A proton of kinetic energy 1.10 107 eV moves in a circular orbit in the magnetic field near the Earth. The strength of the field is 4.40 10-5 T. What is the radius of the orbit?
Added by Lisa J.
Step 1
First, we need to convert the kinetic energy from electron volts (eV) to joules (J). We know that 1 eV = 1.6 x 10^{-19} J. So, the kinetic energy in joules is: K = 1.10 x 10^7 eV * 1.6 x 10^{-19} J/eV = 1.76 x 10^{-12} J Show more…
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