In the figure, an electron with an initial kinetic energy of 4.10 keV enters reg
ID: 1450923 • Letter: I
Question
In the figure, an electron with an initial kinetic energy of 4.10 keV enters region 1 at time t = 0. That region contains a uniform magnetic field directed into the page, with magnitude 0.00690 T. The electron goes through a half-circle and then exits region 1, headed toward region 2 across a gap of 21.0 cm. There is an electric potential difference ?V = 2100 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 0.0220 T. The electron goes through a half-circle and then leaves region 2. At what time t does it leave?
Explanation / Answer
Region 1,
K.E = 1/2*mv^2 = 4.10 keV
Substituing Values,
1/2*9.1*10^-31 * v^2 = 4.10 * 10^3 * 1.6*10^-19
v = 3.80 * 10^7 m/s
Time it takes to leave region 1 t1, = / = (*m)/(q*B) = (*9.11*10^-31)/(1.60*10^-19*0.00690)
t1 = 2.59 * 10^-9 s
Now, electron will accelerate, a = (q*V)/(d*m) = (1.6*10^-19*2100)/(0.21*9.1*10^-31) = 1.76 * 10^15 m/s^2
Time it takes to cross the gap,
s = u*t + 1/2at^2
0.21 = ( 3.80 * 10^7 ) * t + 1/2 * 1.76 * 10^15 * t^2
t2 = 4.96 * 10^-19 s
Time it takes to leave region 2 t3, = / = (*m)/(q*B) = (*9.11*10^-31)/(1.60*10^-19*0.0220)
t1 = 8.13 * 10^-10 s
Total time it takes, t = t1+t2+t3
t = 2.59 * 10^-9 + 4.96 * 10^-9 + 8.13 * 10^-10 s
t = 8.363 * 10^-9 s