A charged particle q = 1.6×10^-19 C moves with velocity v = 2.0×10^6 m/s perpendicular to a uniform magnetic field B = 0.50 T. What is the magnetic force on the particle?

A charged particle q = 1.6×10^-19 C moves with velocity v = 2.0×10^6 m/s perpendicular to a uniform magnetic field B = 0.50 T. What is the magnetic force on the particle?

• A. 1.6×10^-14 N
• B. 1.6×10^-13 N
• C. 1.6×10^-12 N
• D. 3.2×10^-13 N

Answer: (B)

The magnetic force on a moving charge is F = q v B sinθ. Since the velocity is perpendicular to the magnetic field, sinθ = 1, so F = q v B. Plugging in q = 1.6×10^-19 C, v = 2.0×10^6 m/s, and B = 0.50 T gives F = (1.6×10^-19)(2.0×10^6)(0.50) = 1.6×10^-13 N. For a positive charge, the force is perpendicular to both v and B, following the right-hand rule. The other options don’t match this calculation (they differ by factors of 10 or 2), so this is the correct magnitude.

The magnetic force on a moving charge is F = q v B sinθ. Since the velocity is perpendicular to the magnetic field, sinθ = 1, so F = q v B. Plugging in q = 1.6×10^-19 C, v = 2.0×10^6 m/s, and B = 0.50 T gives F = (1.6×10^-19)(2.0×10^6)(0.50) = 1.6×10^-13 N. For a positive charge, the force is perpendicular to both v and B, following the right-hand rule. The other options don’t match this calculation (they differ by factors of 10 or 2), so this is the correct magnitude.