Question

A proton, moving at a velocity of 681 m/s experiences an acceleration of 2.0 x 10¹⁴ m/s² when it enters a magnetic field at an angle of 55 degrees to the field. Calculate the strength of the magnetic field. Assume that the force from the magnetic field is the only force acting on the proton. (Hint: You will need to solve for this force first!)

Answer 1

Ansi = Proton velocity V = 681 mis acceleration of proton Force on proton a= ax 1014 m/s² E = ma 14 267 xlo = 1.6726 x 2 x 10 m = mass of - 13 N proton = 1.678 6 x 10 = 3.3452 x 10-13 Let B be is given magnetic field In magnetic field force is is alv'x') by

a = charge on proton = 1.6022 x 16-1 -190 Magnitude of ude of force F = q VB sino 0 ob 'BE and magnetic is angle between F velocity field qu sino = 550 3.3452 x 10! 13 1•6022 x 10-19 x 681 x Sin (559) o 11 B = 3.743 x 103 T

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