I think its b because when there is unbalanced forces it accelerates.
Hope this helps you out
Answer:
The magnitude of the magnetic force exerted on the moving charge by the current in the wire is 2.18 x 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
= +6.5 x 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 = μI/2πr
μ = 4π x or 1.26 x N/ , thus
B = 4π x x 3 / 2π(0.05)
= 1.2 x 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 x 280 x 1.2 x sin 90°
= 2.18 x N
Using the hand law, the magnetic direction is radially inward
Hey mate!
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Hope it helps