Answer:
The force on the proton is 7.85 x 10⁻¹² N
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;
[tex]K.E = \frac{1}{2} m v^2\\\\v^2 = \frac{2K.E}{m}\\\\v = \sqrt{\frac{2K.E}{m} } \\\\v = \sqrt{\frac{2*3.204*10^{-13}}{1.67 *10^{-27}}} \\\\v = 1.959*10^7 \ m/s[/tex]
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
