A proton (q = 1.6 X 10 -19 C, m = 1.67 X 10 -27 kg) moving with constant velocit

Question

A proton (q = 1.6 X 10-19 C, m = 1.67 X 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 = 0.6 m in the x-direction. The proton leaves the field having a velocity vector (vx, vy) = (3.6 X 105 m/s, 2.2 X 105 m/s).

1)What is v, the magnitude of the velocity of the proton as it entered the region containing the magnetic field?

2)What is R, the radius of curvature of the motion of the proton while it is in the region containing the magnetic field?

3)What is h, the y co-ordinate of the proton as it leaves the region conating the magnetic field?

4)What is Bz, the z-component of the magnetic field? Note that Bz is a signed number.

Explanation / Answer

This is much less of a physics problem as it is a geometry problem. The magnetic field indicates that the proton is moving in a circle (at least while it is in the field). The question is how to get the angles from the geometry.

Draw a complete circle.
Draw the entry and exit vectors of the proton (tangent to the circle).
Make an isosceles triangle connecting the center of the circle (point A), the entry and exit points of the proton (points B and C).
Now draw a smaller right triangle with BC as the hypotenuse. The new point is D.

Now it is just a geometry problem.
1) angleABC = angleACB
2) angleCBD = 90 - angleABC
3) angleCBD + angleBCD = 90

Plug 2 into 3 and get:
4) angleBCD = angleABC

You can solve for angleBCD using the distance and the height given in the problem (x and D) and some trig.

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