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An electron moves with a speed of 5.0 × 104 m/s perpendicular to a uniform magnetic field of magnitude 0.20 T. What is the magnitude of the magnetic force on the electron? (e = 1.60 × 10-19 C). Hint: F = g(Vx B) = gVB Sine, where e is the angle between the direction of the B and V

An electron moves with a speed of 5.0 × 104 m/s perpendicular to a uniform magnetic field of magnitude 0.20 T. What is the magnitude of the magnetic force on the electron? (e = 1.60 × 10-19 C). Hint: F = g(Vx B) = gVB Sine, where e is the angle between the direction of the B and V

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An electron moves with a speed of \(5.0 \times 10^4\) m/s perpendicular to a uniform magnetic field of magnitude 0.20 T. What is the magnitude of the magnetic force on the electron? (\(e = 1.60 \times 10^{-19}\) C). Hint: \[ \vec{F} = q(\vec{V} \times \vec{B}) = qVB \sin \theta, \text{ where } \theta \text{ is the angle between the direction of the } B \text{ and } V \]
Transcribed Image Text:An electron moves with a speed of \(5.0 \times 10^4\) m/s perpendicular to a uniform magnetic field of magnitude 0.20 T. What is the magnitude of the magnetic force on the electron? (\(e = 1.60 \times 10^{-19}\) C). Hint: \[ \vec{F} = q(\vec{V} \times \vec{B}) = qVB \sin \theta, \text{ where } \theta \text{ is the angle between the direction of the } B \text{ and } V \]
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