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.