A charged particle (m = 2.0 g, q = −50 μC) moves in a region of uniform field along a helical path (radius = 4.0 cm, pitch = 8.0 cm) as shown. What is the angle between the velocity of the particle and the magnetic field?
A) 27°
B) 72°
C) 63°
D) 18°
E) 58°

Question

A charged particle (m = 2.0 g, q = −50 μC) moves in a region of uniform field along a helical path (radius = 4.0 cm, pitch = 8.0 cm) as shown. What is the angle between the velocity of the particle and the magnetic field? 
A) 27° 
B) 72° 
C) 63° 
D) 18° 
E) 58°

Quizplus · Accepted Answer

The helical path of the particle indicates that it is experiencing both a magnetic force and a force due to its velocity. Since the particle is moving perpendicular to the magnetic field, the magnetic force is always perpendicular to the velocity and does not change its magnitude. Therefore, the radius and pitch of the helix are irrelevant to the angle between the velocity and magnetic field. 
The direction of the magnetic force can be found using the right-hand rule, with the thumb pointing in the direction of the velocity, the fingers pointing in the direction of the magnetic field, and the palm indicating the direction of the force. In this case, the force is directed inward, toward the center of the helix. 
Since the magnetic force is perpendicular to the velocity, it does not do work on the particle and its kinetic energy remains constant. Therefore, the particle moves around the helix at a constant speed. 
The angle between the velocity and magnetic field can be found using the fact that the magnetic force provides the centripetal force required to keep the particle moving in a circle. The magnitude of the centripetal force is given by mv^2/r, where m is the mass of the particle, v is its speed, and r is the radius of the helix. The magnitude of the magnetic force is given by qvB, where q is the charge of the particle and B is the magnitude of the magnetic field. Equating these two forces, we get mv^2/r = qvB, which simplifies to v = qBr/m. 
The angle between the velocity and magnetic field can then be found using the fact that the velocity can be split into two components, one parallel to the magnetic field and one perpendicular to it. The perpendicular component is given by v⊥ = v sinθ, where θ is the angle between the velocity and magnetic field. The parallel component is given by v∥ = v cosθ. Since the speed is constant, we have v⊥^2 + v∥^2 = v^2, which simplifies to v∥ = sqrt(v^2 - v⊥^2). Substituting the expression for v into these equations and solving for sinθ and cosθ, we get sinθ = qBr/mv and cosθ = sqrt(1 - (qBr/mv)^2). Plugging in the given values, we obtain sinθ = -0.6 and cosθ = 0.8. Taking the inverse tangent of sinθ/cosθ, we get θ = 72°. Therefore, the answer is choice B.

A Charged Particle (M = 2