In the figure, an electron with an initial kinetic energy of 3.50 keV enters reg

Question

In the figure, an electron with an initial kinetic energy of 3.50 keV enters region 1 at time t = 0. That region contains a uniform magnetic field directed into the page, with magnitude 0.00910 T. The electron goes through a half-circle and then exits region 1, headed toward region 2 across a gap of 22.0 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 0.0175 T. The electron goes through a half-circle and then leaves region 2. At what time t does it leave?   

Explanation / Answer

here,

We need the three individual times t1, t2 & t3

the speed of the particle entering region1 comes form its kinetic energy

K = 1/2*m*v^2

K = 3500 eV * 1.60 * 10^-19J/ev

k = 5.6 * 10^-16 J

so v = sqrt(2 * 5.6 * 10^-16/9.11 * 10^-31)

v = 3.51 * 10^7 m/s

The period in the field ,

T = 1/f = 2 * pi * / w = 2*pi*m/(q*B) = 2*pi*9.11 * 10^-31/(1.60 * 10^-19*0.00910) = 3.93 * 10^-9 s

but since it only travels in a semi circle the time t1 = 3.93 * 10^-9s/2 = 1.96 * 10^-9 s

Part two it will e * it region 1 with a speed of 3.51 * 10^7m/s

it will undergo a constant acceleration of E*q/m = (delta V/d)*q/m = 2000*1.60 * 10^-19/(0.22 * 9.11 * 10^-31

a = 1.6 * 10^15 m/s^2

so the time to cross the gap comes from * = v * 0*t + 1/2*a*t^2

or (1.6 * 10^15/2)*t^2 + 3.51 * 10^7*t - 0.22 = 0

or

t2 = 5.56 * 10^-9s

Part three The period in the field T = 1/f = 2*pi/ w

T = 2*pi*m/(q*B)

T = 2*pi*9.11 * 10^-31/(1.60 * 10^-19 * 0.0175)

T = 2.04 * 10^-9 s

but since it only travels in a semi circle the time t3 = 1.02 * 10^-9 s

Therefore total time = 1.96 * 10^-9 + 5.56 * 10^-9 + 1.02 * 10^-9

t = 8.54 * 10^-9 s

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