A proton of kinetic energy 1.0 x 10^-7 eV moves in a circular orbit in the magnetic field near the Earth. The strength of the field is 5.0 x 10^-5 T. What is the radius of the orbit?
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Step 1
Given: Kinetic energy = 1.0 x 10^-7 eV 1 eV = 1.6 x 10^-19 J Kinetic energy in joules = (1.0 x 10^-7) * (1.6 x 10^-19) = 1.6 x 10^-26 J Show more…
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