Answer:
The magnitude of the magnetic force exerted on the moving charge by the current in the wire is 2.18 x [tex]10^{-8}[/tex] N
The direction of the magnetic force exerted on the moving charge by the current in the wire is radially inward
Explanation:
given information:
current, I = 3 A
[tex]q_{0}[/tex] = +6.5 x [tex]10^{-6}[/tex] C
r = 0.05 m
v = 280 m/s
and direction of the magnetic force exerted on the moving charge by the current in the wire, we can use the following formula:
F = qvB sin θ
where
F = magnetic force (N)
q = electric charge (C)
v = velocity (m/s)
θ = the angle between the velocity and magnetic field
to find B we use
B = μ[tex]_{0}[/tex]I/2πr
μ[tex]_{0}[/tex] = 4π x [tex]10^{-7}[/tex] or 1.26 x [tex]10^{-6}[/tex] N/[tex]A^{2}[/tex] , thus
B = 4π x [tex]10^{-7}[/tex] x 3 / 2π(0.05)
= 1.2 x [tex]10^{-5}[/tex] T
Now, we can calculate the magnitude force
F = qvB sin θ
θ = 90°, because the speed and magnetic are perpendicular
F = 6.5 x [tex]10^{-6}[/tex] x 280 x 1.2 x [tex]10^{-5}[/tex] sin 90°
= 2.18 x [tex]10^{-8}[/tex] N
Using the hand law, the magnetic direction is radially inward
