An alpha particle (a He nucleus, containing two protons and two neutrons and having a mass of 6.64×10−27 kg ) traveling horizontally at 36.6 km/s enters a uniform, vertical, 1.10 T magnetic field.
A. What is the diameter of the path followed by this alpha particle? Express your answer in millimeters to three significant figures.
B. What effect does the magnetic field have on the speed of the particle?
C. What is the magnitude of the acceleration of the alpha particle while it is in the magnetic field? Express your answer in meters per second to three significant figures.
D. What is the direction of the acceleration of the alpha particle while it is in the magnetic field?E. Explain why the speed of the particle does not change even though an unbalanced external force acts on it.

A) d = 1.38 10-3 m, B)   the velocity modulus remains constant and the velocity direction changes since the force is perpendicular

c)  a  = 1.94 10¹² m / s

d) the force is perpendicular to these two, that is, it points out of the page (positive z-axis)

e) E) the modulus of velocity remains constant because the force is perpendicular to the movement,

Explanation:

A)  Let's use Newton's second law for this problem where the force is the magnetic force given by

            F = m a

The magnetic force is

          F = q v x B = q v B sin θ

Where the angle between the speed and the magnetite field is ce 90º so the sin 90 = 1

This implies that the acceleration is centripetal, which is given by

           a = v² / R

We substitute

          q v B = m v² / r

          r = m v / qB

 Let's calculate

         r = 6.64 10⁻²⁷ 36.6 10³ / (2 1.6 10⁻¹⁹  1.10)

         r = 6,904 10⁻⁴ m

Diameter is twice the radius

          d = 2 r

          d= 1.38 10-3 m

B) The effect of the magnetic field is that the particle describes a circular motion, where the velocity modulus remains constant and the velocity direction changes since the force is perpendicular

C) the centripetal acceleration is

           F = m a

           q v B = m a

           a = q v B / m

Let's calculate

           a = 2  1.6 10⁻¹⁹ 36.6 10³ 1.10 / 6.64 10⁻²⁷

           a  = 1.94 10¹² m / s

D) Acceleration has the same direction of force toward the center of the circular orbit,

If we assume that the horizontal direction of the alpha particle is on the x-axis, the field is on the positive y-axis, so the force is perpendicular to these two, that is, it points out of the page (positive z-axis)

E) the modulus of velocity remains constant because the force is perpendicular to the movement, but the direction of the velocity does change in the direction of the force

i believe the answer is a hope it and is correct