Final answer:
The radius of the proton's circular path is calculated using the relationship between the magnetic force, acting as the centripetal force, and the velocity of the proton in the uniform magnetic field. By equating the magnetic force to the centripetal force, we can solve for the radius of the path.
Step-by-step explanation:
The student's question involves calculating the radius of the circular path that a proton follows when it moves in the presence of a magnetic field. The force that causes this circular motion is the magnetic force, which can be determined using the formula Fm = qvBsin(θ), where Fm is the magnetic force, q is the charge of the proton (1.602 x 10-19 C), v is the velocity of the proton, B is the magnetic field strength, and θ is the angle between the velocity and the magnetic field.
Because the proton's velocity is perpendicular to the magnetic field, θ = 90° and sin(θ) = 1. The magnetic force in this situation is what provides the centripetal force necessary to keep the proton in a circular path. The centripetal force Fc is given by the equation Fc = mv2/r, where m is the mass of the proton (1.673 x 10-27 kg) and r is the radius of the circular path.
Equating the magnetic force to the centripetal force, we have qvB = mv2/r. Solving for r, the radius of the circular path, yields r = mv/qB. Substituting in the given values (v = 8.8 x 104 m/s, B = 1.270 x 10-3 T), we find r = (1.673 x 10-27 kg) * (8.8 x 104 m/s) / (1.602 x 10-19 C * 1.270 x 10-3 T), which gives the radius of the proton's circular path.