The figure to the right shows the trajectory of a proton (mp = 1.67 times 10^-27

Question

The figure to the right shows the trajectory of a proton (mp = 1.67 times 10^-27 kg, qp = 1.602 times 10^-19 C) in a magnetic field. Suppose the field is of strength of the maximum field of a traditional superconducting magnet B = 15T. The proton travels at a speed of 1 times 10^7m/s, compute the radius and the direction of the orbit. Select One of the Following: 3 times 10^-5m, counter-clockwise 3 times 10^-5m, clockwise 2 times 10^-4m, counter-clockwise 2 times 10^-4m, clockwise 6 times 10^-4m, counter-clockwise 6 times 10^-4m, clockwise 7 times 10^-3m, counter-clockwise 7 times 10^-3m,. clockwise

Explanation / Answer

The magnetic field provides the centripetal acceleration to the proton.

Hence, the centripetal force on the proton is balanced with the magnetic force.

mv2/r = Bvq

Hence, the radius of the path of the proton is,

r = mv/Bq = 1.67x10-27 *1x107 / 15 * 1.602x10-19 = 6.95e-3 = 7x10-3 m, clockwise

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