012 10.0 points What is the radius of the smallest possible circular orbit that

Question

012 10.0 points What is the radius of the smallest possible circular orbit that a proton with the kinetic energy equal to 2.9 MeV can have in a 2 T magnetic field? The mass of a proton is 1.67 x 1027 kg and its charge is 1.609 × 10-19 C. Answer in units of m. 013 10.0 points Two species of singly charged positive ions of masses 1.4x 10 26 kg and 2.5 x 10-26 kg enter a magnetic field at the same location with a speed of 1.5 x 105 m/s. The charge on the ion is 1.60218 x 10-19 C. If the strength of the field is 0.17 T, and they move perpendicularly to the field, find their distance of separation after they com- plete one half of their circular path. Answer in units of cm

Explanation / Answer

here,

magnetic feild , B = 2 T

let the speed of electron be v

kinetic energy = 2.9 MeV = 2.9 * 10^6 * 1.6 * 10^-19

0.5 * m * v^2 = 2.9 * 10^6 * 1.6 * 10^-19

0.5 * 1.67 * 10^-27 * v^2 = 2.9 * 10^6 * 1.6 * 10^-19

solving for v

v = 2.41 * 10^7 m/s

the radius of the smallest circular orbit , r = m * v /( q * B)

r = 1.67 * 10^-27 * 2.41 * 10^7 /( 1.6 * 10^-19 * 2)

r = 0.12 m

the radius of the smallest circular orbit is 0.12 m

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