A 2MeV proton is moving perpendicular to a uniform magnetic field of 2.5 Tesla the force on the proton is ?

Please refer to the solution posted by your friend above.

@Aamir...great job, keep it up!

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You are given the kinetic energy of proton = 2 x 106 e V

You can use this information to calculate the velocity of proton.

1/2 m v2 = 2 x 106 x 1.6 x 10-19 J

v2 = 2 x 2 x 1.6 x 10-13 / m

mass of proton = 1.65 x 10-27 kg

v2 = 2 x 2 x 1.6 x 10-13 / 1.65 x 10-27 = 3.87 x 1014

v = 1.97 x 107 m/s

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Force on a moving charge = q(vXB)

F = qvB sin(90) = qvB = 1.6 x 10-19 x 1.97 x 107 x 2.5 = 7.88 x 10-12 N

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