An electron moving perpendicular to a uniform magnetic field of 3.2 × 10^-2 T moves in a circle of radius 0.40 cm.How fast is this electron moving? (m_el = 9.11 × 10^-31 kg.e = 1.60 × 10^-19C) A) 2.2 × 10⁷ m/s B) 1.9 × 10^-2 m/s C) 1.9 × 10^-30 m/s D) 3.0 × 10⁶ m/s

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The correct answer is A.The magnetic force on the electron is given by:$$F = qvB$$where:* q is the charge of the electron* v is the speed of the electron* B is the magnetic field strengthThe centripetal force on the electron is given by:$$F = mv^2/r$$where:* m is the mass of the electron* v is the speed of the electron* r is the radius of the circleSetting the magnetic force and the centripetal force equal to each other, we get:$$qvB = mv^2/r$$Solving for v, we get:$$v = qBr/m$$Plugging in the given values, we get:$$v = (1.60 × 10^{-19} C)(3.2 × 10^{-2} T)(0.40 × 10^{-2} m)/(9.11 × 10^{-31} kg)$$$$v = 2.2 × 10^7 m/s$$

An Electron Moving Perpendicular to a Uniform Magnetic Field of 3.2