Question

an electron is accelerated from rest by a potential difference of 320 v. it then enters a uniform magnetic field of magnitude 230 mt with its velocity perpendicular to the field. (a) calculate the speed of the electron.

Answer 1

Answer: 1.42 sec

Step-by-step explanation:

The object is dropped from the height. So initial velocity, u = 0 m/s.

Height = 10m and Acceleration = g = 9.8 m/s² (or g can be approximated to 10 m/s²)

Using the Second equation of motion;

s = ut + (1/2) at²

~ 10 = 0 + (1/2) gt²

~ t² = 20/9.8 = 100/49

~ t = 10/7 seconds = 1.42 seconds