... In a nuclear research laboratory, a proton moves in a particle accelator thr
ID: 2244973 • Letter: #
Question
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In a nuclear research laboratory, a proton moves in a particle accelator through a magnetic field of intenseity 0.444 T at speed of 4.60218 times 107 m/s. The charge of a proton is 1.60218 times 10-19 C. If the proton is moving perpendicular to the field, what force acts on it ? Answer in units of N 005(parts 2 of 2) 10.0 points If the proton of mass 1.67262 times 10-27 kg continues to move in a direction that is consistently perpendicular to the field, what is consistently perpendicular to the field, what is the radius of curvature of its path ? Answer in units of m.Explanation / Answer
(a)F = q v B= 1.6 * 10-19 * 4.08 * 10^7 * 0.444
= 2.89 * 10-12 N
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(b) F = m v^2 / r
r = m v^2 / F
= (1.67 x 10-27 * (4.08 x 10^7)^2 ) / (2.89 * 10-12)
= 0.96 meters
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