Question

(a) At what speed (in mVs) will a proton move in a circular path of the same radius as an electron that travels at 7.10 x 10 m/s perpendicular to the Earth's magnetic field at an altitude where the field strength is 135x10-12 m/s () What would the radius (inm) of the path be if the preton had the same speed as the electron? m 19 What would the radius (in my be if the proton had the same kinetic energy as the electron? m (0) What would the radus in m) be if the proton had the same momentum as the electron?

Answer 1

As we know helation of radius of Cewiatore gno mv ZB die me ve le Be ele 3 - Mp Vp ap Bo Zt has same radics of both then equate and me ve re=ap} Mp Vp Yp Bp 2 Be = Bp Ve ble Ve meve Ve = 7.10x10ml Criven. mp = 9.11*10-3187,10x106 1.67x10-27 Vp = 31873x103 mis Hence, Velocity of Protod.

(b) .vp = Ve = 7.108 106 mis Olip mp up { Bp= 6.75x10-5 * 10:57] qp Bp = 1.67x10-27x 7.10X106 1.6*10-14 x125x 10-5 Hence, radius for Orp = 5.928 x 103 m of Pocoton (9) K.EE 12 mua { some K.E} Ki Ep Kife V2 Vo = ve v me mp 7.10 X 100 19.1x1031 1.67x10-27 Vb = 1.66x105 mis p = mp up qp Bp

916 = 1.67x10-27 X 1066x105 1.68 10-19 X 1.25X105 Hence, 946 = 138.61m (d) Pe = pp { same momentum} mè ve Mp Vp vp = meve те x 7.10x106 =9.11x103) 1.6710-27 = 3.873 x 103 mis Ne I" mp Vp qp Bp =1067x10-27% 3.8 73x103 1.6*10*190 X 1,25% 1005 Hence, 1941," = 3.23m