A proton moves with velocity v = 2.0 x 10^6 m/s perpendicular to a uniform magnetic field B = 0.50 T. What is the radius of its circular orbit? (Mass of proton = 1.67 x 10^-27 kg, charge = 1.6 x 10^-19 C)
Answer and explanation
Correct answer: B. 4.18 cm
- Step 1: Cyclotron orbit radius: r = m v / (q B). Step 2. r = (1.67 x 10^-27 2.0 x 10^6) / (1.6 x 10^-19 0.50) = (3.34 x 10^-21) / (0.80 x 10^-19) = 4.175 x 10^-2 m = 4.18 cm. Answer: 4.18 cm. Why not the others: 2.09 cm uses B = 1.0 T; 8.35 cm uses v = 4.0 x 10^6 m/s.
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