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1.6x10C #2 a proton moves IN a circular path perpendicular to a magnetic Field of (0.2.35. T.). If the KE( ANectic Emergy) of the electron 15 3.30 X10-191, find: BAUT a) ('V') a) the speed of the electron. 6.) the radius the Circular path.. c.) the magnetic Force on the charge. b) r=m > mv 2 B V= KE = mv2 3. 30 x 10¹ J = 1/2 (1.67x10 "" kg) 3.30 X 10 - ¹J = 8.35 x 1024 .v² -28 of c) F = q vB sin & -19 KE 1/2 m = (1.67×10² kg)(1.99x10m/s) (1.6×1019 c) (0.235+) ● (3.30x10¹5 8.35x10-28 V F=1.6x10¹°C 1.99 x 10 m/s 0.235 T JF 1.99x104m -4 8.83x10 m -16 ·7₁ Six (90°) = 7.48×10 ¹6 N O

1.6x10C #2 a proton moves IN a circular path perpendicular to a magnetic Field of (0.2.35. T.). If the KE( ANectic Emergy) of the electron 15 3.30 X10-191, find: BAUT a) ('V') a) the speed of the electron. 6.) the radius the Circular path.. c.) the magnetic Force on the charge. b) r=m > mv 2 B V= KE = mv2 3. 30 x 10¹ J = 1/2 (1.67x10 "" kg) 3.30 X 10 - ¹J = 8.35 x 1024 .v² -28 of c) F = q vB sin & -19 KE 1/2 m = (1.67×10² kg)(1.99x10m/s) (1.6×1019 c) (0.235+) ● (3.30x10¹5 8.35x10-28 V F=1.6x10¹°C 1.99 x 10 m/s 0.235 T JF 1.99x104m -4 8.83x10 m -16 ·7₁ Six (90°) = 7.48×10 ¹6 N O

College Physics
College Physics
11th Edition
ISBN:9781305952300
Author:Raymond A. Serway, Chris Vuille
Publisher:Raymond A. Serway, Chris Vuille
Chapter1: Units, Trigonometry. And Vectors
Section: Chapter Questions
Problem 1CQ: Estimate the order of magnitude of the length, in meters, of each of the following; (a) a mouse, (b)...
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' moves IN a circular path perpendicular to a magnetic Field of (0.2.35.-T.). If the KE (ANectic Energy) of the electron 15 3.30x10-191, find: BOUT 1.6x10C #2 a proton a) b). a) the speed of the electrono b.) the radius of the Circular path. c.) the magnetic Force on the charge. KE = 2 mv ² 3.30x10=19J = 2(1.67x10-27 kg) 3.30 X 10 - ¹1 J = 8.35 x 10 tag v² r = mv 2 B KE V= v=√√ — m c) F = q vB sin o -28 v² 2 -27 = (1.67×10² kg) (1.99x10 "/s) (1.6×10 9 c) (0.235+) O 3 de -15 3.30x10¹9J √√² $$13:25 2017031 = (1.99×187% 8.35x10-28/2 L -4 = 8.83x10 m -19 4 F = 1.6x10¹ C +1.99x10 / +0.235 T six (10) = 7.48×10¹ N
Transcribed Image Text:' moves IN a circular path perpendicular to a magnetic Field of (0.2.35.-T.). If the KE (ANectic Energy) of the electron 15 3.30x10-191, find: BOUT 1.6x10C #2 a proton a) b). a) the speed of the electrono b.) the radius of the Circular path. c.) the magnetic Force on the charge. KE = 2 mv ² 3.30x10=19J = 2(1.67x10-27 kg) 3.30 X 10 - ¹1 J = 8.35 x 10 tag v² r = mv 2 B KE V= v=√√ — m c) F = q vB sin o -28 v² 2 -27 = (1.67×10² kg) (1.99x10 "/s) (1.6×10 9 c) (0.235+) O 3 de -15 3.30x10¹9J √√² $$13:25 2017031 = (1.99×187% 8.35x10-28/2 L -4 = 8.83x10 m -19 4 F = 1.6x10¹ C +1.99x10 / +0.235 T six (10) = 7.48×10¹ N
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