A proton is released such that it has an initial speed of 4.1 × 10⁵ m/s from left to right across the page.A magnetic field of 0.8 T is present at an angle of 45° to the horizontal direction (or positive x axis).What is the magnitude of the force experienced by the proton? (q_p = 1.6 × 10−¹⁹ C) A)5.9E-33 N B)9.3E-33 N C)3.7E-14 N D)4.5E-14 N E)3.7E-14 N

Question

Quizplus · Accepted Answer

The magnitude of the force experienced by a charged particle moving through a magnetic field is given by:

F = qvBsinθ

Where q is the charge of the particle (in this case, a proton with a charge of +1.6 × 10−19 C), v is the velocity of the particle (4.1 × 10^5 m/s to the right), B is the magnetic field strength (0.8 T at an angle of 45° with the positive x axis), and θ is the angle between the particle's velocity and the magnetic field (45° in this case).

Plugging in the values, we get:

F = (1.6 × 10−19 C)(4.1 × 10^5 m/s)(0.8 T)sin(45°)
F = 3.7 × 10−14 N

Therefore, the correct answer is C.

A Proton Is Released Such That It Has an Initial