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A proton moves with velocity 10⁶ m/s perpendicular to a uniform B = 0.5 T. Radius of its circular path is approximately (m_p = 1.67×10⁻²⁷ kg, e = 1.6×10⁻¹⁹ C)
AAbout 5 m
BAbout 2.1 cm
CAbout 21 cm
DAbout 0.21 mm
Answer & Solution
Correct answer: B. About 2.1 cm
r = mv/(qB) = (1.67e-27 · 1e6) / (1.6e-19 · 0.5) ≈ 2.09e-2 m ≈ 2.1 cm.
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