use calculus to calculate how the path of the particle changes as the angle and force change (this last one seems a bit more complicated).

Can someone with tons of patience help me visualize an example where I could calculate the changing force?

Sorry, I haven't been clear enough, first off I want to understand the trigonometry part, how the sinØ affects and is determined? So I am trying to understand why sine is used when calculating the force required to pull a boat with a rope from shore when the rope and how its affected by the angle.

 

[berkeman]

I'm reading how a charged particle moving through a magnetic field experiences a force (called magnetic force).
I understand that it moves the particle in a certain direction and because the greatest force is experienced when the magnetic field is perpendicular to the particle's velocity vector, then it keeps moving the particle in a circular motion, acting as a centripetal force.

My interest is specifically to apply my calculus and trigonometry to better understand their application in a related example. So I know that F = qvBsinØ and I know that I can use calculus to calculate how the path of the particle changes as the angle and force change (this last one seems a bit more complicated).

Can someone with tons of patience help me visualize an example where I could calculate the changing force?

Sorry, I haven't been clear enough, first off I want to understand the trigonometry part, how the sinØ affects and is determined? So I am trying to understand why sine is used when calculating the force required to pull a boat with a rope from shore when the rope and how its affected by the angle.

Have you studied vectors yet? This is best done with vectors -- the "Lorentz Force" is a vector equation, which you have listed a simplified version of.

 

[Chandra Prayaga]

I will add to berkeman's post by suggesting that the rope pulling the boat is a simpler problem, and if you have uncertainties with trigonometry, you should start there, and not with the magnetic force. Formulate a rope pulling a boat problem and state, via a diagram, what is bothering you, and we will go from there.

 

[marciokoko]

OK in the rope pulling example, here it is:

Why is the force calculated as F x cosO?

At the top we are told the force is split up into 2 vectors, FcosO and FsinO.

 

[Chandra Prayaga]

The response to that will depend on your background. Are you familiar with Newton's second law, and the technique of finding components of a vector along the axes of a coordinate system?

 

[marciokoko]

Yes I am familiar with Newton's laws and I studied vectors. I am a Biochem Major and I've used vectors in programming a bit.

 

[Chandra Prayaga]

The next steps are:
1. From the diagram, write down Newton's second law: The sum of all the forces = mass x acceleration
2. Take the x and y components of the equation that you wrote down
3. The y component equation will give you the reaction force (usually called the normal force)
4. The x component will give you precisely what you are looking for.

 

[marciokoko]

Ok here is my drawing.

The person is pulling a rope, B.
The boat will move forward along A [because there is an imaginary rail keeping it from crashing into the shoreline :-) ].
There are no vertical forces, at least none that result in vertical displacement. So I need to solve horizontally.
All forces = ma.
Vector B has an x & y component. The x-component could be A.

 

[marciokoko]

I've added some missing forces.

So friction pulls back against cos0, right?

 

[Chandra Prayaga]

cos0 is not how you would describe the direction of friction. You should set up a coordinate system. Call the direction A as the x axis, and the direction along C as the y axis. Then the pull force B has a component B cosθ along the x axis, and B sinθ in the y direction. The friction is acting in the negative x direction. This is a description of the forces. Now, what is the question?

 

[marciokoko]

Sorry, I did mean B*cos0 as the x component. I left out the B part.

OK so on this problem, if the boat is not moving at all or is moving at a constant speed, all forces add up to zero? Why, because in both cases a=0 so F=m*0?

 

[Chandra Prayaga]

That is correct. The total acceleration VECTOR is zero, so the net force VECTOR is zero, which means that each component of the net force vector is zero

 

[marciokoko]

Ok so the problem I am wanting to understand is the effect of a magnetic field on a charged particle along a straight wire, like this:

Where the magnetic field is given by:

B = m(o)I / 2πr
I = 3A
m(o) = 4π x 10-7 T · m / A
r = 0.50m
q = +6.5 x 10-6 C
v = 280 m/s

I get a B of 1.2 x 10^-5 and a Magnetic Force of 2.184E-08 C * m/s * T

This would be the magnetic force experienced by the particle running parallel to the wire. From what I understand, the particle is affected the most by the magnetic force when the field is perpendicular to its velocity vector as in this picture:

So what I am trying to understand is, when the field is perpendicular or at 90deg to the velocity of the particle, sin90 = 1, so at that moment that component is eliminated. So when the field is applied while ⟂ to the velocity, the particle's direction changes. This means that the particle's velocity is no longer exactly perpendicular to the field. This means that to keep the particle moving, we must change the direction of the field to adjust.

What I want to understand is how to use calculus to understand how the direction changes as the field changes?

 

[marciokoko]

bump :-)

 

[Chandra Prayaga]

I do not understand this statement:

"This means that to keep the particle moving, we must change the direction of the field to adjust." Why do you think so? What do you want the particle to do? There are tow types of problems:
1. The field is given, and the initial position and velocity of the particle are given, find the subsequent path of the particle
2. You want the particle to follow a given path, with given initial conditions, find the configuration of the fields, if such a solution is possible.

What do you need?

 

[marciokoko]

What caught my attention was that I believe I read somewhere that this is how the particle colliders work in terms of how they accelerate the particles to move in a circle, is this correct?

We apply a magnetic field ⟂ to the particle's velocity and it changes course, but now the field isn't ⟂ to the particle anymore...We must figure out how the velocity changed in order to change the field in such a way that the particle will keep turning in the direction we want it to such that it goes around in circles, right?

Q1/ How is that done?
Q2/ I would assume it has to do with calculus since it would be the study of how angle changes as the force changes, no?

 

[Nugatory]

We apply a magnetic field ⟂ to the particle's velocity and it changes course, but now the field isn't ⟂ to the particle anymore...We must figure out how the velocity changed in order to change the field in such a way that the particle will keep turning in the direction we want it to such that it goes around in circles, right?

Suppose we apply a constant magnetic field along the z axis and give the particle an initial velocity in the x-y plane The force on the particle will be perpendicular to both the direction of travel and the field, so the particle will remain in the x-y plane and the magnetic field will remain perpendicular to the direction of travel. The effect is that no matter what direction the particle is travelling, as long as it remains in the x-y plane, it will be deflected to one side of its direction of travel. The result is uniform circular motion in the x-y plane even though the magnetic field is constant.