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Position of a Proton in a Magnetic Field

  • Thread starter ZeroSum
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Homework Statement


At t=0, a proton is moving with a speed of [tex]5.8\times 10^5[/tex] m/s at an angle of 30° from the x-axis, as shown in the figure. A magnetic field of magnitude 1.7 T is pointing in the positive y-direction.

What will be the y-coordinate of the proton 15 [tex]\mu s[/tex] later?

Homework Equations



[tex]F = |q| v \vec{B} \sin{\alpha}[/tex]
[tex]F = \frac{m v^2}{r}[/tex]

The Attempt at a Solution



Hello everyone! I'm a first time poster, so hopefully I supplied all of the relevant info correctly. Thank you in advance for any help offered.

First I used the right hand rule to figure out which direction and path the proton will take. It seems like it should be taking an upward helical path in the counterclockwise direction (viewed from above).

Since they're asking what the y-coordinate is, is this really as simple as taking the y-component of the velocity given and multiplying by the time?

If I do that I get:

[tex]5.8\times 10^5\; m/s \cdot sin(30^{\circ}) \cdot 15\times 10^{-6}\; s = 4.35 m[/tex]

I feel as if I should be using the equation for the force on the moving proton in the magnetic field, but I don't think that affects it in the +y direction, since that component of the velocity is parallel to the field. Did I do this right?
 

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  • #2
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Welcome to PF :D

thought I could help you with your problem, then realised I'd forgotten basic electromagnetism haha... I'll have a look :)
 
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  • #3
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think about it, if the magnetic field is causing the particle to change velocity, yet its speed remains constant... how can your answer be correct? You've assumed that it continues moving at 30 degrees from x axis at a constant speed while still somehow accelerating upward?
I'm not sure I understand what you mean. The direction of the magnetic field is upward. That tells me that the only path along which the particle will feel a force is along the x axis, which is perpendicular to it. I am not assuming that there's any acceleration upward, but that the velocity upward remains constant and is the y-component of the original velocity. My assumption stems from the fact that the y-component of the velocity lies in the plane of the field (parallel to it), and that component should not be experiencing any forces or accelerations.
 
  • #4
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on second thoughts i think you're right. magnetic field in the y direction only influences components of velocity perpendicular. thus, yes it'd go in some sort of spiral around the y axis but its y velocity would remain unchanged.

If you want to check this you can do the lorentz force cross product:

[tex]\vec{F}=q\vec{v} \times \vec{B}[/tex]

you should find that force is only experienced in the x and z directions.
 
  • #5
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If you want to check this you can do the lorentz force cross product:

[tex]\vec{F}=q\vec{v} \times \vec{B}[/tex]
I never learned how to do this since I'm taking an algebra based physics course instead of the calculus based one (I wish I'd thought more about that, but C'est la vie).

I know how to do the actual calculations using matrices, but I don't know how to represent a problem like this in matrix form. Can you point me to a resource which will help me understand how to do this?

Thank you for the suggestions, by the way.
 
  • #6
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Well the lorentz force is really a vector problem rather than a calculus/matrix one. Magnetic field, velocity and force are all vectors. Do you know about splitting vectors into their components?

For example I could write the force equation instead like this:

[tex]\begin{pmatrix} F_x \\ F_y \\ F_z \end{pmatrix}=q \begin{pmatrix} v_x \\ v_y \\ v_z \end{pmatrix} \times \begin{pmatrix} B_x \\ B_y \\ B_z \end{pmatrix}[/tex]

When doing a cross product of two vectors, as is shown in the Lorentz force it can be nice to write this as the determinant of a matrix... makes the maths tidier, but its not a proper matrix in the mathematical sense! It's just a convenient notation.

I'm not really sure where to point you except maybe the beginning of a maths book in the vector section? I'm sure if you wade through the internet for a little while you'll find a nice explanation somewhere!
 
  • #7
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I decided to submit my answer and determined that I was correct. The problem appeared to be more complicated than it was (because it's helical).

The reason I mentioned it was the algebra based physics course is that they don't do matrix mathematics to determine anything in the algebra based course (I have no idea why - it's not related to calculus). We use trig for everything. :grumpy:

I know my matrix mathematics, but I don't know how to connect that knowledge to the physics values just yet. When I have more time (finals week coming up) I'll look over what you wrote and see if I can puzzle out the meaning with some web resources.

Thank you for taking the time to look at this problem for me!
 

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