Question

ANALYSIS Present your results in tabulated form as shown below. Here you are given the distances from the filament to each crossbar/indicator mark, these representing the diameters of the circular orbits of the electrons. Diameter of beam orbit (m) B (eqn. 3) (T) e/m (eqn. 2) (C/kg) (A) 2.25 1.41 1.65 0.05 0.06 0.07 0.08 0.09 0.10 0.11 (V) 200 200 200 200 200 200 200 1.45 1.20 1./> Loy Calculate the average value of e/m and the average deviation textbook Compare your result with that published in your

Answer 1

Raellu. . ri I Bleau") ] v. elmenen (m) I 1917. I v. City) Dia of sem orbitim 0.05 0.06 0.07 0.025 0.030 0.035 0.040 0.045 0050 0-055 2.25 B - 188154 200 4.448108 I 1.91 02-121x1067 200 5.24x102 1.65 8-9.4X106T 200 1606x 100 1045 Bus 7.25X166 2006.09X 100 1.30 B6-1l-5X1061 200 3.86X100 1.17 B6 =4.60X1oT) 2001 0.54x102 1.04 B = 3.781567 goo 19.62x100 0.00 0.09 0:10 0.11 calculate and teu there aurgge value of elm acurage deviation calculate the magnetic field (B) Mo Il Qar (47x107T MAX(2.25) Á 875 X 0.025 m = 180x10T 18.0X10-6 = 10 4T BL Scanned with CamScanner

Date: Page No: B2 = Mo Iz 4*X167 X 1.9/ 8tX 0.030 2 -27133440-7I 25₂ B2 = 12.7 8106 T - B2 = 12.7 YT . MOI - QTV 147 X10-7) x (2005)(1.65) de (47810 12AX (0.035) -9.42 X10-6 94.20 X10-7 T = 9.428 YT Mo I 4 (4xx107) (1.45). 2TYG QTX 0.040) & 72.58167 T 27.25 X 10°6T 3725 XHT B.5= - 115.55X10 Mo In 2005 147 X107) (130) (275 X 0.045) 1. 2 .11.5 = 11.58TL Scanned with CamScanner

LO Date: Page No: 147 X10-7) *(117) @ X 0.050) B6 = Mo If atro = 46.0 X10 IT = 464 4.68x10-6 T 4.60 MT MOI for 140X10-7) (1.04) 127 X 0.055) QTry - 37. 012 X100T 3-7 RIQ X10-6 T = 3.7010 HT Determine the elm VI Bir 900 til 18X106 x 0.025 201X 906 bon 9300X01 PODlxhhin t a tonttat oelito hin 200 Va B2 r2 m2 12.7x10-6 X 0.030 5:24 X100 Scanned with CamScanner

Date: Page No: for Third V3 m /3 - B3 83 LE2 2 00___ 9.420X106 x 10.035) y = 6.06 Xiou 6068100% cling or fourth 900 elemente transport to Bury 7.25X1096 X 0.040 be for fifth - V5 m5 200 11.5X1055X0.045 3.064109 cling Sixth 200 BENG 46*166*0-050 - - 8.548700 [cling Scanned with CamScanner

hayu No add up all the values you foemd. = 0 (-93X100 +11013x100 +10:31X100) + 10.52X100 Lil( 2.50X109 +12.17X100)+ ( 3.25 xioa) => 1 11.02 X100 in data diuided Them by all the number pitems of items so. => 11.0218108 7 ľ de 1060015x100 The awrage deviation is = 1068 X1000 el Scanned_with C Camsdahierdi! Xinbell