An electron moving perpendicular to a magnetic field of 3.2 × 10^-2 T moves in a circle of radius 0.40 cm. How fast is this electron moving?

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Quizplus · Accepted Answer

The correct answer is A.Explanation:The centripetal force on the electron is provided by the magnetic force:$$F_c = F_m$$$$ma_c = qvB$$$$a_c = \frac{qvB}{m}$$$$v = \frac{a_cr}{qB}$$$$v = \frac{(qvB/m)r}{qB}$$$$v = \frac{vr}{m}$$$$v = \frac{(0.40 	imes 10^{-2} 	ext{ m})(3.2 	imes 10^{-2} 	ext{ T})}{9.11 	imes 10^{-31} 	ext{ kg}}$$$$v = 2.3 	imes 10^7 	ext{ m/s}$$

An Electron Moving Perpendicular to a Magnetic Field of 3