Question

An electron in a TV camera tube is moving at 7.20×106 m/s in a magnetic field of strength 65 mT. Without knowing the direction of the field, what can you say about the greatest and least magnitude of the force acting on the electron due to the field?

Maximum force? Submit Answer Tries 0/99

Minimum force? Submit Answer Tries 0/99

At one point the acceleration of the electron is 7.509×1016 m/s2. What is the angle between the electron velocity and the magnetic field? (deg)

Answer 1

q = charge on electron = 1.6 x 10^-19 C

V = speed = 7.20 x 10⁶ m/s

B = magnetic field = 0.065 T

magnetic force is given as

F = q VB Sin$\theta$

maximum force when $\theta$ = 90

F_max = q VB = (1.6 x 10^-19 ) (7.20 x 10⁶ ) (0.065) = 7.5 x 10^-14 N

minimum force when $\theta$ = 0

F_min = q VB Sin0 = 0 N