A proton is projected with a velocity of 7.0 × 10³ m/s into a magnetic field of 0.60 T perpendicular to the motion of the proton. What is the force that acts on the proton? A) 0 N B) 3.4 × 10^-16 N C) 4.2 × 10^-16 N D) 13 × 10^-16 N E) 6.7 × 10^-16 N

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Quizplus · Accepted Answer

The force on a charged particle moving in a magnetic field is given by F = qvB sin(θ). Here, q = 1.6 × 10^-19 C (charge of a proton), v = 7.0 × 10^3 m/s, B = 0.60 T, and θ = 90° (since the field is perpendicular to the motion). Thus, F = (1.6 × 10^-19 C)(7.0 × 10^3 m/s)(0.60 T) = 6.72 × 10^-16 N, which is approximately 6.7 × 10^-16 N.

A Proton Is Projected with a Velocity of 7