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 + |

Answer #1

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.

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answer. Question A group of particles is traveling in a magnetic
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