Question

(a)At what speed (in m/s) will a proton move in a circular path of the same radius as an electron that travels at 7.15 x 10 m/s perpendicular to the Earth's magnetic field at an altitude where the field strength is 1.15 x 10-5 T? m/s (b) What would the radius (in m) of the path be if the proton had the same speed as the electron? m (c) What would the radius (in m) be if the proton had the same kinetic energy as the electron? m (d) What would the radius (in m) be if the proton had the same momentum as the electron? m

Answer 1

b ) v=7.15XL04 qVB = mes El meve qB 8 - 9.1 X10-31 X 7.15X109 1.6810-19x 1.15X20-5 35.36 8103 m Aswer (r = a) VBY P VP mp 1.6860-19% 1.15X20-% x 35.36X 10-13 1.67810-27 V 38.959 x 100 Vp = 38.96 m/s Aswer C) 16.E of P K. E of ezme ve ² The va- mere 2 I up mp up = 9.1 x 10-31 1.67X10-27 [up mp rp2 me ve x 715 8104 1 669.04 m/s ] CS Scanned with Cam SLAME

Soj more B 1.67 X10-2?x 1669.09 1.6 X 10-19 x 1.15 X10-5 15 14.835X10-3 8= 7= 1.514m Aswer ed) = ما le = pp merea mp up 9.1410131 x 7:15X104 1.67X1027 vp = 38.96X100 [Up = 38.964s mp up Sol r = 1.678402?x 38-96 1.68 (0-19 X HSXIOS 35. 36x10 -3 m CS Sned with CamScammer