# The proton moves in a simultaneous magnetic field with induction B = 50mT. Along a circle with a diameter

**The proton moves in a simultaneous magnetic field with induction B = 50mT. Along a circle with a diameter of d = 20 cm. Determine the kinetic energy K of the conductor.**

Data: B – field induction (B = 50 mT = 0.05 T); d is the diameter of the circle (d = 20 cm; in the SI system d = 0.2 m).

Const: qp – proton charge (qp = 1.6 * 10 ^ -19 C); mp is the mass of a proton (m = 1.673 * 10 ^ -27 kg).

To find out the kinetic energy of the indicated proton, consider the equality: Fц (centripetal force) = Fl (Lorentz force).

mp * ac = qp * V * B.

mp * V ^ 2 / R = qp * V * B; * from this equality we express: mp * V / R = qp * B and V = qp * B * R / mp.

0.5 * mp * V ^ 2 = 0.5 * qp * V * B * R.

Ek = 0.5 * qp * V * B * R = 0.5 * qp * (qp * B * R / mp) * B * R = 0.5 * qp ^ 2 * B ^ 2 * R ^ 2 / mp = 0.5 * qp ^ 2 * B ^ 2 * (0.5 * d) ^ 2 / mp.

Let’s perform the calculation: Ek = 0.5 * qp ^ 2 * B ^ 2 * R ^ 2 / mp = 0.5 * (1.6 * 10 ^ -19) ^ 2 * (0.05) ^ 2 * (0 , 5 * 0.2) ^ 2 / (1.673 * 10 ^ -27) ≈ 1.913 * 10 ^ -16 J.

Answer: The kinetic energy of the indicated proton is 1.913 * 10 ^ -16 J.