Question

A group of particles is traveling in a magnetic field of unknown magnitude and direction. You observe that a proton moving at 1.60km/s in the +x-direction experiences a force of 2.06

Answer 1

F = qv cross B, and |F| = |qvB| sin theta,
where theta is either positive angle between v and B.
In the case of the proton,
(1.602 x 10^(-19) C)(1600 m/s) B sin theta-p = 2.10 x 10^(-16) N
Here "theta-p" is measured from the x axis, whichever part of it is nearer.
In the case of the electron,
(1.602 x 10^(-19) C) (4300 m/s) B sin theta-e = 8.50 x 10^(-16) N
Here "theta-e" is measured from the z axis, whichever part of it is nearer.

I'm going to finish part (B) before part (A):
The magnetic force in the y direction guarantees that the direction of B is in the x-z plane.
The proton's moving towards +x and being forced towards +y shows that the z component of B is negative, since "i cross k" is -j.
The electron's moving towards -z and being forced towards +y shows that the x component of B is positive, since "k cross i" is +j.

We also know that theta-p + theta-e = 90 degrees, so
sin theta-p = cos theta-e.
If we now simply DIVIDE the electron equation by the proton equation, we get
(43/16) tan (theta-e) = 4.05
tan (theta-e) = 1.507
theta-e = 56.433 degrees
That answers part B, but you can round off the answer to 3 digits
EXCEPT when using it to answer part A...

For part (A), plug theta-e into the electron equation above.

For part (D), note that the direction of the force will be exactly perpendicular to B,
so it will be 56.433 from the x-axis.
Bearing in mind that B = + some i - some k,
and qv = + some j,
F must be + some i + some k;
it is directed 56.433 "above" the +x axis, so it's 123.6 degrees from the -ve x direction.
Why they didn't ask for the angle "from the +ve x direction" is, I don't know.

For part (C),
|F| = |qvB| since "sin theta" is 1.
|F| = (1.602 x 10^(-19) C) (3400 m/s) times the answer from part A.