a proton moving with a speed of 1.0 x 106ms-1 through a magnetic field of 1.0T experiences a magnetic force of magnitude 8.0 x 10-14 N. The angle between th...

Answer Details

When a charged particle moves through a magnetic field, it experiences a magnetic force perpendicular to both the velocity of the particle and the direction of the magnetic field. The magnitude of the magnetic force is given by the equation F = qvB sin(theta), where F is the force, q is the charge of the particle, v is the velocity of the particle, B is the magnetic field, and theta is the angle between the velocity of the particle and the magnetic field. In this problem, a proton with a speed of 1.0 x 10^6 m/s is moving through a magnetic field of 1.0 T and experiences a magnetic force of magnitude 8.0 x 10^-14 N. We can use the equation F = qvB sin(theta) to solve for the angle between the proton's velocity and the field. Rearranging the equation, we get sin(theta) = F / (qvB). Substituting the given values, we get sin(theta) = (8.0 x 10^-14 N) / [(1.6 x 10^-19 C) x (1.0 x 10^6 m/s) x (1.0 T)] = 0.005. Taking the inverse sine of both sides, we get theta = sin^-1(0.005) = 0.29°. Therefore, the angle between the proton's velocity and the magnetic field is approximately 0.29° or very close to 0°, which is almost parallel to the field. Thus, the correct answer is option (C) 30°.