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Course Contents >> ... > Assignment 5 >> Alpha Particle in Magnetic Field Timer Notes Evaluate Feedback Print Info Alpha Particle in Magnetic Field Due on Tuesday, May 1 at 03:00 pm (EDT) An alpha particle (the nucleus of a helium atom formed by 2 protons and 2 neutrons with charge q = +2(1.6 x 10-19 C) is moving perpendicular to a magnetic field of 0.650 tesla at a velocity of 3.80E+6 m/s ; calculate the magnitude of the magnetic force exerted on it. 7.90x10-13 N You are correct. Your receipt no. is 156-8394 8204 a Previous Tries Calculate the acceleration of the alpha particle. 183 x 10M14 m/s2 Submit Answer this question expects a numeric answer. Tries 0/10 Previous Tries What will be the radius of curvature of the path followed by the alpha particle in the magnetic field? 0.112 m Submit Answer Incorrect. Tries 1/10 Previous Tries What would be the force on the alpha particle if it were traveling parallel to the magnetic field lines? 0.000 N You are correct. Your receipt no. is 156-4486 Previous Tries

Explanation / Answer

charge q = 2 x 1.6 x 10-19 = 3.2 x 10-19 C

magnetic field B = 0.650 T

velocity v = 3.80 x 106 m/s

a) Force on the particle F = q v B sin90

= 3.2 x 10-19 x 3.80 x 106 x 0.650

= 7.90 x 10-13 N

b) mass of proton mp = 1.6727 x 10-27 Kg

mass of neutron mn = 1.6750 x 10-27 Kg

total mass M = 2mp + 2mn

= (3.3454 + 3.35 ) x 10-27

= 6.7 x 10-27 Kg

accelaration a = F/M = 7.90 x 10-13 / 6.7 x 10-27

= 1.18 x 1014 m/s2

c) from q v B = mv2 / r

radius r = mv/qB

=  6.7 x 10-27 x 3.80 x 106 / (  3.2 x 10-19 x 0.650)

= 0.12 m

d) for this F = q v B sin 180 = 0

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