What is the speed of a proton moving in a magnetic field with an induction of 2 mT if the radius of the circle it describes is 2.7 cm?

m = 1.67 * 10-27 kg.

q = 1.6 * 10-19 Cl.

B = 2 mT = 2 * 10-3 T.

R = 2.7 cm = 0.027 m.

∠α = 90 °.

V -?

The electric charge q, which moves at a speed V in a magnetic field with induction B, is affected 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 the magnetic induction vector IN.

Let’s write 2 Newton’s law for the proton: m * a = q * V * B * sinα.

We write the centripetal acceleration a according to the formula: a = V2 / R, where R is the radius of the circle along which the proton moves.

m * V2 / R = q * V * B * sinα.

m * V / R = q * B * sinα.

The proton’s velocity V will be determined by the formula: V = q * B * sinα * R / m.

V = 1.67 * 10-19 C * 2 * 10-3 T * sin90 ° * 0.027 m / 1.6 * 10-27 kg = 5630 m / s.

Answer: the speed of the proton is V = 5630 m / s.