In the figure, an electron accelerated from rest through potential difference V_
ID: 1600202 • Letter: I
Question
In the figure, an electron accelerated from rest through potential difference V_1 = 1.00 kV enters the gap between two parallel plates having separation d = 20.0 mm and potential difference V_2 = 100 V. The lower plate is at the lower potential. Neglect fringing and gravity, 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?Explanation / Answer
Electron energy E1 = 1 keV = 1000*1.6022E-19 = 1.6022E-16J
Velocity v1 = sqrt(2E1/me)
Felec = EQ (E is the electric field)
Fmag = Qv1 X B; assuming v1 and B are perpencicular, F = Qv1B
Thus EQ = Qv1B ==> B = E/v1
E = V2/d
B = E/v1 = V2/(v1d) = V2/(d*sqrt(2E1/me))
= 100/(0.020*sqrt(2*1.6022E-16 / 9.10938215E-31))
= 4.77E-4 T
In terms of vector notation
E is along +y axis, so magnetic field is along +x axis
B = .(4.77E-4 )i T