Charged Particle Motion in a Magnetic Field
Question:
A 1 g particle carrying a charge of 65 μC enters a uniform 2 T magnetic field at a speed of 32 m/s and at an angle of 22° with respect to the magnetic field lines.
For each of the following statements, determine whether it is True (T) or False (F):
- A) The particle’s speed is unchanged as it passes through the magnetic field.
- B) The work done by the field on the particle is zero as the force is normal to the displacement.
- C) The force on the particle is in the x-direction.
- D) Since the force is normal to the velocity, the path of the particle will be a circle.
- E) The x-component of the particle’s velocity is unchanged as it passes through the magnetic field.
Answer with Detailed Explanation:
A) True
The magnetic force acts perpendicular to the velocity, which means it does no work. Hence, the **speed (magnitude of velocity) remains constant** even though the direction may change.
The magnetic force acts perpendicular to the velocity, which means it does no work. Hence, the **speed (magnitude of velocity) remains constant** even though the direction may change.
B) True
The magnetic field does no work on a moving charged particle because the **Lorentz force is always perpendicular** to the displacement. Therefore, the **work done is zero**.
The magnetic field does no work on a moving charged particle because the **Lorentz force is always perpendicular** to the displacement. Therefore, the **work done is zero**.
C) False
The magnetic force vector is given by:
The magnetic force vector is given by:
F = q (v × B)
The velocity vector has x and y components, while the magnetic field is along the x-direction:
v = (32cos(22°) î + 32sin(22°) ĵ) ≈ (29.67 î + 11.99 ĵ) m/s
B = 2 î T
Using cross-product:
B = 2 î T
F = 65×10⁻⁶ C × ((29.67 î + 11.99 ĵ) × 2 î) = 65×10⁻⁶ C × (−1.56 × 10⁻³ k̂) N
So the force is actually **in the −z direction**, not x.
D) True
Since the force is always perpendicular to the velocity, the resulting path of the charged particle in a uniform magnetic field will be **circular or helical** depending on the initial angle. In this case, the circular motion occurs in the plane perpendicular to the magnetic field.
Since the force is always perpendicular to the velocity, the resulting path of the charged particle in a uniform magnetic field will be **circular or helical** depending on the initial angle. In this case, the circular motion occurs in the plane perpendicular to the magnetic field.
E) True
The magnetic field does not affect the component of velocity that is **parallel to the field lines**. Since the x-direction is parallel to the field, the **x-component of velocity remains unchanged**.
The magnetic field does not affect the component of velocity that is **parallel to the field lines**. Since the x-direction is parallel to the field, the **x-component of velocity remains unchanged**.
Final Answer:
Correct True/False Sequence: TTFTT
Key Concepts:
- Lorentz Force: F = q(v × B)
- Work by Magnetic Field: Always zero (F ⊥ displacement)
- Effect on Velocity: Only direction changes, not magnitude
- Component Behavior: Parallel component (vx) unaffected; perpendicular component causes circular motion