Calculating Magnetic Force on Charged Particle in a Magnetic Field

[warnexus]

Homework Statement

Homework Equations

magnetic force = charge * velocity * magnetic field


they ask for distance but this was the only relevant equation for this problem to corresponding section

The Attempt at a Solution

veloctity = 8.4* 10 ^ 6 m/s
magnetic field = 190 Gauss
perpendicular = sin (90) = 1

magnetic force = (1.6 * 10 ^ -19 C) (8.4 * 10 ^ 6 m/s )(190 G)

 

[mfb]

Which path do the electrons describe in the field?
This will help to solve the problem.

 

[rude man]

I would start by writing F= ma equations for all three axes. (Might two suffice?).
Good problem.

 

[warnexus]

I would start by writing F= ma equations for all three axes. (Might two suffice?).
Good problem.

all 3 axes, even if z axis is not mentioned. we have to resolve components using a force diagram?

okay here goes:
F_x = m * a_x
F_y = m * a_y
F_z = m * a_z

 

[rude man]

all 3 axes, even if z axis is not mentioned. we have to resolve components using a force diagram?

What I alluded to is that perhaps one of the axes never sees any motion. Since mag force is always perpendicular to the B field, which axis do you suppose that might be?

PS certainly the z axis is in on this.

 

[warnexus]

What I alluded to is that perhaps one of the axes never sees any motion. Since mag force is always perpendicular to the B field, which axis do you suppose that might be?

PS certainly the z axis is in on this.

magnetic force perpendicular to the magnetic field. the y -axis is perpendicular!

 

[mfb]

Z-axis is the most important one...
And please, do not use cartesian coordinates to solve equations of motions. This is just overkill for this problem.

 

[rude man]

all 3 axes, even if z axis is not mentioned. we have to resolve components using a force diagram?

okay here goes:
F_x = m * a_x
F_y = m * a_y
F_z = m * a_z

OK, now what are F_x, F_y and F_z in terms of q, m, velocities, etc? Also, write x_double-dot in lieu of a_x, etc.

 

[rude man]

Z-axis is the most important one...
And please, do not use cartesian coordinates to solve equations of motions. This is just overkill for this problem.

Request not granted :grumpy:. And the answer is in x, not z.

 

[rude man]

magnetic force perpendicular to the magnetic field. the y -axis is perpendicular!

Right. So there are only two equations, in x and in z.

 

[warnexus]

OK, now what are F_x, F_y and F_z in terms of q, m, velocities, etc? Also, write x_double-dot in lieu of a_x, etc.

F_x is force in the x or force sub x(probably should have done that)

you mean like this:

F_subx = m* a..
F_subz = m* a..

 

[rude man]

F_x is force in the x or force sub x(probably should have done that)

you mean like this:

F_subx = m* a..
F_subz = m* a..

No. Look at what I wrote.
You need to introduce q, B, and velocity components in lieu of forces F_x etc.

For a_x you should write d²x/dt² etc.

 

[warnexus]

No. Look at what I wrote.
You need to introduce q, B, and velocity components in lieu of forces F_x etc.

For a_x you should write d²x/dt² etc.

oh I see.

charge* velocity* magnetic field = mass * (d^2 * x)/(d*t^2)

may you explain how you got acceleration to be d^2 * x)/(d*t^2). I went through the chapter called amperes law where this question was associated with and don't see the equation like that of (d^2 * x)/(d*t^2)

d as in distance, t as in time, and what is x and we need to find time?

 

[rude man]

oh I see.

charge* velocity* magnetic field = mass * (d^2 * x)/(d*t^2)

may you explain how you got acceleration to be d^2 * x)/(d*t^2). I went through the chapter called amperes law where this question was associated with and don't see the equation like that of (d^2 * x)/(d*t^2)

d as in distance, t as in time, and what is x?

Question: have you had calculus incl. elementary differential equations? I never know what level of student I am dealing with.

This is a pretty sophisticated problem, involving coupled motions. I.e. the initial x velocity generates a z velocity, and the z velocity in turn generates an x velocity. I wouldn't try to solve this in a non-calculus manner. Maybe mfb knows how ...

 

[warnexus]

Question: have you had calculus incl. elementary differential equations? I never know what level of student I am dealing with.

This is a pretty sophisticated problem, involving coupled motions. I.e. the initial x velocity generates a z velocity, and the z velocity in turn generates an x velocity. I wouldn't try to solve this in a non-calculus manner. Maybe mfb knows how ...

i had two semester of calculus: differential and integral. taking 3rd and last semester of calculus in the spring

 

[rude man]

i had two semester of calculus: differential and integral. taking 3rd and last semester of calculus in the spring

OK. Then I can say that I meant d²x/dt² to be "the second derivative of x with respect to time". Aka "acceleration in the x direction". Etc.

My way you wind up with two second-order linear differential equations in x and z. Solve for x and there's your answer. The equation in x is the same you get from a spring-mass oscillator.

(You're not asked for the z solution which is similar but not exactly the same as a spring-mass oscillator).

 

[warnexus]

OK. Then I can say that I meant d²x/dt² to be "the second derivative of x with respect to time". Aka "acceleration in the x direction". Etc.

My way you wind up with two second-order linear differential equations in x and z. Solve for x and there's your answer. The equation in x is the same you get from a spring-mass oscillator.

(You're not asked for the z solution which is similar but not exactly the same as a spring-mass oscillator).

oh i see. interesting.

btw how come magnetic field is given in Gauss, the previous questions that involved with magnetic field were given in Tesla? Wouldn't Tesla have to be used?

 

[rude man]

oh i see. interesting.

btw how come magnetic field is given in Gauss, the previous questions that involved with magnetic field were given in Tesla? Wouldn't Tesla have to be used?

1 Tesla = 10,000 Gauss. Your teach is trying to make life even more miserable than it has to be ... tell him you won't do any more problems in any other than SI units...

Actually, pure physicists tend to use the cgs system which includes the Gauss. Applied physicists and engineers gravitate towards the SI system, aka 'rationalized mks" system ("meter, kilogram, second").
.

 

[warnexus]

1 Tesla = 10,000 Gauss. Your teach is trying to make life even more miserable than it has to be ... tell him you won't do any more problems in any other than SI units...

Actually, pure physicists tend to use the cgs system which includes the Gauss. Applied physicists and engineers gravitate towards the SI system, aka 'rationalized mks" system ("meter, kilogram, second").
.

he did teach us how to convert units super fast in the first physics lecture. he really want us to convert units in some assigned problems. but it's for practice which I totally needed but am getting better at. for this one it would be :

90 Gauss * (1 Tesla/10,000 Gauss) = .009 Tesla.

okay now to plug these values in:

charge * velocity * magnetic field = (mass of an electron) (2nd derivative of position respect to time)

(1.6 * 10 ^ -9 C) * (8.4 * 10 ^ 6 m/s) (.009 T) = (9.1 * 10 ^ -31 kg) (2nd derivative of position respect to time.

well I am left with acceleration given as a constant value of 1.3 * 10 ^ 26 m/s^2 . I could integrate this twice if this was a function so I can get distance but its just a constant value

 

[rude man]

he did teach us how to convert units super fast in the first physics lecture. he really want us to convert units in some assigned problems. but it's for practice which I totally needed but am getting better at. for this one it would be :

90 Gauss * (1 Tesla/10,000 Gauss) = .009 Tesla.

okay now to plug these values in:

charge * velocity * magnetic field = (mass of an electron) (2nd derivative of position respect to time)

Right!

(1.6 * 10 ^ -9 C) * (8.4 * 10 ^ 6 m/s) (.009 T) = (9.1 * 10 ^ -31 kg) (2nd derivative of position respect to time.

well I am left with acceleration given as a constant value of 1.3 * 10 ^ 26 m/s^2 . I could integrate this twice if this was a function so I can get distance but its just a constant value

Don't mess with numbers now. You need to get your equations first.

So why not start with the z-axis. What does F = ma look like for it? Keep in mind that the velocity is in x and the B field is in y direction, so use the appropriate velocity component ...

 

[mfb]

Request not granted :grumpy:. And the answer is in x, not z.

Right, but the deflection in z-direction is important.

This is a pretty sophisticated problem, involving coupled motions.

It is not. There is a very simple solution if you know some basics about that type of motion involved here.

 

[warnexus]

Right!Don't mess with numbers now. You need to get your equations first.

So why not start with the z-axis. What does F = ma look like for it? Keep in mind that the velocity is in x and the B field is in y direction, so use the appropriate velocity component ...

well if we start with the z axis, the z axis is the axis that is towards you. there is no velocity in the z, so its zero

charge * velocity * magnetic field = m ( acceleration)
0 = m(acceleration)

mass * acceleration is also zero

 

[rude man]

well if we start with the z axis, the z axis is the axis that is towards you. there is no velocity in the z, so its zero

Oh no? If the initial velocity is in the x direction and the B field is in the y direction, you don't expect to see any velocity buildup in the z direction?

Write your F = ma for the z axis.

 

[warnexus]

Oh no? If the initial velocity is in the x direction and the B field is in the y direction, you don't expect to see any velocity buildup in the z direction?

Write your F = ma for the z axis.

is that because the direction of velocity is perpendicular to the magnetic field that's why there is velocity build-up?

i will try to write one out.

(charge) (velocity)(magnetic field) = (mass of an electron) (acceleration_sub z)

 

[rude man]

is that because the direction of velocity is perpendicular to the magnetic field that's why there is velocity build-up?

i will try to write one out.

(charge) (velocity)(magnetic field) = (mass of an electron) (acceleration_sub z)

That's good, but which velocity component is the appropriate one?
Remember, F = qv x B is a vector equation.

 

[warnexus]

That's good, but which velocity component is the appropriate one?
Remember, F = qv x B is a vector equation.

(charge)(velocity sub_z)(magnetic field) = (mass of an electron) (acceleration sub_z)

 

[rude man]

(charge)(velocity sub_z)(magnetic field) = (mass of an electron) (acceleration sub_z)

Nope. What direction does the velocity have to be in order for the B field, which is in the y direction, to give motion in the z direction?

 

[warnexus]

Nope. What direction does the velocity have to be in order for the B field, which is in the y direction, to give motion in the z direction?

velocity would need to be in the y direction

 

[rude man]

velocity would need to be in the y direction

No. Velocity has to be at right angles to the B field and to the z direction in order to produce motion along the z direction, right? And don't write the full names, use q for charge, B for mag field, v for velocity ...

BTW do you use vectors in your physics course?

 

[warnexus]

No. Velocity has to be at right angles to the B field and to the z direction in order to produce motion along the z direction, right? And don't write the full names, use q for charge, B for mag field, v for velocity ...

BTW do you use vectors in your physics course?

Yes, Vectors was covered... Oh I see now! Kinematic Equations! Now I know.

 

[rude man]

Yes, Vectors was covered... Oh I see now! Kinematic Equations! Now I know.

OK, so back to writing the equation for the z motion ...

 

[warnexus]

OK, so back to writing the equation for the z motion ...

q v sin(90) B = m * a..