Question

Question 7 /1 View Policies Current Attempt in Progress In the figure, an electron with an initial kinetic energy of 3.60 keV enters region 1 at time t 0. That region contains a uniform magnetic field directed into the page, with magnitude 0.0130 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 AV 1900 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 0.0242 T. The electron goes through a half-circle and then leaves region 2. At what time t does it leave? Region 1 Region 2 On Number Units

Answer 1

The answer for above problem is explained below.

Case cis In Region 1, By Applying Newton's second lace, Fuet = qu.B my2 And Fuet = mun z) quB = mya v z qoy m 2T8 And Ta a And 2 I'm qB time, to = 2 time,

zs t = [21m /aR] as IT m t,= qB 9.1x10 39 kg * 3.14 z t = 106x1619C x 0.0130T = t; = 1 10 3737 x 109 sec In case (ii) In Region 2, Pole PH wo 1 = 9.1x10-%'kg x 3014 106x1019 x 0.0242 T

Case (lij In the gap, Kinetic energy, KE = Ź mv2 3.60 x103 x 106 x 1619 C-va ã x 9.1x10 31 kg x 5.76x1516 = 4.55 x163l xV2 5.76 x 15 16 z) v = 4.55 x 10-31 = v=13.55199 x10* m/s) And Acceleration, a = 1 h as a = q E - m q CAVA) z a = m > a = 1.6x16'90 x 1900V 9.1x173 kg 2a xiom

3 az 1.518 x 185 m/s time across the gap is, x= vt + hat? 0.22m = (3.55799 x 10m15]t +ă (1.518x107m15] ? os (7.5924 x1054 m/s] + +[3.55799 x10°ms]+ - 0.aam-o By Solving above equation, we get, [4, 5.53 x 109 see? 2 Total time T= t + tz + tz T = [1.3137 x10^s«c] +[7.3796 xlo“see] +(5.53 x1ā” sec] T= 7.6416 x 10 9 sec