- #1
thunderjolt
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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).