- #1

thunderjolt

- 9

- 0

## Homework Statement

At a particular instant, charge q_1 = 4.80×10^−6 C is at the point (0, 0.250m , 0) and has velocity v_1=(9.2*10^5 m/s) i_hat. Charge q_2 = −2.50×10^−6 C is at the point (0.150m , 0, 0) and has velocity v_2 = (-5.3x10^5 m/s) j_hat.

At this instant, what is the magnetic force that q_1 exerts on q_2?

## Homework Equations

The equations that I used are:

the X in both equations represents the cross product of the components

magnetic field of a moving point charge:

B=(μ_0/4pi)*(qvXr_hat)/r^2

Force of a magnetic field:

F_B=qvXB

## The Attempt at a Solution

Using the right hand rule, I found that the magnetic field coming from q_2 would be in the +k_hat direction; and I found that the force from q_1 with that magnetic field would be in the -j_hat direction.

Furthermore, using the first equation, I found the B-field from q_2 to be 8.02*10^-17 (+z direction) using B_2=10^-7*q_2*v_2*r_x/r^3 (where r_x is the x-component of r, the distance between the 2 charges). With that B-field, I used the second equation to get a force of 3.54*10^-16 N (-y direction) using F_B=q_1*v_1XB. So, my final answer is 3.54*10^-10 μN (-j_hat).