A proton travels at 1.0 × 10⁶ m/s perpendicular to a uniform magnetic field of 0.10 T. Take mₚ = 1.67 × 10⁻²⁷ kg, e = 1.6 × 10⁻¹⁹ C. What is the radius of its circular path?
Correct answer B. 0.10 m
Explanation
r = mv/(qB) = (1.67 × 10⁻²⁷ × 1.0 × 10⁶) ÷ (1.6 × 10⁻¹⁹ × 0.10) = 1.67 × 10⁻²¹ ÷ 1.6 × 10⁻²⁰ ≈ 0.10 m. Inverting the ratio gives 9.6.
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