An electron has a kinetic energy of 1.40 10-17 J. It moves on a circular path that is perpendicular to a uniform magnetic field of magnitude 7.80 10-5 T. Determine the radius of the path?
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You know that KE = (1/2)mv^2. Therefore, the velocity of the electron is 5.54E6 m/s.
You also know that F = qvB, because the electron is in circular motion, the centripetal force = (mv^2) / r. Solving for r gives you .40416 meters
You also know that F = qvB, because the electron is in circular motion, the centripetal force = (mv^2) / r. Solving for r gives you .40416 meters