In the figure, an electron with an initial kinetic energy of 4.40 keV enters reg
ID: 1468099 • Letter: I
Question
In the figure, an electron with an initial kinetic energy of 4.40 keV enters region 1 at time t = 0. That region contains a uniform magnetic field directed into the page, with magnitude 0.00620 T. The electron goes through a half-circle and then exits region 1, headed toward region 2 across a gap of 26.0 cm. There is an electric potential difference ?V = 2000 V across the gap, with a polarity such that the electron's speed increases uniformly as it traverses the gap. Region 2 contains a uniform magnetic field directed out of the page, with magnitude 0.0234 T. The electron goes through a half-circle and then leaves region 2. At what time t does it leave?
Explanation / Answer
Here ,
the electron completes half a circle in each region ,
as the time period of electron to complete one circle in the magnetic field B is given as
T = 2*pi*m/(B*e)
for half circles in A and B
time taken is
T = TA/2 + TB/2
T = 2pi*9.109*10^-31/(0.0062 * 2 * 1.602 *10^-19 ) + 2pi*9.109*10^-31/(0.0234 * 2 * 1.602 *10^-19 )
T = 3.644 *10^-9 s
Now , for the time taken to travel between region 1 and 2
initial speed is u
0.5 * 9.1 *10^-31 * u^2 = 4.4 *10^3 * 1.602 *10^-19
u = 3.93 *10^7 m/s
acceleration in the region , a = q * E/(m)
a = 1.602 *10^-19 * (2000/.26)/(9.1 *10^-31)
a = 1.35 *10^15 m/s^2
let the time taken is t
0.26 = 3.93 *10^7 * t + 0.5 * 1.35 *10^15 * t^2
solving , t = 6*10^-9 s
total time taken = 6 *10^-9 + 3.644 *10^-9
total time taken = 9.644 *10^-9 s
the electron leaves at 9.644 *10^-9 s