Question

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

Answer 1

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;

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

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