In the figure below, an electron with an initial kinetic energy of 4.0 keV enter
ID: 2192738 • Letter: I
Question
In the figure below, an electron with an initial kinetic energy of 4.0 keV enters region 1 at time t = 0. That region contains a uniform magnetic field directed into the page, with magnitude 0.010 T. The electron goes through a half-circle and then exits region 1, headed toward region 2 across a gap of 29.0 cm. There is an electric potential difference ?V = 2600 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 0.030 T. The electron goes through a half-circle and then leaves region 2. At what time t does it leave?Explanation / Answer
K1 = 4e3 eV = 4e3 * 1.6e-19 = 6.4e-16 J
v1 = (2K/m) = (2*6.4e-16/9.109e-31) = 3.7486e7 m/s
m v1^2/R1 = q v1 B1
==> R1 = m v1/q B1
==> R1 = 9.109e-31*3.7486e7/1.6e-19/0.010 = 0.021341
==> w1 = v1/R1 = 3.7486e7/ 0.021341 = 1.7565e9 rad/s
==> T1 = 2/w1 = 2*3.1416/1.7565e9 = 3.5771e-9 s
==> t1 = T1/2 = 1.78855 ns
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a = F/m = qE/m = q(V/d)/m = 1.6e-19*2000/0.26/9.109e-31 = 1.3512e15 m/s2
vf^2 = 2 a d + vi^2 = 2*1.3512e15*0.26 + 3.7486e7*3.7486e7
vf = 4.5911e7 m/s
t2 = V/a = (4.5911e7-3.7486e7)/1.3512e15 = 6.2352 ns
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==> R3 = m v3/q B3
==> R3 = 9.109e-31*4.5911e7/1.6e-19/0.030 = 0.0087126
==> w3 = v3/R3 = 4.5911e7 /0.0087126 = 5.2695e9 rad/s
==> T3 = 2/w3 = 2*3.1416/5.2695e9 = 1.1924e-9 s
==> t3 = T3/2 = 0.5962 ns
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t = 1.78855 + 6.2352 + 0.5962 = 8.62 ns