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? (e = 1.60 × 10^-19 C,m_electron = 9.11 × 10^-31kg) A)2.2 × 10⁷ m/s B)1.9 × 10⁷ m/s C)1.9 × 10⁶ m/s D)3.0 × 10⁷ m/s E)0.80 × 10⁷ m/s

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Quizplus · Accepted Answer

The speed of the electron can be calculated using the formula for the radius of the circular path in a magnetic field: r = mv/(qB). Solving for v gives v = r*q*B/m. Substituting the given values: r = 0.40 cm = 0.004 m, q = 1.60 × 10^-19 C, B = 3.2 × 10^-2 T, and m = 9.11 × 10^-31 kg, we find v ≈ 2.2 × 10^7 m/s.

An Electron Moving Perpendicular to a Uniform Magnetic Field of 3.2