Final answer:
The magnitude of the magnetic field causing a force on a proton is approximately 5.8×10-9 T, and its direction is downward or negative, as per the right-hand rule based on the proton's velocity and the force experienced.
Step-by-step explanation:
The question asks about the magnitude and direction of the magnetic field that causes a force on a proton moving due east at a velocity of 5.8×106 m/s. To find the magnetic field (B), we use the formula for the magnetic force on a moving charge: F = qvBsin(θ), where F is the force, q is the charge of the proton (1.6×10-19C), v is the velocity, B is the magnetic field strength, and θ is the angle between the velocity and the magnetic field. Since the force is maximum, we infer that the angle θ is 90 degrees, making sin(θ) equal to 1.
To find the magnitude:
F = qvB
5.4×10-14N = (1.6×10-19C)(5.8×106 m/s)B
B = 5.4×10-14 N / [(1.6×10-19C)(5.8×106 m/s)]
B ≈ 5.8×10-9 T.
For the direction, the force on positively charged particles due to a magnetic field is given by the right-hand rule. Since the force is directed due south and the proton's velocity is due east, it indicates that the magnetic field is directed downward. If we represent up as positive, then down would be negative, so the direction of the magnetic field causing the force is down (less than zero).