Particle moving in an electric field

[dwn]

Homework Statement

A particle with charge q = 5.0 nC and mass m = 3.0 µg moves in a region where
the magnetic field has components B_x= 2.0 mT, B_y=3.0 mT, and B_z= -4.0mT. At an instant when the speed of the particle is 5.0 km/s and the direction of its velocity is 120°relative to the magnetic field, what is the magnitude of the acceleration of the particle?

ans: 38.86 m/s²

Homework Equations

I assume it is: F_m= qv χ B

The Attempt at a Solution

5*10^-6(5*10³) χ (0.02i + 0.03j - 0.04k)sin(120)

this was my initial setup, but then I proceeded to do the following:

5*10^-6(5*10³)sin(120)(.02) = c
5*10^-6(5*10³)sin(120)(.03) = d
5*10^-6(5*10³)sin(120)(-.04) = e

√(c²+d²+e²)

and calculate the magnitude of the product --- then using this value in F=ma to determine the acceleration.

 

[voko]

5*10^-6(5*10³) χ (0.02i + 0.03j - 0.04k)sin(120)

This formula makes no sense. You have a number to the left of the cross, and a vector to the right.

A cross product requires two vectors.

 

[dwn]

That's what i thought, which left be quite perplexed as to how i need to solve this. I wasn't sure what to do given that we were presented with a vector in the question.

 

[voko]

If you know the magnitudes of two vectors, and the angle between them, can you compute the cross product?

What prevents you from doing this here?

 

[dwn]

I see where my error was, but my order of operations was not in the correct order. Thank you.