Question

A proton moving at a speed of 3.8 ´ 106 m/s cuts across the lines of a magnetic field at an angle of 70°. the strength of the field is 0.25 ´ 10-4 t. what is the magnitude of the force acting on the proton? (qp = 1.6 ´ 10-19 c)

Answer 1

To calculate the magnetic force on a proton, use the formula F = qvBsin(θ). Using the provided charge, velocity, magnetic field strength, and angle, the force on the proton can be computed accordingly.

Step-by-step explanation:

The question pertains to the magnetic force experienced by a proton moving through a magnetic field. To determine the magnitude of the force acting on a charged particle moving in a magnetic field, you can use the equation F = qvBsin(θ), where F is the force, q is the charge of the particle, v is the velocity of the particle, B is the magnetic field strength, and θ is the angle between the velocity and the magnetic field direction.

Given the proton's charge (q = 1.6 × 10^-19 C), speed (v = 3.8 × 10^6 m/s), magnetic field strength (B = 0.25 × 10^-4 T), and angle (θ = 70°), the magnitude of the force F can be calculated as follows:

F = (1.6 × 10^-19 C) × (3.8 × 10^6 m/s) × (0.25 × 10^-4 T) × sin(70°)

Perform the calculation to find F, the force in newtons acting on the proton.