An electron is moving at 3.0 $\times$ 10⁶ m/s at an angle of 40$^{\circ}$ to a 0.80 T magnetic field.What is the magnitude of the acceleration of the electron? A) 2.7 $\times$ 10¹⁷ m/s² B) 1.4 $\times$ 10¹⁷ m/s² C) 5.2 $\times$ 10¹⁷ m/s² D) 7.8 $\times$ 10¹⁷ m/s² E) 9.8 $\times$ 10¹⁷ m/s²

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Quizplus · Accepted Answer

The force on the electron is given by F = qvBsinθ. The acceleration is a = F/m. Using q = 1.6 * 10^-19 C, m = 9.11 * 10^-31 kg, v = 3.0 * 10^6 m/s, B = 0.80 T, and θ = 40°, the acceleration is approximately 2.7 * 10^17 m/s^2.

An Electron Is Moving at 3 ×\times× 106 M/s at an Angle of 40

An Electron Is Moving at 3 $\times$ 10⁶ M/s at an Angle of 40