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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?