Question

A proton enters a region of a uniform magnetic field with velocity v = 5.0×10⁶ m/s in +Y direction. The magnitude of the field is 3.0 T and is in −Z direction. What is the magnetic force (magnitude and direction) exerted on the proton from the field?

A. - (2.4e⁻¹² N) x with hat on top
B. - (2.4e⁻¹² N) y with hat on top
C. (2.4e⁻¹² N) y with hat on top
D. (2.4e⁻¹² N) x with hat on top
E. There is no force acting on the proton

Answer 1

The magnetic force exerted on a proton moving in the +Y direction through a 3.0 T magnetic field in the -Z direction is 2.4 x 10^-12 N in the +X direction, determined by the right-hand rule and calculated using F = qvB.

Step-by-step explanation:

The question revolves around the concept of a magnetic force acting on a charged particle moving through a magnetic field. According to the right-hand rule, the direction of the magnetic force on a positively charged particle, such as a proton, is perpendicular to both the velocity of the particle and the magnetic field. Using the formula F = q(v x B), where F is the magnetic force, q is the charge of the proton, v is the velocity, and B is the magnetic field, we can find the magnitude and direction of the force on the proton.

In this specific case, since the proton's velocity is in the +Y direction, and the magnetic field is in the -Z direction, the force will be in the +X direction. The magnitude of the force is given by F = qvB sin(θ), where θ is the angle between the velocity and the magnetic field vectors. Since the velocity is perpendicular to the magnetic field, sin(θ) equals 1. Hence, F = qvB.

The charge of a proton is about 1.6 x 10-19 C. Using this value, the magnitude of the force on the proton can be calculated as F = (1.6 x 10-19 C)(5.0 x 106 m/s)(3.0 T) = 2.4 x 10-12 N in the +X direction, which corresponds with answer D provided in the question. Therefore, the direction of this force is along the +X axis, as indicated by the right-hand rule.