Expression of radius and helical motion in a magnetic field

[thee qs]

Homework Statement

charged particule reaches a uniform magnetic B field with a velocity V , and at angle theta with the magnetic field .
What is the expression of the relation , expressed as a ratio, of the radius and helical step of the trajectory of the particule .

a- 2pi/tan a
b- 2pi tan a
c- 2pi tan a
d- tan a / 2pi

Homework Equations

r= mv/qb
p = (v cos theta 2 pi m) / qB
F= m * ( v^2 / r )
period = 2pim / qB
ffrquency and angular frequency
qB / 2pim qB / m

The Attempt at a Solution

[/B]
after algebraic manipulations, i ended up with ( m tangent theta / 2 pi ) , so correct answer is D ?

i posted the pic and rewrote retranslated the problem

 

[thee qs]

 

[Nathanael]

after algebraic manipulations, i ended up with ( m tangent theta / 2 pi ) , so correct answer is D ?

There isn’t any m in part D. Maybe you should show your steps.

If v is meant to be the speed of the object then I disagree with your first three “relevant equations.” (This is helical motion not circular motion.)

All those relevant equations aren’t needed though. If you find the ‘helical step’ in terms of the radius r, then it reveals the answer nicely.

 

[BvU]

after algebraic manipulations

You don't show them, so we can't comment.

r= mv/qb 1
p = (v cos theta 2 pi m) / qB 2

hard to read: v in 1 is not v in 2 but b in 1 is B in 2
And: do you mean /qB or do you mean /(qB) ? Physics is an exact science...

F= m * ( v^2 / r ) 3
period = 2pim / qB 4
ffrquency and angular frequency
qB / 2pim qB / m 5

Where do you use 3 ?
Who is pim in 4 ? Ah ! under the $\sum$ button there is a π waiting for you !
5 qB/2πm or qB/(2πm) ? Please understand that PF is not for stamp-approving your answers; that way anyone can score ten out of ten and we would get in trouble with teachers, parents and who else. Please check the guidelines (again?). Best we can do is ask guiding questions when we find a mistake in the working, or give hints when someone is stuck halfway.

In your case, I don't know what to do: you are smart enough to come up with answers, you just seem to be a little sloppy in checking them.

PS what happened to the 8V, 2A resistor network thread ?

 

[thee qs]

no parenthese . / qB

reposting the thread for the 8v 2a circuit.

as the reasons why i registered on this forum
- indeed i do intend to get 10 out of 10, but i intend to earn it .
- i did not register to get spoonfed the answers. but as you will see with time, i tend to get deep into the questions i fail to get at first, all the way to the sub-atomic level if need be. The exams is next week so i need to finish these before .

Also, i do not have access to people who can help me with the questions , so i wasnt able to check my answers , but i did try, which is why each question i ask on the forum has already been completed , i just haven't submitted them for grading yet, as i need to good average to get into my program in chemistry

back to the thread .

question only asks for the expression for raduis and helical step . .
the expression with i used that ended up with mass was the formula for radius in circular motion , and the other formula is for the helical step. i know that charged particle will travel in circular motion, not helical, if the magnetic field is orthogonal to the particle velocity's direction.
helical motion happens when the field is not 90 degree to the velocity ..

so i should use the radius from the helical motion and the helical step . so this question is only about helical motion then,

velocity vector has 2 components ;

v cos phi is for the helical step
v sin phi is for the helical radius

parallel velocity * period = ( v cos phi )*2pim /qB

book says the orthogonal velocity determines the radius , and should be substituted for v in the equation r = mv /qB

the only equation i have which does not have the mass is for the period T= 2pir / v or 2r = 2/B square root (
i have no equations for the radius or helical step that do not take the mass into account. ...
well tangent theta = Vy/Vx = D-y / L or
y= 1*eE/2m (l/V0)^2
i have another questions which could help us clarify this .( which is basically the same question but with given values instead of variables
let's put numbers in play

question states , charged particule moving at velocity v , arrives at a region , forming a theta angle , with the uniform magnetic field B . question then ask for the expression of the ratio ( rapport ) of the radius of the motion , with the helical step.

so if i guess from your previous posts, the radius i used was not the correct one, which would make sense as it is the formula for circular radius

so let's go step by step .

if particle is at rest at t=0 , and is subject to a potential difference of 15000 vols resulting in a acceleration over 1 distance of 1 cm .

so to calculate the velocity in the x-axis (i) if 1/2 mv^2 - qV = 0 then velocity = square root ( 2qV/m)
so square root of ( 2(1.602*10^-19)(15000))/9.109*10^-31 = 7.26 *10^7 which should be the electron velocity with a 15000

the electron then enters the magnetic field B = 0.01 which is ortogonal to the velocity, meaning at a 90 degree angle . resulting in a change of direction of the particle from a straight line to circular motion .

so the radius and helical step has to only be dependent on the parallel and perpendicular velocity components.

 

[BvU]

no parenthese . / qB

Yes parenthesis: you mean .../(qB) and not .../q * B
Or, much more legible: you mean $\mathstrut...\over qB$ and not ${\mathstrut...\over q} \, B$

reposting the thread for the 8v 2a circuit

Seen it. Many pages illegible stuff for such simple problem. You are in good hands with @haruspex
My advice: re-order the drawing to resemble a standard circuit: battery to the left, four parallel branches to the right.

as the reasons why i registered on this forum
- indeed i do intend to get 10 out of 10, but i intend to earn it .
- i did not register to get spoonfed the answers. but as you will see with time, i tend to get deep into the questions i fail to get at first, all the way to the sub-atomic level if need be. The exams is next week so i need to finish these before .

Also, i do not have access to people who can help me with the questions , so i wasnt able to check my answers , but i did try, which is why each question i ask on the forum has already been completed , i just haven't submitted them for grading yet, as i need to good average to get into my program in chemistry

Already noted and duly recognized in

$\$

And PF is a good place ( especially if you are not in a hurry ) . We've steered clear from the stamp-approving and you are clearly committed.
​

question only asks for the expression for raduis and helical step . .

Even stronger: it asks for the ratio !

parallel velocity * period = ( v cos phi )*2pim /qB

book says the orthogonal velocity determines the radius , and should be substituted for v in the equation r = mv /qB

True, but not efficient: you want r/p_h so it is much smarter to write $v_\parallel * T = 2\pi r$
and then write a comparable expression for p_h
Divide the two and you are done.

lets put numbers in play

This is not a good idea:
In general you want to work in symbols:

1. That way often a lot of factors cancel (as above)
2. You have the opportunity to check the dimensions -- a very useful way to find mistakes in reasoning and in math handling

Only at the latest possible moment and only if a numerical answer is required do you bring numbers into play.

so if i guess from your previous posts, the radius i used was not the correct one

That was indeed what I meant: only the component $\perp \vec B$ counts.

so the radius and helical step has to only be dependent on the parallel and perpendicular velocity components.

No: m, q and B also come in. But for the ratio r/p_h everything cancels, except $\theta$.

Best of 'luck'with your exam -- I'm going on holiday (in France !)

B

 

[thee qs]

ohhhhhh , is the answer C !?

i think it is...

 

[Nathanael]

ohhhhhh , is the answer C !?

i think it is...

I don’t think it is.
BvU and I have said that things cancel out nicely to give the answer. To exemplify this, let me re-ask the question, but without the magnetism:

“If I travel in a helix of radius R, with a ‘circular’ speed Vsinθ and ‘transverse’ speed Vcosθ then what is the ratio of helical step to R?”

Remember, the helical step is how far you move transversely during one circular loop.

 

[BvU]

ohhhhhh , is the answer C !?

i think it is...

Don't think. Work it out carefully. Physics is not a guessing game.

 

[thee qs]

tangent

sin / cos = tangent , so both v cancel out , leaving tangent in the numerator, thus eliminating 2 choices , no?

 

[BvU]

Oh, and if you are as smart as I think you are, you also check for some extreme cases: how would r/p_h behave for $\theta \uparrow 90^\circ$ ? and for $\theta \downarrow 0\circ$ ?

 

[thee qs]

sin 90 = 1
cos 90 = 0 , so 1/0 no good

from 0 to pi ,( 0 to 180 ) mouvement is upwards and from pi to 2 pi, goes downward...? ? hmmm

wait , cos 180 is -1 , sin 180 is 0

from 0 to 90 , sine increases, , cosine decreases
and from 90 to 180 , since decreases, cosine increases

 

[thee qs]

so helical step increases and radius decreases from 0 to 90 degrees,
and from 90 to 180 degrees , helical step decreases and radius increases
no?

 

[BvU]

tangent

sin / cos = tangent , so both v cancel out , leaving tangent in the numerator, thus eliminating 2 choices , no?

correct

sin 90 = 1
cos 90 = 0 , so 1/0 no good

So if $\theta = 90^\circ$, r is maximum and p_h=0. Why 'no good' ?

so helical step increases and radius decreases from 0 to 90 degrees,
and from 90 to 180 degrees , helical step decreases and radius increases
no?

Come on ! For $\theta \downarrow 0^\circ$, r is quasi non-existent ($v_\parallel$ very small) and p_h grows and grows. Why 'no ?' ?

 

[thee qs]

1/0 doesn't exist . having cos 90 at denominateur will make the entiere fraction divided by 0 . even then , it would mean C would be the answer which it isn't , and since it isn't D . it has to be A.

 

[thee qs]

well the correct answer was D.

which is the answer i withe method i had used.

and this question counted for 1% of my final grade...

well thx

 

[thee qs]

god, i had the right answer all along, and been trying for 3 days to understand people explaining to me the incorrect answer

so apparently i was right all along folks, should of trusted my judgement first

 

[BvU]

1/0 doesn't exist . having cos 90 at denominateur will make the entiere fraction divided by 0 . even then , it would mean C would be the answer which it isn't , and since it isn't D . it has to be A.

Point is that at 90 degrees r/p_h is also '1/0', so that is a nice agreement -- excludes answers a and b, just like you reasoned (#10), but (ambiguously?) described 'no good' in #13.

well the correct answer was D.
which is the answer i withe method i had used.
and this question counted for 1% of my final grade...

Of course it is D . It's not the 1% that counts here: you can get top grades if you can fine-tune your method of work to let careful steps and meticulous math replace impression and judgement.

god, i had the right answer all along, and been trying for 3 days to understand people explaining to me the incorrect answer
so apparently i was right all along folks, should of trusted my judgement first

This hurts and feels a bit unfair. I don't think you've been 'explained the wrong answer' (where, precisely?).

My subjective impression is that there is a culture issue: here at PF we ask with emphasis that people post their working and show detailed steps (not as colourful artwork, by the way ). Your 3 days were caused by

after algebraic manipulations, i ended up with ( m tangent theta / 2 pi )

which is wrong and leads us homework helpers to asking questions to check if you understand properly, or just follow some intuition. After all, our telepathic capabilities are near-zero so we don't know if you just happened to pick D while ignoring the m in your result !

There is nothing wrong with your intelligence and self-consciousness (I find myself presumptious writing this -- sorry) and perhaps there is some room to help you improve your approach and scientific attitude. That should yield you a lot more than 1%.

What about #7 ? Triggered me to state physics is not a guessing game...

Bonne 'chance ' with your exams.

B