Question

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

Answer 1

The force on the proton is 7.85 x 10⁻¹² N

Step-by-step explanation:

Given;

kinetic energy of the proton, K.E = 2MeV = 2 x 10⁶ x 1.602 x 10⁻¹⁹ J

= 3.204 x 10⁻¹³ J

magnitude of the magnetic field, B = 2.5 T

The kinetic energy of the proton is given by;

`K.E = (1)/(2) m v^2\\\\v^2 = (2K.E)/(m)\\\\v = \sqrt{(2K.E)/(m) } \\\\v = \sqrt{(2*3.204*10^(-13))/(1.67 *10^(-27))} \\\\v = 1.959*10^7 \ m/s`

The force on the proton moving perpendicular to magnetic field is given by;

F = qvB

F = 1.602 x 10⁻¹⁹ x 1.959 x 10⁷ x 2.5

F = 7.85 x 10⁻¹² N

Therefore, the force on the proton is 7.85 x 10⁻¹² N