A charge q = 2 μC moves with velocity v = 3×10^5 m/s perpendicular to a uniform magnetic field B = 0.5 T. What is the magnetic force on the charge?

A charge q = 2 μC moves with velocity v = 3×10^5 m/s perpendicular to a uniform magnetic field B = 0.5 T. What is the magnetic force on the charge?

• A. 0.6 N
• B. 0.15 N
• C. 0.9 N
• D. 0.3 N

Answer: (D)

Magnetic force on a moving charge depends on the charge, speed, and magnetic field through F = q v B sin(theta). Since the velocity is perpendicular to the field, theta = 90°, so sin(theta) = 1 and F = q v B. Convert the charge: q = 2 μC = 2×10^-6 C. Then q v = (2×10^-6 C)(3×10^5 m/s) = 0.6 C·m/s. Multiply by B: 0.6 × 0.5 T = 0.3 N. So the magnetic force is 0.3 newtons, directed perpendicular to both v and B (for a positive charge, by the right-hand rule).

Magnetic force on a moving charge depends on the charge, speed, and magnetic field through F = q v B sin(theta). Since the velocity is perpendicular to the field, theta = 90°, so sin(theta) = 1 and F = q v B. Convert the charge: q = 2 μC = 2×10^-6 C. Then q v = (2×10^-6 C)(3×10^5 m/s) = 0.6 C·m/s. Multiply by B: 0.6 × 0.5 T = 0.3 N. So the magnetic force is 0.3 newtons, directed perpendicular to both v and B (for a positive charge, by the right-hand rule).