A proton of kinetic energy 1.0x10^7 eV moves in a circular orbit in the magnetic field near the Earth. The strength of the field is 5.0x10^-5 T. What is the radius of the orbit?
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Step 1
First, we need to convert the kinetic energy from electron volts (eV) to joules (J) because the standard unit of energy in physics is joules. We know that 1 eV = 1.6x10^-19 J. So, the kinetic energy of the proton in joules is 1.0x10^7 eV * 1.6x10^-19 J/eV = Show more…
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