Determine the radius of the circle along which a proton with an energy of 4500 keV moves in a cyclotron

Determine the radius of the circle along which a proton with an energy of 4500 keV moves in a cyclotron perpendicular to the lines of force of a magnetic field with an induction of 1 T. Proton charge 16 * 10 ^ -20 cells, mass 1.6 * 10 ^ -27 kg. 1 eV = 16 * 10 ^ -20 J.

W = 4500 eV = 7200 * 10 ^ -19 J.

q = 16 * 10 ^ -20 Cl.

∠α = 90 °.

B = 1 T.

m = 1.6 * 10 ^ -27 kg.

R -?

A proton with an electric charge q, which moves at a speed V in a magnetic field with induction B, is acted upon by the Lorentz force Fl, the value of which is determined by the formula: Fl = q * V * B * sinα, where ∠α is the angle between the direction of motion of the charge V and vector of magnetic induction B.

m * a = Fl – 2 Newton’s law.

We express the centripetal acceleration a by the formula: a = V2 / R, where V is the speed of the proton, R is the radius of the proton’s circle.

m * V ^ 2 / R = q * V * B * sinα.

R = m * V ^ 2 / q * V * B * sinα = m * V / q * B * sinα.

We express the proton energy W by the formula: W = m * V ^ 2/2.

V = √ (2 * W / m).

R = m * √ (2 * W / m) / q * B * sinα.

R = 1.6 * 10 ^ -27 kg * √ (2 * 7200 * 10 ^ -19 J / 1.6 * 10 ^ -27 kg) / 16 * 10 ^ -20 C * 1 T * sin90 ° = 0 , 00945 m.

Answer: the proton moves in a circle with a radius of R = 0.00945 m.