Aproton moves through a magnetic field at 26.7% of the speed of light. at a location where the field has a magnitude of 0.00649 t and the proton's velocity makes an angle of 135∘ with the field, what is the magnitude of the magnetic force acting on the proton?

A magnetic force of 58.22 fN is acting on the proton

Explanation:

The magnetic force of a particle moving through a magnetic field B is defined from the Lorentz Force Law:

F=q*v*B*sin()

q ⇒ charge ⇒ proton electric charge ⇒ 1.6021766208* C ≈ 1.6* C

v ⇒ velocity ⇒ 26.7% of the speed of light ⇒ 0.267* 299 792 458 m/s ≈ 0.267*3* m/s

B ⇒ magnetic field ⇒ 0.00649 T

⇒ angle in degrees between the velocity and the magnetic field ⇒ 135°

F = 1.6* C * 0.267*3* m/s * 0.00649 T * sin(135°)

T = N*s/C*m

C* (m/s) * T= C* (m/s) * (N*s/C*m) = N (Newton, force unit)

sin(135°) = 0.70710678118 ≈ 0.7

F = 1.6* C * 0.267*3* m/s * 0.00649 T * 0.7

F = 58.22* N

F = 58.22 fN

answer; how far out of place which the object finished relative to the starting location;

it's a b and