How to Determine Proton Movement in Electric and Magnetic Fields?

[Astronuc]

so since the magnetic field is into the page (-z) then the opposite would be +z ??

Careful. Do not confuse magnetic field with magnetic force.

The magnetic force on a moving charge is perpendicular to the plane formed by the velocity and magnetic field vectors by virtue of v x B.

Please follow Doc Al's guidance.

 

[physicsbhelp]

wait, how am i confusing magnetic force and field?

and i really don't get the perpendicular part.

is doc al back yet?

 

[Doc Al]

wait astronuc, what do you mean.

doc al and the other person said that the magnetic force is in the -x direction and the electric force is in the +x direction.

and you said that:
"the electric field is equal and opposite the force imposed by the magnetic field on the moving proton."

so since the magnetic field is into the page (-z) then the opposite would be +z ??

Read carefully. Equal and opposite the magnetic force, not the magnetic field. (Astro and I are saying the same thing, of course.)

okay so i am trying to understand this electric field stufff by the websites. and this is what i get "The electric field from a positive charge points away from the charge; the electric field from a negative charge points toward the charge. " and since we are talking about a proton, it is a positive charge. so the electric filed will point away from the charge.


so is the electirc filed in the positive x direction? since its pointing away?

Those sites are talking about the field of a point charge--but that's not related to this problem. You want to understand the force on a charge due to an external field. See my previous post (#32).

 

[Astronuc]

The electric field (a vector field) imposes a force (a vector) which is parallel with the field.

A magnetic field (a vector field) imposes a force which is perpendicular to the magnetic field. The magnetic force is proportional cross (vector) product of the velocity vector and magnetic field vector.

One is trying to find the magnitude and direction of the electric field such that the force on the proton is equal in magnitude, but opposite in direction, to the force imposed by the magnetic field ($$\vec{B}$$).

 

[physicsbhelp]

i am not trying to find the magnitutde of the elctric field.



and now i understand you doc al.

but where do i find ben crowell's txt books.

i don't have that much time on my hands. its true.

 

[physicsbhelp]

so if the electric field is not in the +z or +x direction, it must be in the +y direction right?

 

[Astronuc]

so if the electric field is not in the +z or +x direction, it must be in the +y direction right?

Go back to post #9, where you are correct on the direction of the magnetic force. The electric force must be in the opposite direction in order to be able to cancel the magnetic force.

 

[physicsbhelp]

i know that Astronuc, i know that the electric Force is in the +x but i am looking for the electic field.

 

[physicsbhelp]

wiat astronuc said that electirc filed is parallel to magnetic field?
so is it also in the -z diretion??

 

[Doc Al]

i am not trying to find the magnitutde of the elctric field.

But you do need the direction of the electric field.

but where do i find ben crowell's txt books.

I gave his link in post #26.

so if the electric field is not in the +z or +x direction, it must be in the +y direction right?

Who said that it's not in one of those directions? Answer each of these questions:

Which way does the magnetic force point?

Which way must the electric force point to balance the magnetic force?

So which way must the electric field point to produce such a force? (See post #32.)

 

[physicsbhelp]

Who said that it's not in one of those directions? Answer each of these questions:

Which way does the magnetic force point?

Which way must the electric force point to balance the magnetic force?

So which way must the electric field point to produce such a force? (See post #32.)

1) -x
2) +x
3) +x

 

[physicsbhelp]

3) ?

 

[Doc Al]

1) -x
2) +x
3) +x

Exactly!

 

[Astronuc]

1) -x
2) +x
3) +x

Correct.

If the velocity vector direction is +y, and the magnetic field is +z (into the page), then the cross (vector) product must be -x, by the right hand convention. Therefore the electric field vector must point in the opposite (+x) direction, in order to impose a force on a 'positive' charge in the +x direction.

 

[physicsbhelp]

THANKS!


now for part b) ;)

 

[physicsbhelp]

now i have to : Determine the speed of the proton while in the box if it continues to move vertically upward. Express your answer in terms of the fields and the given quantities.

to find velocity do i use the formula Fm=qv*B

 

[physicsbhelp]

but how can i express myh anwer in terms of the fields when i am not given anything about charge.
do i somehow need to incorporate the proton.

or should i use a different formula.

 

[Doc Al]

You need to solve for the speed in terms of both fields. Set up an equation for the net force on the charge in terms of the fields. (What's the net force on the charge?)

 

[Astronuc]

Fm=Fe

qvB=qE??

Correct. Note that the charge cancels. Charge by the way is a scalar quantity.

Now solve for v.

 

[physicsbhelp]

thanks

 

[physicsbhelp]

this is for part c) On the figure above, sketch the path of the proton after it leaves the box.

 

[physicsbhelp]

which one is it? i drew all three of them, i just don't know which way I am supposed to draw it.

 

[Astronuc]

What is the trajectory of charge moving in a constant magnetic field?

Hint - http://hyperphysics.phy-astr.gsu.edu/Hbase/magnetic/maspec.html

 

[physicsbhelp]

too complicated words? i am not sure what you mean, i looked at the hint.
wait why can't you tell me because i drew those three image.s

 

[Astronuc]

too complicated words? i am not sure what you mean, i looked at the hint.
wait why can't you tell me because i drew those three image.s

trajectory is the path. What path does a charged particle of constant speed take in a uniform magnetic field?

 

[Astronuc]

yes but the particle gets curved becuase it leaves that box thingy. and is moved by the force in a circular path

Correct. The path is circular, and it would travel in a circular path until it hit the side of the box.

The radius of the path is the cyclotron radius.


Since we established that the magnetic force is initially in the -x direction, the path must be circular in the counterclockwise direction.

 

[physicsbhelp]

exactly isn't that what i drew!?

and i have no clue what the cyclotron radius is haha

it is the first picture i drew, but more circular?

 

[Astronuc]

The proton moves in a circular path immediately upon leaving the box. If it had a small radius of curvature, then the proton would impact the top of the box. If it had a large rad. of curv. it would hit the side.

See hint I provided, and look at the figure.

 

[physicsbhelp]

but can you draw a sample because i can't really understand.

 

[Astronuc]

Draw it like figure #1, but as a circle.

Something like - http://hyperphysics.phy-astr.gsu.edu/Hbase/magnetic/imgmag/maspec.gif

but in the direction of your figure #1.

 

[physicsbhelp]

so i draw three lines? to the left

 

[Astronuc]

so i draw three lines? to the left

The three lines could be three different speeds, mass or charges in the same magnetic field.

The radius is the ratio of the linear momentum (mv) and the product of q and B, i.e.

r = (mv)/(qB).

If v, q, and B are constant then r is simply a linear function or m and that is how one can separate particles of different mass (but with same v and q)

 

[physicsbhelp]

h/o
let me draw another figure and post it so you can see if it is right

 

[Astronuc]

h/o
let me draw another figure and post it so you can see if it is right

Just draw a circle, such that the circumference is tangent at the exit of the box and goes counterclockwise to the top or side of the box.

 

[Astronuc]

so even theoughtthe problem says draw it after the proton leaves the box it can go three ways.

What do you mean three ways?

This is just a sketch!

If you drew it to scale, then you could use the knowledge of m, q, v, B to draw an exact path.

Your sketch is more or less correct. If it was a fast proton, or if B was very strong and the box relatively large, the path would be a semi-circle into the top of the box. If the proton was moving very slow or the B was very weak, then the proton would travel a path between a semi-circle and 3/4 of a circle into the side of the box, or perhaps more.