Magnetic Force and Two Moving Charges

[Soaring Crane]

Homework Statement

A pair of point charges, q = 7.60 microC and q' = -4.50 microC, are moving in a reference frame, as shown in the figure, with speeds 9.00×10^4 m/s and v' = 6.40×10^4 m/s.

a. When the point charges are at the locations shown in the figure, what magnetic force does q' exert on q?


b. What is its direction?

Homework Equations

I came across a formula for the magnetic force F_a on b:

F_ab = q_b*v_b[(mu_0/(4*pi)]*[(q_a*v_a*u_ab)/r^2]

where u_ab = unit vector directed from q_a to q_b

Shortened F_ab = q_b*v_b*B_a??

The Attempt at a Solution

I really don’t know if I used the formula properly, so:

Magnitude B_q’ = [mu_0/(4*pi)]*[(q_a*v_b*u_ab)/r^2 = [mu_0/(4*pi)]*[(q_a*v_b*r_ab)/r^3

= (1.0*10^-7)*(32000 m^2/s)*(4.50*10^-6 C)/[0.500m]^3
= 819.2 T ??

Magnitude F_q’ on q = q*v_q*B_q’ = ((7.60*10^-6 C)*(9.00*10^4 m/s)*819.2 T) = 560.33 N ?

Will the force’s direction be in the positive(+) y-direction?

Thanks.

 

[Soaring Crane]

Can anyone please inform me if the method/direction is correct?

Any advice is appreciated.

Thanks again.

 

[m2003]

Hello,

Thank you for sharing your attempt at solving this problem. However, I believe there are some errors in your calculations. The formula you have used is correct, but you have not properly accounted for the velocity of the charges in the calculation of the magnetic field. Also, the direction of the force will not necessarily be in the positive y-direction.

To solve this problem, you can use the formula F = qvBsinθ, where θ is the angle between the velocity and the magnetic field. In this case, the velocity and magnetic field are perpendicular, so θ = 90 degrees and sinθ = 1. Plugging in the values, we get:

F = (7.60*10^-6 C)*(9.00*10^4 m/s)*(1.0*10^-7 T)*1 = 0.684 N

Therefore, the magnitude of the magnetic force exerted by q' on q is 0.684 N. To determine the direction, we can use the right-hand rule. If we point our fingers in the direction of v' and curl them towards B, our thumb will point in the direction of the force. In this case, the force will be in the positive z-direction.

I hope this helps and clarifies any confusion. Keep up the good work in your studies!