A particle with charge q = 2 μC moves through a

Solution

Given:

• Charge: q = 2 μC = 2 × 10⁻⁶ C
• Magnetic field: B = 0.5 T
• Velocity: v = 4 m/s
• Angle between v and B: θ = 60°

Formula for magnetic force: The magnitude of the magnetic force on a moving charged particle is: F = qvB sin θ Calculation: F = qvB sin θ F = (2 × 10⁻⁶)(4)(0.5) sin(60°) F = (2 × 10⁻⁶)(4)(0.5)(√3/2) F = (2 × 10⁻⁶)(4)(0.5)(0.866) F = (2 × 10⁻⁶)(1.732) F = 3.464 × 10⁻⁶ N Converting to μN: Since 1 μN = 10⁻⁶ N: F = 3.464 × 10⁻⁶ N = 3.464 μN