Find the Velocity given Magnetic field and Magnetic Force

[MedEx]

Homework Statement

A proton moves through a uniform magnetic field given by B=(13.7i−22.9j+23.5k) mT. at time t1 v=(vxi +vyj+1.68k)km/s and the magnetic force is given by F= (4.22e-17i + 2.52e-17j). What are vx and vy

Homework Equations

F=q(v x B)

The Attempt at a Solution

First I converted into SI units so Bi=.0137T, Bj=.0229T, Bk=.0235T and vz= 1680 m/s
Then I set up (aybz-byaz)i + (azbx-bzax)j + (axby-bxay)k (v as b, B as a)
multiplied everything by the charge (1.6e-19)
Now I am at (61.5552-.0376vy)e-19i + (.0376vy-36.8256)e-19j + (.02192vy-.03664vx)e-19k = (4.22e-17)i + 2.52e-17)j
Set the is equal to each other and the js equal to each other and I get vy as -9586.3 and vx as 7681.53

Is it something simple like switching v and B in which is a and which is B? I don't want to waste my guesses. Thanks.

 

[PeroK]

Homework Statement

A proton moves through a uniform magnetic field given by B=(13.7i−22.9j+23.5k) mT. at time t1 v=(vxi +vyj+1.68k)km/s and the magnetic force is given by F= (4.22e-17i + 2.52e-17j). What are vx and vy

Homework Equations

F=q(v x B)

The Attempt at a Solution

First I converted into SI units so Bi=.0137T, Bj=.0229T, Bk=.0235T and vz= 1680 m/s
Then I set up (aybz-byaz)i + (azbx-bzax)j + (axby-bxay)k (v as b, B as a)
multiplied everything by the charge (1.6e-19)
Now I am at (61.5552-.0376vy)e-19i + (.0376vy-36.8256)e-19j + (.02192vy-.03664vx)e-19k = (4.22e-17)i + 2.52e-17)j
Set the is equal to each other and the js equal to each other and I get vy as -9586.3 and vx as 7681.53

Is it something simple like switching v and B in which is a and which is B? I don't want to waste my guesses. Thanks.

You should try learning some Latex:

https://www.physicsforums.com/help/latexhelp/

Also, what about using some algebra. For example, you should be able to get an equation relating $v_x$ and $v_y$, given that the force in the z-direction is 0. It would make it a lot easier to see where you might be going wrong. For you and us!

 

[MedEx]

Also, what about using some algebra. For example, you should be able to get an equation relating vxvxv_x and vyvyv_y, given that the force in the z-direction is 0. It would make it a lot easier to see where you might be going wrong. For you and us!

I can put $.02192 v_y-.03664 v_x=0$ that means $.02192 v_y=.03664 v_x$ and therefore $v_y=1.67153 v_x$ which obviously doesn't work for the numbers got because my $v_y$ was negative and not 1.67153 times my $v_x$ value

So we definitely know I'm wrong, but I already knew that

 

[PeroK]

I can put $.02192 v_y-.03664 v_x=0$ that means $.02192 v_y=.03664 v_x$ and therefore $v_y=1.67153 v_x$ which obviously doesn't work for the numbers got because my $v_y$ was negative and not 1.67153 times my $v_x$ value

So we definitely know I'm wrong, but I already knew that

What about getting a formula for $v_y$ in terms of $v_z$, which you know? And take it from there.

 

[MedEx]

What about getting a formula for $v_y$ in terms of $v_z$, which you know? And take it from there.

I just don't really know how I would relate the two of them outside of how I'm using them to try to get the cross product, and that isn't working. Sorry.

 

[PeroK]

I just don't really know how I would relate the two of them outside of how I'm using them to try to get the cross product, and that isn't working. Sorry.

What I would do is this:

$F_z = q(v_xB_y - v_yB_x) = 0$

$v_x = \frac{B_x}{B_y}v_y$

And, I can see immediately that your numbers are wrong. You may be doing the cross product wrong.

Anyway, then I would do:

$F_x = q(v_yB_z - v_zB_y)$

And get an expression for $v_y$.

Then plug in the numbers.

 

[MedEx]

Fx=q(vyBz−vzBy)Fx=q(vyBz−vzBy)F_x = q(v_yB_z - v_zB_y)

so that's $4.22e-17 = 1.6e-19(.0137v_y - 38.472)$ multiply both sides by 1.6e-19 and you get $263.75 = .0137v_y - 38.472$ add 38.472 to both sides and divdie by .0137 and you get $v_y$ = 22060.
Then you relate $v_y$ to $v_x$ with

vx=BxByvyvx=BxByvyv_x = \frac{B_x}{B_y}v_y

that means that $v_x= .598253v_y$ and therefore $v_x$ is 13197.5
does that make sense?

 

[PeroK]

so that's $4.22e-17 = 1.6e-19(.0137v_y - 38.472)$ multiply both sides by 1.6e-19 and you get $263.75 = .0137v_y - 38.472$ add 38.472 to both sides and divdie by .0137 and you get $v_y$ = 22060.
Then you relate $v_y$ to $v_x$ with

that means that $v_x= .598253v_y$ and therefore $v_x$ is 13197.5
does that make sense?

From your OP, $B_x$ is +ve and $B_y$ is -ve. So, $v_x$ and $v_y$ should have opposite signs.

$v_x= -0.598253v_y$

Can you do the other bit now?

PS I don't get that for $v_y$.

Do you know the numerical answers you're supposed to get?

 

[MedEx]

From your OP, BxBxB_x is +ve and ByByB_y is -ve. So, vxvxv_x and vyvyv_y should have opposite signs

I've been using $B_y$ as .0229 this whole time! no wonder none of the math made sense. Okay so I made 2 mistakes in my last one. The first was that I had .0229 as positive and the second was that I wrote in the value of $B_x$ where I should have put in the value for $B_z$. I got $v_y$ as 9586.3 and $v_x$ as -5735.03 which is the correct answer. Thanks so much for your help!

 

[PeroK]

I've been using $B_y$ as .0229 this whole time! no wonder none of the math made sense. Okay so I made 2 mistakes in my last one. The first was that I had .0229 as positive and the second was that I wrote in the value of $B_x$ where I should have put in the value for $B_z$. I got $v_y$ as 9586.3 and $v_x$ as -5735.03 which is the correct answer. Thanks so much for your help!

1) I'm not connected with any university, so I don't know how these things are marked, but it seems to me that if you'd got:

$v_y = \frac{1}{B_z}(\frac{F_x}{q} + B_yv_z)$

$v_x = \frac{B_x}{B_y}v_y$

Then, you'd get maximum credit in an exam even if you then plugged in the wrong numbers.

2) It certainly would be a lot easier for anyone on here to see whether I've done this right. Anyone can easily check my expressions and see whether I've gone wrong.

3) Working with algebra as far as is possible seems easier (to me) than manipulating lots of multi-digit numbers.

4) The more advanced the physics the more of a handicap the "plug-and-chug" approach becomes!