A proton high above the equator approaches the
Earth moving straight downward with a speed of
365 m/s.
Find the acceleration of the proton, given that the magnetic field at its altitude is 4.10*10^-5 T.

Respuesta :

We can see from the calculations that the acceleration of the proton is 1.43 * 10^7 ms-2.

What is acceleration?

Acceleration is the rate of change of velocity with time. We know that the magnetic force is obtained by;

F = qvB

q= 1.6 * 10^-19 C

v = 365 m/s

B =  4.10*10^-5 T

F = 1.6 * 10^-19 C * 365 m/s *  4.10*10^-5 T

F = 2.39 * 10^-20 N

Recall that the mass of the proton is 1.67 × 10−27 kg hence

a = F/m = 2.39 * 10^-20 N/1.67 × 10−27 kg  = 1.43 * 10^7 ms-2

Learn more about acceleration: https://brainly.com/question/12134554