A proton ( , ) enters a uniform magnetic field of magnitude with a speed of

PHYSICS · Single correct · CBSE · Physics · Moving Charges and Magnetism · Moving Charges and Magnetism

Problem

A proton (m=1.67×10−27 kgm = 1.67 \times 10^{-27}\,\text{kg}, q=1.6×10−19 Cq = 1.6 \times 10^{-19}\,\text{C}) enters a uniform magnetic field of magnitude 0.20 T0.20\,\text{T} with a speed of 4.0×106 m s−14.0 \times 10^{6}\,\text{m\,s}^{-1}, the velocity being perpendicular to the field. Determine the radius of the circular path followed by the proton. (a) 2.1 m2.1\,\text{m} (b) 0.21 m0.21\,\text{m} (c) 0.104 m0.104\,\text{m} (d) 0.021 m0.021\,\text{m}

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