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## Homework Statement

An electron that has a velocity with x component 2.3 x 106 m/s and y component 2.9 x 106 m/s moves through a uniform magnetic field with x component 0.040 T and y component -0.12 T. (a) Find the magnitude of the magnetic force on the electron. (b) Repeat your calculation for a proton having the same velocity.

## Homework Equations

F = q vXB

## The Attempt at a Solution

Since both magnetic field and velocit were given in component form I figured they could be combined into one (meaning one for velocity and one for magnetic field). After doing

a = sqrt[ (ax)^2 + (ay)^2] for both V and B I got V =3.7 x10 ^6 m/s, angle = 5.158 degrees; B = 1.2649 x 10^-1, angle = -71.565.

Then I added the two angles (I figured their sum is the total difference b/w the B and V vectors?) and plugged in:

F = q v X B

F= (1.6 x 10^-19)(3.7 x 10^6)(1.2649 x 10^-1)sin (-66.407)

F = -6.862 x 10^-14 N

For part b) it should be the same # as the charge of the proton and electron is the same (only sign is different and we use absolute values in that equation).

I hope that my approach is ok, though, assuming my approach is ok I think I might have messed up the combining of the two angles. Any help/guidance would be greatly appreciated. Thank you in advance.

P.S. I thought I should mention that no diagram is provided. Thanks again.