In the figure, an electron accelerated from rest through potential difference V1
ID: 1442669 • Letter: I
Question
In the figure, an electron accelerated from rest through potential difference V1=0.825 kV enters the gap between two parallel plates having separation d = 20.9 mm and potential difference V2= 164 V. The lower plate is at the lower potential. Neglect fringing and assume that the electron's velocity vector is perpendicular to the electric field vector between the plates. In unit-vector notation, what uniform magnetic field allows the electron to travel in a straight line in the gap?
Need answer in mT
Explanation / Answer
q = charge on electron = 1.6 x 10-19 C
V1 = potential difference = 825 V
v = speed gained
m = mass of electron = 9.1 x 10-31 kg
using the conservation of energy
Kinetic energy gained = electric potential energy
(0.5) m v2 = qV1
(0.5) (9.1 x 10-31) v2 = (1.6 x 10-19) (825)
v = 1.7 x 107 i^ m/s
E = electric field between the plates = V2/d = 164 / 0.0209 = 7846.88 in down direction
electric force acts in upward direction , so magnetic force has to act in downward direction. for that magnetic field has to be inward to the page
for the electron to pass in straight line
- q (V x B) = -q E
((1.7 x 107 i^ ) x B) = 7846.88 j^
B = - 4.62 x 10-4 k^ = - 0.462 k^ mT