Diagram for Charge Directions-Electron Path-Magnetic Fields

[julianwitkowski]

Homework Statement

A magnetic field of 0.0200 T [up] is created in a region.

a) Find the initial magnetic force on an electron initially moving at 5.00 E+6 m/s [N] in the field.

b) What is the radius of the circular path?

c) Make a sketch showing the path of the electron.

Electron Charge = q = 1.6⋅10^-19
Coulomb Constant = k = 9⋅10⁹

Homework Equations

F = qvB

F = k ⋅Q ÷ r² → r = √ k ⋅Q ÷ E ?

F = qvB = mv²/r

a=qE/m

The Attempt at a Solution

[/B]
a) F = qvB = 1.6⋅10^-19⋅5⋅10⁶⋅2⋅10^-2 = 1.6⋅10^-14

b) r = √ 9⋅10⁹ ⋅1.6⋅10^-19 ÷ 1.6⋅10^-14 = 314.6

I'm kinda confused as to the distinction between F = k ⋅q ⋅q ÷ r² and E = k ⋅Q ÷ r².
I hope I used this right, but I'm finding lots of different answers... My answer seems way to large unless it is not in meters... I'm getting 314 so I think this is wrong.

If it's wrong am I supposed to use F = qvB = mv²/r ? If this is true how am I supposed to use the rest mass of the electron?

a=qE/m and F=qvb? F=ma ... ?

For the diagram, I'm having trouble visualizing this.
If you can help me understand what the diagram should look like...
Extra kudos for a half tried paint brush image because I'm more a visual learner.

 

[BvU]

Hello Julian,

Your second and fourth relevant equations have to do with electrostatic forces. They don't appear in this exercise.
First and third are fine: magnetic force = centripetal force.

If you can, try to fix the vector version of the Lorentz force in your memory, and you'll be well equipped for a very long time !

$$\vec F = q\;(\vec E + \vec v \times \vec B) $$

One other tip: always -- always -- write dimensions after quantities.

distinction between F = k ⋅q ⋅q ÷ r2 and E = k ⋅Q ÷ r2.

One is a force, the other is a force per unit of charge. Confusion eliminated with ##\vec F = q\vec E## (you see it works already ! )

use F = qvB = mv2/r

yes. m you'll have to look up. In decent units: kilograms.

a=qE/m and F=qvb? F=ma ... ?

A definite yes! for all three

Re visualization: sorry, no time. But there's lots of stuff all over, e.g. here

Oh, and we had a useful (I hope) thread about right-hand rules etc here (but my advice is to remember one rule only: the right-hand rule. And to forget about Fleming, left hand and what have you until after your PhD )

 

[julianwitkowski]

One other tip: always -- always -- write dimensions after quantities.

Thank you very much for your time! Do you mean with the S/I units for algebra I'm assuming? I can imagine when they really pour this stuff on to me it might have to factor with the dimensions of the units? That scares me.

Anyway, that was insightful... I'm trying to figure out exactly what you mean with everything there, particularly the vector version of the Lorentz force equation, as I do want to be equipped for a long time...

But I digress; in the meantime, is this close for the diagram or am I way off? Should (up) point to the top of circle or off of the x,z plane? Should I even include a 3rd dimension... So much confusion. I think I have this wrong in the image for sure... I've been looking through pics and I've been trying to absorb stuff from hyper physics and alike. I'm not sure if it's supposed to have all the magnetic field lines or not or maybe lines to show the force...

Another thing that makes me think is how do you state the direction, if it's centripetal force, I think of it revolving around an axial point, but the question is vague I don't know... If that is right, do you state something like 90° perpendicular to the velocity? I'm probably way off.

 

[BvU]

Re units: I mean instead of writing F = qvB = 1.6⋅10^-19⋅5⋅10⁶⋅2⋅10^-2 = 1.6⋅10^-14 one writes
F = qvB = 1.6⋅10^-19C ⋅ 5⋅10⁶ m/s ⋅ 2⋅10^-2 T

= 1.6⋅10^-14 C m/s kg/s²/A
= 1.6⋅10^-14 kg m/s²
= 1.6⋅10^-14 N

(making good use of [A] = [C/s])​

And your F = qvB = mv²/r gives r = mv/q/B = 9.11 10^-31 kg ⋅ 5⋅10⁶ m/s ##\Big /## (1.6⋅10^-19C ⋅ 2⋅10^-2 T )

= 1.4 10^-3 kg m/s ##\Big /## (C kg/s²/A )
= 1.4 10^-3 kg m/s ##\Big /## ( kg/s )
= 1.4 10^-3 m​

I know I'm exaggerating, but it really helps to check dimensions painstakingly.

Re direction of the force: You've got it quite correctly (apart from typo like E (up) instead of B). The + in the center of the circle marks the center I assume (and not some positive charge...). And the exercise mentions N, so you want to make clear that F points West initially.

Vector product: turn v towards B over the smallest angle. The direction where the corkscrew goes is the direction of ##\vec v \times \vec B##. To the left in your picture. The sign of q (electron) makes it point the other way -- as you drew it.

If you prefer handedness or don't like corkscrews: thumb is v index finger is B middle finger bent points in the direction of ##\vec v \times \vec B##.

x,y,z is somewhat vague -- and you draw a left-handed coordinate system. My advice (see above): never ever before your PhD unless forced. Stck to right-handed !The exercise doesn't give a size for the 'region', so you can't tell if the electron comes out again or keeps circling. I'd draw a circle too, assuming the region extends over more than 3 mm.

 

[julianwitkowski]

It really helps to check dimensions painstakingly.

Will do, I think I stopped because I'm really OCD about being neat and having all those units and dimensions makes it look more confusing.

About the right hand rule, I think I see what you mean but still may have some work to do? Is this better?

Is it okay to have north in South's negative? Normally I imagine South is the negative of North. Or should I flip the circle's N and S?

And with F = qvB = mv² / r ... Why is r = mv / q/B better than r = mv² / F

One thing I need to understand is why the initial force is west?

How does an initial magnetic field directed up, an initial velocity directed north equal an initial force directed West?

 

[BvU]

Picture was good already. Now it's near perfect.
N and S: just fine.
B is up: given.
Force is West: ##\
\vec F = q\;(\vec E + \vec v \times \vec B)\ =\
q\; \vec v \times \vec B##

Vector product: turn v towards B over the smallest angle. The direction where the corkscrew goes is the direction of ## \ \vec v \times \vec B## . To the east. The sign of q (electron) makes it point the other way -- as you drew it.

 

[julianwitkowski]

Vector product: turn v towards B over the smallest angle. The direction where the corkscrew goes is the direction of v⃗ ×B⃗ . To the east. The sign of q (electron) makes it point the other way -- as you drew it.

Ok, I think it's time to watch some videos and do an indepth investigation on the right hand rule.
Thank you for getting me here, I really appreciate this because I didnt have a clue before.

Thanks :)