I've tried several different ways, with a tutor, and with other colleagues. One of my colleagues has even gone to the professor with whom he worked out an equation for the component, and it was later confirmed to be incorrect.

Here is the question:

A proton ($q = 1.6 \times 10^{-19}$ C, $m = 1.67 \times 10^{-27}$ kg) moving with constant velocity enters a region containing a constant magnetic field that is directed along the $z$-axis at $(x,y) = (0,0)$ as shown. The magnetic field extends for a distance $D = 0.55 m$ in the $x$-direction. The proton leaves the field having a velocity vector $(v_x, v_y) = (4.8 \times 10^5, 1.8 \times 10^5)m/s$.

This is the diagram:

enter image description here

The question we have is how to determine $h$, the $y$ coordinate of the proton as it leaves the magnetic field.

I've already determined that the velocity $v$ is equal to 512640.2247 m/s, and that the radius $R$ is 1.533 m.

The trajectory of the proton is curved on the inside, so we can't use anything involving any derivation of similar triangles. I've tried solving for $v$ as a function of time, but then realized that $v_y$ is always changing anyway, which lead me to calculate the acceleration so the time term that I used incorporated the change, but that didn't work either.

Edit: Really sorry folks, I wrote this in a hurry to get to work on time. Yes, the radius I mentioned is indeed the radius of the circular trajectory of the proton after it enters the magnetic field.

Also, I do understand to keep things variable, and I mentioned the radius and velocity only as to sort of show what I have available to work with.

I've drawn the imagined circle suggested below, and I've made the realization that it would be helpful to relate the radius, the angle, and the velocity, but I'm not sure how to do so accurately.


closed as off-topic by user10851, John Rennie, Abhimanyu Pallavi Sudhir, jinawee, tpg2114 Feb 22 '14 at 17:37

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  • 1
    $\begingroup$ What is $R$ the radius of? $\endgroup$ – David Z Feb 22 '14 at 2:55
  • 2
    $\begingroup$ @DavidZ I'm guessing not the proton. But in all seriousness user125342, you should try not to plug in numbers until the end. Keep your work in terms of symbols. This way it will be easier to follow along with what you did (for example, we would probably have an easier time seeing what the meaning of $R$ is), and it should be easier for you to check your own work later. $\endgroup$ – Brian Moths Feb 22 '14 at 4:01
  • $\begingroup$ I think it is radius of the circular trajectory...Now, wasn't that obvious? $\endgroup$ – evil999man Feb 22 '14 at 4:16
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    $\begingroup$ @Awesome well what matters is what user125342 is using it to mean, and that's not obvious. The question doesn't say anything about a circular trajectory, or anything about $R$, other than that it is a radius and that user125342 has determined it's equal to 1.533 m. $\endgroup$ – David Z Feb 22 '14 at 4:34
  • $\begingroup$ Your diagram also suggests that the velocity vector at which the proton enters the magnetic field is perpendicular to the x-axis. Is this true, since otherwise the magnetic field strength should be given. $\endgroup$ – fibonatic Feb 22 '14 at 4:54

Hint : Speed of particle remains constant as magnetic field is always perpendicular to it.

Also try to find a relation between the theta , D and h. You can also include R if you want.

Try to draw the circle and its centre by using that the fact that normal at a point of a circle passes through the centre. Feel free to leave a comment below if you have any problems.

Use $\frac{mv^2}R = qvB$

In the image that D in diagram should be h. Everything else seems fine.

enter image description here


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