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

Question

Quizplus · Accepted Answer

The magnitude of the acceleration of an electron moving in a magnetic field is given by the formula $a = \frac{q|vB\sin(\theta)|}{m}$, where $q$ is the charge of the electron ($1.6 \times 10^{-19} C$), $v$ is the velocity of the electron ($3.0 \times 10^{6} m/s$), $B$ is the magnetic field strength ($0.80 T$), $\theta$ is the angle between the velocity and the magnetic field ($40^\circ$), and $m$ is the mass of the electron ($9.11 \times 10^{-31} kg$). Plugging in the values and solving for $a$ gives a value in the order of $10^{17} m/s^2$, which matches the magnitude of choice A.

An Electron Is Moving at 3