In the figure below, an electron with an initial kinetic energy of 4.0 keV enter

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 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 0.030 T. The electron goes through a half-circle and then leaves region 2. At what time t does it leave? 1 _______ . ns

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

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