A proton enters a uniform 0.32 T magnetic field perpendicular to the field at 3.20 × 10^6 m s⁻¹. Find the circular-path radius and orbital period.

0.104 m; 2.05 × 10^-7 s Magnetic force supplies centripetal force: qvB = mv²/r. Thus r = mv/(qB) = 0.104 m. T = 2πr/v = 2πm/(qB) = 2.05 × 10^-7 s, independent of speed in the nonrelativistic model.

AP Physics · APCEM-5 · structured

Proton in a uniform magnetic field 6

AP Physics C: Electricity and Magnetism · Magnetic fields · Original GioPhysics question with a detailed, mark-by-mark answer guide.

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Marks: 5
Topics: 2
Answer: Complete

Question5 marks

A proton enters a uniform 0.32 T magnetic field perpendicular to the field at 3.20 × 10⁶ m s⁻¹. Find the circular-path radius and orbital period.

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Demand: routine

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