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23-07-2020
Physics

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What is the acceleration of a proton moving with a speed of 9.5 m/s at right angles to a magnetic field of 1.8 T

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nobleswift nobleswift

23-07-2020

Explanation:

F = qvBsin(angle)

F = 1.6 x 10^-19 x 9.5 x 1.8 x Sin (90)

F = 273.6 x 10^-20N

a = F/M

M of a proton = 1.67 x 10^-27kg

a = 273.6 x 10^-20/1.67 x 10^-27

a = 1638323353.29m/s

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