Question

A proton travels with a speed of 5.02×10⁶ m/s in a direction that makes an angle of 60.0° with the direction of a magnetic ficld of magnitude 0.180T in the positive x direction. What are the magnitudes of (b) the proton's acceleration?

Answer 1

Answer: The magnitude of the proton's acceleration is 0.748 ×10^14 m/s²

Explanation:
the velocity ,v, of ththe proton = 5.02×10^6 m/s
Magnitude , B , of the magnetic field = 0.180 T

First , we need to find the magnitude of the Force on the proton. This is given by the relation :
F = q(v x B) = qvBsinθ

where 'q' is the charge of proton , q= 1.6×10^-19 C
θ is the angle the proton makes with the direction of the magnetic field

Putting the respective values of v, B ,θ in the above equation, we get:

F = (1.6×10^-19 C)(5.02×10^6 m/s)(0.180T) sin60°
∴ F = 1.25 ×10^-13 N

Now , from Newton's second law we know that ,
F=m×a

∴ a = F/m

Mass of a proton = 1.67×10^27 kg
a= 1.25 × 10^-13 N / 1.67 × 10^27 kg

a= 0.748 × 10^14 m/s² =acceleration of the proton

(To know more about Magnetic Fields : brainly.com/question/9095546)