Answer:
A
Explanation:
space
11.43 cm
yes
rolling friction
Magnitude of the force on proton = F = 1.1085 × 10^-15 N
Charge on proton = q = 1.60 × 10^-19 C
Velocity of proton = V = 4.0 × 10^4 m/s
Magnetic field = B = 0.20 T
Angle between V and B = θ = 60
We know that,
F = qVBsin θ = (1.60 × 10^-19)( 4.0 × 10^4)( 0.20)sin(60)
F = 1.1085 × 10^-15 N
a is the answer
