Answer:
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
