# Homework Help: Magnetic field of a moving charge

1. Mar 2, 2016

### squeak

1. The problem statement, all variables and given/known data
A negative charge q = −3.20×10-6C is located at the origin and has velocity υ⃗ =(7.50×104m/s)ι^+((−4.90)×104m/s)j^.
At this instant what is the magnetic field produced by this charge at the point x = 0.230 m , y = -0.300 m , z= 0? Give the x, y and x components

2. Relevant equations
B = u0/4π. (qv x r)/r2

3. The attempt at a solution
So to find Bx i've tried using r = 0.230 and v = 7.5x104 giving 10-7 x 3.2 x 10-6 x 7.5 x 104/0.2302. However i have a feeling I can't just take the x/y components individually. I know that a cross product goes to 1 if the components are perpendicular and 0 if they are parallel and feel like this will help but i just can't put it all together to give me the correct answer. any hints would be very much appreciated!

2. Mar 3, 2016

### Simon Bridge

You are correct - you cannot just take the components individually like that. You have a vector equation, why not use it?
Do you know how to evaluate a cross product between two arbitrary vectors?