At time t = 0, an electron with kinetic energy 14 keVmoves through x = 0 in the

Question

At time t = 0, an electron with kinetic energy 14 keVmoves through x = 0 in the positive direction of anx axis that is parallel to the horizontal component ofEarth's magnetic field. The field's vertical component is downwardand has magnitude 64.5 µT. (a) What is themagnitude of the electron's acceleration in meters per secondsquared, due to the magnetic field? (b) What isthe electron's distance from the x axis in centimeterswhen the electron reaches coordinate x = 21 cm? I figured out a) the answer is 7.94E14 m/s^2 but I can'tfigure out b?? Thanks for the help At time t = 0, an electron with kinetic energy 14 keVmoves through x = 0 in the positive direction of anx axis that is parallel to the horizontal component ofEarth's magnetic field. The field's vertical component is downwardand has magnitude 64.5 µT. (a) What is themagnitude of the electron's acceleration in meters per secondsquared, due to the magnetic field? (b) What isthe electron's distance from the x axis in centimeterswhen the electron reaches coordinate x = 21 cm? I figured out a) the answer is 7.94E14 m/s^2 but I can'tfigure out b?? Thanks for the help

Explanation / Answer

K.E. =14 KeV = 14*1.6*10-16 J                        = 2.24 *10-17 J So, Vx= (2K.E./M)            =7.01*106 m/s Now, force due to magnetic field = BeVx                                                          =55*10-6*1.6*10-19*7.01*106  N                                           = 6.17 *10-17 N Thus,      ma= 6.17*10-17         =>    a=(6.17*10-17)/(9.1*10-31 )                      = 6.78 *1013 ms-2 From x axis, x=V*t =>t=x/V       = 0.21/(7.01*106)       =2.996*10-8 s Now, at y axis, s=0.5at2 =0.5*6.78*1013*(2.996*10-8)2 =0.0304 m =3.04 cm

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