Question

An electron moves in uniform magnetic field with a speed of 5·106 m/s along the x axis. The magnetic field strength is 0.050 T, directed at an angle of 45° to the x axis and lying in the xy plane. Calculate the magnetic force 𝐹 on and acceleration 𝑎 of the electron.

Answer 1

(a) We can find the magnetic force on the electron as follows:

$$F=qvBsin\theta,$$ $$F=1.6\cdot10^{-19}\ C\cdot5\cdot10^6\ \dfrac{m}{s}\cdot0.05\ T\cdot sin45^{\circ}=2.83\cdot10^{-14}\ N.$$

(b) We can find the acceleration of the electron from Newton's Second Law of Motion:

$$F=ma,$$ $$a=\dfrac{F}{m}=\dfrac{2.83\cdot10^{-14}\ N}{9.11\cdot10^{-31}\ kg}=3.1\cdot10^{16}\ \dfrac{m}{s^2}.$$