An electron is moving at 30 106 ms at an angle of 40 to a 0

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?

Solution

F = q v B = 1.6 * 10^-19 * 3.0 * 10⁶ * 0.80

= 3.84 * 10^-13 N

F = m a

a = F / m = 3.84 * 10^-13 / 9.11 * 10^-31

= 4.22 * 10¹⁷ m/s²