Methods in Calculating with Vector Components

[wjdgone]

<Moderator's note: Moved from a technical forum and therefore no template.>

Homework Statement

A proton (q = 1.60 x 10^-19) is in a uniform, 0.500 T magnetic field. This proton has velocity components v_x = 1.50 x 10⁵, v_y = 0, and v_z = 2.00 x 10⁵ m/s. Find the force on the proton at t=0.

2. Homework Equations
F= qv x B

The Attempt at a Solution

I was wondering specifically about how to go about calculating the v in F = qv x B. Now, I understand the method using vectors (F = q(v_xi + v_zk) x Bi = qv_zBj).

But this method isn't my first instinct when solving this problem. I'd want to go about graphing the velocity components, draw a parallelogram, and use a² + b² = c² to avoid using the component method. What I got using this method was v = 2.50 x 10⁵, instead of the 2.00 x 10⁵ that you get when using the first, component method above. Plugging 2.50 x 10⁵, I got F = 2.00 x 10^-14, compared to 1.60 x 10^-14 using the first method.

Is this second method legit? Why is there a difference in the results?

 

[Cthugha]

You do not have to calculate the v. [itex]\vec{F}=q\vec{v}\times\vec{B}[/itex] is an all out vector equation. The cross product in there acts on vectors, not on numbers, so you do not need to calculate the magnitude of the velocity vector. It does not enter at any point. The velocity vector you need is already given in the problem statement.

Your equation F=qv x B (without vectors and just considering numbers) also does not make sense as the cross product in there is defined for vectors and not for numbers. There is no way to apply this equation to magnitudes only. Also, the vector component version you wrote does not make any sense to me.

It should read: [itex]\vec{F}=q\vec{v}\times\vec{B}=q\begin{pmatrix}
v_y B_z-v_z B_y\\
v_z B_x-v_x B_z \\
v_x B_y-v_y B_x
\end{pmatrix}[/itex]

Note that you also need to know the direction of the magnetic field to get the force. If you know that, you should be able to do the math.

 

[wjdgone]

Oh, I'm sorry I forgot to add the arrows to show that the equation is dealing with vectors. Pretty new to the site, trying to figure out formatting.

When you wrote

It should read: [itex]\vec{F}=q\vec{v}\times\vec{B}=q\begin{pmatrix}
v_y B_z-v_z B_y\\
v_z B_x-v_x B_z \\
v_x B_y-v_y B_x
\end{pmatrix}[/itex]

is this the same notation as
( i j k )
(v_x v_y v_z )
(B_x B_y B_z )?

Again, sorry for the untidiness.

 

[Cthugha]

To be honest, I do not fully get your notation. What is (i j k) supposed to be?

Anyway, the cross product only yields non-zero results for the orthogonal components of the vectors involved. That should tell your that your parallelogram method, which gives you the magnitude and not the vector components cannot yield correct results.

 

[wjdgone]

I use i, j, and k as vector components - they can be substituted with x, y, and z, respectively. Normally they also have arrows or hats, but I'm still figuring out formatting haha. Otherwise, I got to the answer from your explanation. Thank you!

 

[Cthugha]

A guide to using LaTex for formatting equations on these forums can be found here: https://www.physicsforums.com/help/latexhelp/

The BBCodes used for showing Tex math mode are also given on the end of that page. You can also right click on any equation such as the one posted above and choose Show math as -> Tex Commands in order to see the LaTex code used. I hope that helps.